halacha - Government subsidies or tzedoko

it is preferable not to take charity from non-Jews (maybe only if it is not in private)

Source Yore daiya 254

In a case of need (someone needs money for housing and/or food),
which is more preferable:
1. To take/collect tsedoko from Jews
or
2. To get subsidies from the non-Jewish government (e.g. Section 8 or food stamps) not a private non-Jew, but probably considered in public, maybe also Jewish since Jews pay taxes

Sources on subject would be greatly appreciated.

EDIT To fulfill the request of the Honorable @IsaacMoses bellow (sorry for being so slow)

254.1

אָסוּר לְיִשְׂרָאֵל לִטֹּל צְדָקָה מִן הָעוֹבְדֵי כּוֹכָבִים בְּפַרְהֶסְיָא. וְאִם אֵינוֹ יָכוֹל לִחְיוֹת בַּצְּדָקָה שֶׁל יִשְׂרָאֵל, וְאֵינוֹ יָכוֹל לִטְּלָהּ מֵהָעוֹבֵד כּוֹכָבִים בְּצִנְעָא, הֲרֵי זֶה מֻתָּר. It is forbidden to take tsedoko publicly from a non Jew, unless you can't other ways, from a Jew or privately.

So it would seem it is preferable to collect tzedoka from Jews (option one)
254.2

מֶלֶךְ אוֹ שַׂר עוֹבֵד כּוֹכָבִים שֶׁשָּׁלַח מָמוֹן לְיִשְׂרָאֵל לִצְדָקָה, אֵין מַחֲזִירִין אוֹתוֹ מִשּׁוּם שְׁלוֹם מַלְכוּת, אֶלָא נוֹטְלִין מִמֶּנּוּ וְיִנָּתֵן לַעֲנִיֵי נָכְרִים בַּסֶתֶר, כְּדֵי שֶׁלֹּא יִשְׁמַע הַמֶּלֶךְ. הַגָּה: וְיֵשׁ אוֹמְרִים דְּיַעֲשֶׂה בָּהֶן מַה שֶּׁצִּוָּה לוֹ הַמּוֹשֵׁל...
If the king sends money for Jews for tzedoko we should not return it for peace with the government but we should give is secretly to poor non-Jews, so that the kind will not hear about it haga there are those that say that you should do with them (the money) as the king says (give it to poor Jews)

It still seems to me that option one is preferred since the government is not sending the money but offering you to request it (by filling out a form)

But I am asking this question because I know of Orthodox Jews that do it and do not collect tzedoka from Jews at all, so maybe there is something I am not understanding correctly

meaning - What does 氏 mean after a name, how is it different from さん or 様?

If you look in the dictionary for the definition of 氏 you'll find it defined as: family name; lineage; birth.

However I have seen it used in such a way that it is doubtful that it means any of those things in certain contexts.

Here's an example sentence:

今日はマオ氏のアーバンカーで送ってもらった♪(´ε｀ )

What does 氏 mean here?

Note: Mao is a very famous Japanese singer.

choshen mishpat civil law - Are parents responsible if their children damage in a store?

If a parent brings a child (defined as under 13 for a boy and under 12 for a girl) into a store, and that child damages an item, is the parent responsible to pay?

Would it matter if the child simply damaged an item (e.g. pushed a crystal vase, and it shattered into ∞ pieces), or if the child actually benefited from the damaged item (e.g. consuming a candy)?

(answers should focus on the Jewish law aspect and ignore issues of civil law)

physical chemistry - What is the difference between ΔG and ΔrG?

Consider the reaction

$$\ce{A -> B}$$

The reaction Gibbs free energy, $\Delta_\mathrm{r} G$ is given by the following equation $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT \ln Q$$

Now what is the difference between $\Delta G$ and $\Delta_\mathrm{r} G$ ?

Which one of the above represents $G_{\text{products}} - G_{\text{reactants}}$ ?

I read in my book $\Delta_\mathrm{r} G$ is the slope of $G$ plotted against the extent of reaction at different instants. And at equilibrium, it is zero. So can we say that it has value at a particular instant while $\Delta G$ is for a process ($\ce{A}$ to $\ce{B}$ in my question) i.e. is for a time interval?

Does $\Delta G$ for the above reaction represent change in Gibbs Energy when 1 mole of $\ce{A}$ reacts completely to form $1$ mole of $\ce{B}$ ?

Answer

Not for the faint-hearted: There is an excellent, but very mathsy, article here: J. Chem. Educ. 2014, 91, 386 describing the difference.

The Gibbs free energy change, $\Delta G$

You are quite right in saying that $\Delta G$ represents a change in a system over a time interval. The notation $\Delta G$ itself implies that it is a difference in $G$ between two things: an initial state and a final state. Let's use a real example in order to make things clearer. Consider the thermal decomposition of ammonium nitrate:

$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$

If someone were to ask you "what is $\Delta G$ for this reaction?", you should really, technically, be telling them: it is not well-defined. That is because of two things. Firstly, the reaction conditions, e.g. temperature and pressure, are not specified.

But more importantly, a balanced equation does not tell you exactly how much ammonium nitrate is reacting. In your beaker, you could have $1~\mathrm{mol}$ of $\ce{NH4NO3}$ reacting according to the above equation, but you could equally well have $2~\mathrm{mol}$, or you could have $5000~\mathrm{mol}$ (I must say that's one huge beaker though), or you could have $0.001~\mathrm{mmol}$.

The stoichiometric coefficients in the equation do not indicate the amounts of compounds reacting.

So, a more proper question would read like this:

$\pu{2 mol}$ of ammonium nitrate fully decompose according to the balanced equation $\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. Calculate $\Delta G$ for this process.

Great. Because the decomposition is complete, we now know that our final state is $2~\mathrm{mol}~\ce{N2O(g)} + 4~\mathrm{mol}~\ce{H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$, and our initial state is $2~\mathrm{mol}~\ce{NH4NO3(s)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. So,

$$\begin{align} \Delta G &= [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot G_\mathrm{m}(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{NH4NO3(s)})] \end{align}$$

We can't quite find the absolute molar Gibbs free energies, so the best we can do is to use Gibbs free energies of formation.

$$\begin{array}{c|c} \text{Compound} & \Delta_\mathrm{f}G^\circ / \mathrm{kJ~mol^{-1}}\text{ (at }298~\mathrm{K}\text{)} \\ \hline \ce{NH4NO3(s)} & -183.87 \\ \ce{N2O(g)} & +104.20 \\ \ce{H2O(g)} & -228.57 \end{array}$$ $$\scriptsize \text{(data from Atkins & de Paula, }\textit{Physical Chemistry}\text{ 10th ed., pp 975-7)}$$

(Note that the use of standard formation Gibbs free energies is only because of the conditions specified in the question, which conveniently corresponds to the standard state. If we specify different conditions, we can still find $\Delta G$, but we would have to use different data.) So:

$$\begin{align} \Delta G &= [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{NH4NO3(s)})] \\ &= [(2~\mathrm{mol})(+104.20~\mathrm{kJ~mol^{-1}})] + [(4~\mathrm{mol})(-228.57~\mathrm{kJ~mol^{-1}})] - [(2~\mathrm{mol})(-183.87~\mathrm{kJ~mol^{-1}})] \\ &= -1073.62~\mathrm{kJ} \end{align}$$

Note that we have units of kJ. Since $\Delta G$ is the difference between the Gibbs free energy of one state and another, $\Delta G$ has to have the same units as $G$, which is units of energy.

Now, this does not necessarily mean that $\Delta G = G_\text{products} - G_\text{reactants}$. For example, if I changed my question to be:

In the Haber process, 100 moles of $\ce{N2}$ and 300 moles of $\ce{H2}$ are reacted at $800~\mathrm{K}$ and $200~\mathrm{bar}$ according to the equation $\ce{N2 + 3H2 -> 2NH3}$. Only 10% of the starting materials are converted under these conditions. Calculate $\Delta G$ for the process. _{(These numbers are made up.)}

then, your final state would not be the pure products. Your final state is not 200 moles of $\ce{NH3}$. Your final state is $90~\mathrm{mol}~\ce{N2}$, $270~\mathrm{mol}~\ce{H2}$, and $20~\mathrm{mol}~\ce{NH3}$.

In general, one could write, for a chemical reaction,

$$\Delta G = \sum_i (\Delta n_i) G_{\mathrm{m},i}$$

where $\Delta n_i$ is the change in the amount of compound $i$ (in moles), and $G_{\mathrm{m},i}$ is the molar Gibbs free energy of the pure compound $i$, under the $T$ and $p$ conditions specified. Going back to our ammonium nitrate example, we would have $\Delta n_{\ce{NH4NO3}} = -2~\mathrm{mol}$, $\Delta n_{\ce{H2O}} = +2~\mathrm{mol}$, and $\Delta n_{\ce{H2O}} = +4~\mathrm{mol}$.

The Gibbs free energy change of reaction, $\Delta_\mathrm{r} G$

As you have correctly stated, $\Delta_\mathrm{r}G$ is the slope of a graph of $G_\mathrm{syst}$ against the extent of reaction, commonly denoted $\xi$. This is the easiest way of interpreting $\Delta_\mathrm{r}G$. This question contains a slightly fuller derivation and explanation of what $\Delta_\mathrm{r}G$ means.

However, $\Delta_\mathrm{r}G$ does indeed, somewhat, refer to the instantaneous difference between the "molar Gibbs free energies" of the products and reactants. This is different from $\Delta G$ in three main ways.

First, $\Delta G$ is the difference between the Gibbs energy of the entire system at two points in time. Here, we are interpreting $\Delta_\mathrm{r}G$ as the difference between the "molar Gibbs free energies" of two components of the system: reactants and products. In other words, the system generally contains both reactants and products, and $\Delta_\mathrm{r}G$ may be thought of as the difference between the Gibbs energies of the product part of the system, and the reactant part of the system, even though you cannot separate them in the laboratory.

Secondly, it is the difference between the chemical potentials, not the molar Gibbs free energies. The molar Gibbs free energy is simply defined by $G_i/n_i$; the chemical potential is a partial derivative and is defined by $\mu_i = (\partial G/\partial n_i)_{T,p,n_j}$. In the case where there are no other species present (i.e. species $i$ is pure), then the chemical potential is identical to the molar Gibbs free energy.

Lastly, it is weighted by the stoichiometric coefficients $\nu_i$ instead of the change in the amount $\Delta n_i$. The stoichiometric coefficient is a dimensionless quantity, which is negative for reactants and positive for products. So, in the ammonium nitrate decomposition as written at the very top, we have

$$\nu_{\ce{NH4NO3}} = -1; \qquad \nu_{\ce{N2O}} = +1; \qquad \nu_{\ce{H2O}} = +2$$

and our expression for $\Delta_\mathrm{r} G$ is

$$\begin{align} \Delta_\mathrm{r} G &= \sum_i \nu_i \mu_i \\ &= \nu_{\ce{N2O}}\mu_{\ce{N2O}} + \nu_{\ce{H2O}}\mu_{\ce{H2O}} - \nu_{\ce{NH4NO3}}\mu_{\ce{NH4NO3}} \\ &= \mu_{\ce{N2O}} + 2\mu_{\ce{H2O}} - \mu_{\ce{NH4NO3}} \end{align}$$

Now, note the units again. The chemical potential is a partial derivative of the Gibbs free energy (units $\mathrm{kJ}$) with respect to the amount of $i$ (units $\mathrm{mol}$), and so it must have units $\mathrm{kJ~mol^{-1}}$. And from our above expression, $\Delta_\mathrm{r}G$ must also have units of $\mathbf{kJ~mol^{-1}}$, since the stoichiometric coefficients are dimensionless.

Does it matter if you start with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ of ammonium nitrate? The answer is now, no. The amount of starting material does not affect $\mu_i$, nor does it affect $\nu_i$. Therefore, $\Delta_\mathrm{r}G$ is independent of the amount of starting material.

Why is $\Delta_\mathrm{r} G$ an instantaneous difference between the chemical potentials? Well, that is because the chemical potentials of the reactants and products, $\mu_i$, are changing continuously as the reaction occurs. Therefore, if we want to calculate $\Delta_\mathrm{r} G$, we have to take a "snapshot" of the reaction vessel: otherwise it makes absolutely no sense to speak of $\mu_i$ because we wouldn't know which value of $\mu_i$ to use.

Compare this with $\Delta G$ above: the quantities of $G_{\mathrm{m},i}$ are constants that do not vary depending on the extent of reaction. Therefore, we do not need to specify a particular extent of reaction to calculate $\Delta G$.

A diagram to sum up

• Note that, to adequately define what $\Delta G$ is, you need a starting point and an ending point. I showed two possibilities for $\Delta G$; there are infinitely many more.


• Let's say the reaction goes to completion. If you double your starting material and double your product, the difference between $G_\text{products}$ and $G_\text{reactants}$ will also be doubled. So, it's important to specify!


• Likewise, to adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$.


• If you double your starting material and double your product, the curve is stretched by a factor of 2 along the y-axis, but it is also stretched by a factor of 2 along the x-axis because $\xi$ is also doubled. (For a mathematical explanation see the definition of $\xi$ given in the linked question earlier.) So, while the difference $\Delta G$ is doubled, the gradient $\Delta_\mathrm{r} G$ remains unchanged.


• The units should also be clear from this. Since $\Delta G$ is a difference between two values of $G$, it has to have units of $\mathrm{kJ}$. On the other hand, $\Delta_\mathrm{r}G$ is a gradient and therefore has units of $\mathrm{kJ/mol}$.

A caveat

Unfortunately, the notation $\Delta G$ is often loosely used and treated as being synonymous with $\Delta_\mathrm{r}G$. You will therefore see people give $\Delta G$ units of $\mathrm{kJ~mol^{-1}}$. For more information refer to Levine, Physical Chemistry 6th ed., p 343. I would personally recommend making a distinction between the two.

Are coordination numbers of elements fixed?

Can we say that the coordination number of a particular element is fixed? If not please give an example where an element exhibits variable coordination numbers.

meaning - 小さな vs 小さい/大きな vs 大きい

Is there any difference in meaning between 小さな and 小さい / 大きな and 大きい? I know that they have different syntactical rules, but semantically is there any difference? I was once told that 大きな was more appropriate for non-physical things, e.g. 大きな声.

Answer

Shogakukan's Ruigo Reikai Thesaurus Dictionary has this to say about differences in meaning and usage:

「大きい」は、「夢が大きい」「大きく夢みる」のように述語として、また 連用修飾語としても使うが、「大きな」は「大きな…(名詞)」という言い方でしか使わない。
Ōkii is used as both predicate and as a continuative modifier as in yume ga ōkii, ōkiku yume miru, but ōkina is only used in the format of ōkina...(noun).

物事の程度や、関わる範囲などが大であるという意で名詞を修飾する場合、その名詞が「問題」「影響」など抽象名詞のときは、「大きな」を用いることが多い。
When modifying a noun to mean that a thing's degree or relevant scope is large, ōkina is used more often when the noun is abstract, such as a "problem" or "effect".

There's also a table indicating different valid and invalid constructions, which I've clumsily recreated below. The - indicates an invalid use, while the △ indicates an acceptable but less-common use.

            ～声 | 声が～ | ～問題に発展する | ～影響がある | ～拍手で迎えられる
大きい     〇 |    〇     |               △            |           △         |                〇
大きな     〇 |    －     |               〇           |           〇         |                〇

As shown above, both 大きい声 and 大きな声 would be equally valid. The person who told you that 大きな should be used more for non-physical things might have used the wrong word; given the description above, the distinction appears to be abstract vs. concrete, rather than physical vs. non-physical.

Grammatically, 小さい and 小さな have the same usage constraints as 大きい and 大きな. The thesaurus does not, however, mention any differences in abstract vs. concrete between 小さい and 小さな.

employment - How to avoid helping someone sin?

Are there any tips anyone would care to provide to help one avoid (or get out of) a workplace situation in which one has been asked to assist someone do something that violates Halachah?

Examples that come up in the workplace:

• 
  

  Assisting with setting up non-Kosher food for a gathering, where some attendee(s) may be Jewish but not keep Kosher.

  
  


• 
  

  Assisting a disabled, Jewish individual get her non-Kosher lunch out of her hard to reach backpack. (This inspired the question, actually.)

  
  



• 
  

  Helping Jewish colleagues (or customers) prepare for business trips that involve work and/or travel on Shabbath and Yom Tov.

  
  

Answer

The first thing I would do in that situation is to try to provide the halachic options. For example, in setting up the lunch, I would order in some kosher food and set up a separate table for it, making sure it's well-labelled. This might be enough to entice the people who don't keep kosher to eat the kosher food, because they didn't have to make special arrangements. (If you're not already confident in your observance or aspirations of observance, it can be very hard to make that request for accommodation. But if the food's right there in front of you...) For the travel arrangements, I would first present itineraries that don't involve travel on Shabbat/Yom Tov and see if they're acceptable. There's not much you can do if the timing of the meeting/conference/whatever itself violates halacha, unfortunately, but in that case you're not the one placing the stumbling-block.

(I was once sent to a summer conference that started on a Sunday morning. At that company employees proposed itineraries for approval. My submission of a flight on Friday afternoon raised eyebrows, but there were no Saturday-night flights that would get me there in time. I pointed out that the extra two hotel nights were less expensive than the difference in airfares (Saturday stay-over being a big deal at the time) and they agreed. I went to Chabad for Shabbat meals. I mention this to illustrate that even if candidate travel arrangements seem like they would be rejected by the corporate powers that be, it's worth asking. For persuading the employee who'll be doing the travel, you'll probably need to point out some other reasons that spending extra time at the destination is attractive; nobody wants to just sit in a hotel room for an extra day.)

tefilla - Specific things to do when a Gadol passes away

Are there are any Segulot, Tefilot, Limudim brought down in the Sefarim for the passing of the Sadik HaDor?

verbs - Meaning of 欲ばっから and help with the following sentence

Context: Riku, the protagonist boxer of this manga, is fighting against an evasive opponent that is really good at dodging punches and who always tries to win on points rather than KOs. Riku tries to punch him without success, while his opponent manage to hit him with punches that are not so powerful but let him gain points. Riku's trainer tells him:

欲ばっからつけいられるんだよ！！狙うのはもっと…一番小っせーのだ。最小のカウンタに力を集約させろ！

What is the meaning of 欲ばっから? Is it 欲【よく】ばる+から? Or has it something to do with ばっかり? In any case, could you please explain the word formation?

Another thing I don't understand is the させろ at the end. Why wasn't しろ used instead? Is he ordering Riku to do something or to make to opponent do something? Anyway, here's my translation attempt:

He is taking advantage of your greed!! You should focus on the least powerful punch. Concentrate your strength against the least powerful counterpunch!

Here you can see the whole page and the one after it for more context. Thank you for your help!

Answer

• Yes, this 欲ばっから is 欲張るから ("because you're greedy"). In this context, this 欲張る refers to Riku's attitude of throwing too many ineffective punches whenever possible. What's going on here is that Riku's careless attacks are giving the opponent many chances to counterattack. What the second is saying at this point is that Riku should be more careful and wait for a good chance for 最小のカウンター.


• 最小のカウンター is a bit confusing, but seems to mean "a (small) counter-punch thrown right after his smallest punch" here. Riku should not wait for a "big chance" because the opponent is very good at defending. Instead, he should find a way to counterattack after the opponent's small jabs.


• 集約する can be used both intransitively and transitively. In this causative sentence, it's used as an intransitive verb, and 力 is the "causee". 最小のカウンターに力を集約させろ literally means "make your power gather to 最小のカウンター", which simply means "Focus on 最小のカウンター!". See also this question: Causative, causative-passive and particles. After this advise, Riku starts to observe the opponent more carefully and try to find a small chance of counterattack.

Can you use plastic cable-ties to hold down your schach in your sukkah?

I've seen people use plastic cable-ties to hold down the schach in their sukkah (as they're not "makabel tuma"), but others who don't like it, I think because they're not plant matter? Can anyone clarify who holds what, and why?

Answer

There are two issues:

The first is the more commonly accepted issue. There is a question as to whether you are allowed to hold up your schach with something that is pasul for schach. The common example was to use material with a bais kibul, so it is referred to as "maamid b'davar hamekabel tumah". The issue, according to those who assur, is that you may come to use these materials for the schach itself. An exception to this rule is to hold the schach up with something that people would not use for schach (shading). For this reason it is mutar acc. to everyone to rest the schach on a stone wall. Plastic is something that people use as shade and would be part of this issue.

I cannot say that the minhag haolam is to be machmir since I have seen many who rest their schach on metal and plastic poles or sheets, but bnei yeshiva are machmir and Rabbonim are machmir for those that ask, unless, like Dave said, this is your only option for holding the schach down. You also needn't turn down an invitation for a meal for this issue as halachically it is considered a chumra (maase rav).

The second issue is from the Chazon Ish (I did not see this inside, but was told by a well known posek who told me that he is noheg like the Chazon Ish), which is not to build the entire succah with a davar hamekabel tumah because of Tahara. This is not an issue for your plastic ties. Nor have I heard of any one else machmir for this except the aforementioned.

physical chemistry - Why is the rate of a reaction proportional to the concentrations of reactants raised to their stoichiometric coefficients?

Consider a gaseous state elementary reaction $$\ce{aA(g) + bB(g)} \overset{k_\mathrm{f}}{\underset{k_{\mathrm{b}}}{\ce{<=>}}}\ce{ cC(g) + dD(g)}$$

I know that for this reaction, $$\Delta G = \Delta G^{\circ} + RT \ln \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) \tag1$$

Now, if $$K = \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) ,$$

then $\Delta G = \Delta G^{\circ} + RT \ln K$.

Now, how can we say that $$K = \frac{k_{\mathrm{b}}}{k_{\mathrm{f}}}$$

and $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{f}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$ $\,$where $r_{\mathrm{f}}$ and $r_{\mathrm{b}}$ are rates of forward and backward reaction as assumed in this answer.

I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

Answer

I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

I might try to give you some intuition to back up that, for a given (elementary) reaction $\ce{A ->[k] B}$, the reaction rate $r$ can be written as

$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} \propto P_\ce{A}\text{.}$$

(Observe that the second equality above is true due to $P_\ce{A} + P_\ce{B} = \text{constant}$.) First, for an ideal gas, $P_\ce{A} = \frac{n_\ce{A}}{V} RT$. This means that $P_\ce{A} \propto n$ for fixed temperature and volume.

Let's say the reaction happens as a random process. That is to say that, for every time interval $\Delta t$, we have a probability per unit time $p$ of having a single molecule $\ce{A}$ turning into $\ce{B}$. If we wait longer, proportionally more molecules will turn. We'll thus have, for initially $n_\ce{A}$ molecules of $\ce{A}$, after $\Delta t$ seconds,

$$\Delta n_\ce{B} = p n_\ce{A} \Delta t\text{.}$$

This means that, in the time interval $\Delta t$, the population of $\ce{B}$ goes from 0 to $\Delta n_\ce{B}$ (assuming no $\ce{B}$ initially). From the stoichiometry of the reaction, $\Delta n_\ce{B} = -\Delta n_\ce{A}$ (i.e., there's conservation of moles).

Thus,

$$\frac{\Delta n_\ce{B}}{\Delta t} = -\frac{\Delta n_\ce{A}}{\Delta t} = p n_\ce{A}\text{.}$$

After taking the limit of $\Delta t \rightarrow 0$ and multiplying by $\frac{1}{V}$ (i.e., dividing by volume), we get

$$\frac{d \frac{n_\ce{B}}{V}}{dt} = \frac{d [\ce{B}]}{dt} = p \frac{n_\ce{A}}{V} = p [\ce{A}]\text{,}$$

where $[\ce{A}]$ and $[\ce{B}]$ stand for the concentrations of $\ce{A}$ and $\ce{B}$, respectively. Since, again for an ideal gas, $[\ce{A}] = \frac{P_\ce{A}}{RT}$, we get

$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} = p P_\ce{A}\text{,}$$

where the proportionality constant is the probability of reaction per unit time.

culture - Why are some words written backwards on trucks

I was driving the other day and saw a truck with 一般 written as 般一 on the drivers side door. My wife was telling me that this is often the case with trucks, where it is actually written from right to left in 縦 style, as if there was only 1 row. Just was wondering if there was a particular reason for it or if this is limited to just trucks or can this be found elsewhere.

Of note, it was NOT written flipped so as to be seen correctly when looked at through a mirror like ambulances and such in the U.S.

Note: This question is on the border in terms of being off-topic, but since it is referring to language use within Japanese culture, I felt it was appropriate to ask.

Examples of text
http://septieme-ciel.air-nifty.com/nikubanare/2005/09/post_5b58.html

Answer

Matt's answer is right enough, and Axioplases's description does have historical accuracy, but I felt differently enough to propose another answer.

First, here is the truck in question, with the words カンガルー便 written on the side, "backwards".

Note, though, that the text for the parent company, Seino, is the "right" way round, presumeably because it's in romaji, which is not as flexible as Japanese kanji and kana in terms of direction. If this were simply a matter of viewing the car in a "front is top" concept, then why not also include the romaji? It's only the Japanese text that is flexible enough about direction.

How flexible is Japanese in terms of direction? A little. The root of the issue lies in the fact that Japanese text is traditionally written vertically. The choice to go left or right when writing horizontally is, or was, therefor somewhat arbitrary. Back in the day, one would have come across writing right to left more than today, as in this old train sign:

Or how the car brand "Ford" is written ドーォフ (or ドーオフ) on this Taishō era building:

Or in this awesome political map:

Note in all these pictures that the romaji offered goes left-to-right, but both kanji and kana are right-to-left.

Why do these historical examples matter? Because it's evidence that there is a cultural basis for accepting text right-to-left that makes the truck sign possible in Japanese culture.

What I'm driving at is the contrast with English culture where the direction of text is 100% locked in to be left to right. If a hypothetical delivery company in an English country wrote "Yreviled Ooragnak" on the side of their trucks, because they wanted to follow a "front is top" logic, or any other rational, no one would have any doubt they were doing something wrong. The horizontal direction in English is non-negotiable, with the only exception being deliberabe subversion of the norm in some kind of artistic context.

Of course, the massive exposure to English text has influenced textual presentation in Japan, to the point where you don't see this much anymore. But it's not a rule that writing must go left to right, and some people will still go right to left, like this racist jerk here, who probably did it precisely because he doesn't want to play by the west's rules (Just so its clear, although the text direction supports my point, the content of this jerk's sign is unredeemable, and I strongly oppose the sentiment):

So the "backward" phenomena is not just about vehicles or flags, or a front-as-top logic. Japanese text, insofar as it has escaped English cultural hegemony, can be flexible about which horizontal direction it goes in. Why exactly Seino opted to go right-to-left on the right side of their trucks, I'm not sure, I just know that the option to do so existed for them because the language permits it in a general sense.

Maybe right-to-left writing will die out, as it does seem to be the exception these days. Or maybe not. Consider this nostalgic retro-branding of a biscuit box, made available in 2014, preserving the right-to-left writing, for the Japanese text only, to give an old-timey feel. Maybe one day a retro trend will spark a revival...

What I'm pretty sure you won't see, though, is bottom-to-top. The text direction is not that flexible.

matlab - Different way to separate a particular frequency from a signal

Let's say I have a signal which is composed of 2Hz,10Hz,17Hz,19Hz and 25Hz discrete time sinusoidal waves and I need to allow only 17Hz component to pass.

As far as I know, the standard way to approach the problem is implementing a Band Pass Filter.( Analog Prototype Filter -> Analog Filter -> Digital filter using Bilinear Transformation).

We can also implement DFT on the signal and use the information from the spectrum and then reconstruct the 17Hz component,right? If yes, which method is preferred and why?

Answer

If the frequencies to be rejected are the discrete frequencies given, and the sampling rate can be commensurate with these frequencies, then a moving average filter with the coefficients complex rotated to shift the response to 17 Hz can be done to reject all other integer frequencies as in the plot below.

This is the concept of "homodyning the coefficients" rather than homodyning the signal. When we homodyne a signal, the signal is multiplied by a complex rotator ($e^{-j\omega t})$ which will in the process shift it in frequency according to the rate of rotation. In similar fashion, rotating your coefficients will shift the response of the filter. So you can either move your signal to the filter as in a downconversion followed by a low pass filter... or move your filter to the signal as in this case!

Here is the Matlab/Octave code that implemented this filter.

In this case the sampling rate is 26 Hz with a complex (I and Q) signal path.

z=exp(j*2*pi/26);
coeff=z.^([0:25]*17);
[h,w]=freqz(coeff,26,2048,'whole');
plot(w/pi*13,20*log10(abs(h)))
grid on
axis([0 26 -50 0])

(Note, this is the same frequency response as one bin in the DFT)

The coefficients themselves follow the pattern as shown in the following complex diagram, where the first 5 coefficients are labeled, and the rest can be followed along the trajectory lines shown. Observe that the magnitude of each coefficient is one (as in a moving average filter) but the phase is continuously rotated which causes the frequency response to shift at the rate of rotation.

Update: As further suggested by @MarcusMüller in the comments, actually doing the signal homodyne followed by a low pass filter may likely be simpler in this case given that the OP has variable conditions for the signal of interest. Assuming the interference to be rejected is still discreet tones with an integer spacing relationship to the signal of interest (the signals need not be integer frequencies, but the multiplies of their spacing must be, as the nulls of a moving average filter will be located at n/T Hz where n is any integer and T is the time duration of the moving average in seconds).

So in this case we would first multiply the signal by $e^{-j2\pi 17 t}$ and then follow the (complex) output with a moving average filter (which means two filters, one for the I (in phase or real) path and one for the Q (quadrature or imaginary) path. In this case the moving average length as a minimum must be:

2Hz,10Hz,17Hz,19Hz and 25Hz

Move 17 to 0:

-12Hz, -7Hz, 0 Hz, 2Hz, 8Hz

And the least common multiple is 1 so a 1 second moving average is required with a commensurate sampling rate. Further as @MarcusMüller has suggested, this moving average structure could be implemented as a CIC decimating filter. The output would be the "DC Equivalent" of the signal of interest, with the same amplitude and phase as the signal at 17 Hz. Further if the signal that is desired changes in frequency, the phase rotator can be changed in one place rather than updating the taps as I had done.

For further reading see:

Phase rotator: Numerically Controlled Oscillator (NCO) for phasor implementation?

CIC: CIC Cascaded Integrator-Comb spectrum

Note also to be considered as a filtering approach is the 2nd order resonator (or cascade of two of these structures), based on a rotated exponential averaging filter. (see FFT analysis for Vibration Signal). The image of this filter structure is duplicated below (see link for details on what the variables mean and how to compute them):

sampling - Nonnegative or positive band-limited interpolation

Given samples of an everywhere non-negative or positive-valued continuous-time signal band-limited to half the sampling frequency, is there some practically applicable way to interpolate it so that there is no danger of producing negative values? The approach should be scalable to arbitrary accuracy so basic linear interpolation won't do. By arbitrary accuracy I mean an arbitrary small difference between the band-limited continuous-time signal and the interpolation.

Perfect sinc interpolation would work because negative contributions would get cancelled by positive contributions. Ideal sinc interpolation is not possible in practice so other filters are used. This could give negative values because usually the filter impulse response has negative values. For example cubic Hermite spline piece-wise polynomial interpolation gives this impulse response:

Figure 1. Cubic Hermite spline impulse response has negative values.

Lagrange polynomials used as splines would also have negative values in the impulse response. B-splines have everywhere non-negative impulse responses, but their pass-band frequency responses deviate increasingly from the brick-wall frequency response of sinc interpolation with increasing B-spline order. The B-spline impulse response tends to a Gaussian function as the order is increased. A Gaussian function in time domain is a Gaussian function in frequency domain, so not a perfect brick-wall filter.

The obvious way to fix the problem would be to 1) take the absolute value of the interpolation output or 2) set all negative output values to zero, but I'm wondering if there is some better (inherently non-negative) method, because both 1) and 2) would create discontinuities in the derivative of the signal. This also creates spectral images.

I don't know how bad the problem stated in this question really is, because if the band-limited signal is non-negative, this also rules out some contradictory discrete signals like $[\dots, 0, 0, 0, 1, 0, 0, 0, \dots].$ The corresponding non-negative continuous signal would then have to be sinc, which is partially negative, which is the contradiction.

---

Shay Maymon has investigated "bandlimited square-roots" in his 2011 thesis Sampling and Quantization for Optimal Reconstruction. The idea there is to find (not easy) another band-limited signal (complex and at half the bandwidth) the squared magnitude of which equals the wanted band-limited signal. Interpolating that other signal and multiplying the result by its complex conjugate is never negative. That reminds me of this question about interpolation of the magnitude of discrete Fourier transform (DFT).

grammar - How does は apply itself to によっては?

From the question: "How does the use of いかんによっては in this question determine one answer over another?", it is observed that:

• The use of によって means to change state or behaviour depending on, or according to something. It expresses variety.


• The use of によっては pinpoints one outcome from a range of possible outcomes.

While the previous question deals with appropriate usage determination, this question seeks to understand the underlying principle governing it.

(Question) Why does the presence of は do this, and how does this link back to what we know of は?

What I typically know of the particle は, is that it is a disambiguative particle and has thematic and contrastive roles.

Answer

I think you've picked up on my explanation, which was taken from 新完全マスター文法N3, 日本語表現文型辞典 so, as I do find the expression with は　intuitive/natural, I should expand on my previous answer.

The definitions were:

によって：means to change state or behaviour depending on something or according to something. It expresses variety and is often used with さまざまだ　and かえる.

によっては：pinpoints one outcome from a range of possible outcomes.

Sentences given in the book applying によって　are:

国によって習慣が違う

感じ方は人によって様様だ。

One sample sentence in the N1 book pinpointing a single outcome　using によっては　was:

私の帰宅時間は毎日違う。日によっては夜中になることもある。

I get home at a different time every day. Some days I get even home in the middle of the night.

A similar sentence to this last sentence applying the same principles used in the first two sentences to express variety using によって　would be:

私の帰宅時間は日によって違う。 The time I get home varies from day to day.

If we work backwards, taking the last last sentence as the starting point and then seek to pin point one result, taking "as for" as a crude translation of は　then we would say:

Every day I get home at a different time. "As for some days, depending on the day itself", I even get home in the middle of the night.

Which takes us back to the sample sentence of によっては with what I hope is a crude but reasonable explanation of how は　is used not just in this expression but in a manner that is consistent with some of its normal uses (raising/drawing attention to/ emphasising a topic)

Consider also other ～ては structures:

• ～については


• ～に関しては

BによってA is used in sentences to make statements such as "A varies, depending on B".
BによってはC is used to make statements such as "As for all possible results dependent on B, C can happen", C being a member of set A.
は selects Bによって as a topic for further discourse, similar to ～については and ～に関しては.

grammar - How does the passive function in this sentence

The sentence is an example taken from the grammar section of my textbook. 先生が手紙を書いてくださったおかげで、大きい病院で研修を受けられることになった。 => "Because the teacher kindly sent a letter, it happened that I could absolve an internship in a big hospital."

In the section where this sentence comes from "...おかげで/おかげだ" is explained and how it puts the positive outcome X into causal relation to Y. However, this shall not be the focus of the question here. What confuses me here is the use of the bold part 研修を受けられる.

Just before I also learnt about various new applications for the passive mood in japanese. I must say now that this part wasn't very well done by my textbook since it felt like it would mix up some grammatical categories or at least it didn't explain its point well. In this section, I was taught two new ways of using the passive mood:

1) indirect passive (intransitive verbs) 2) indirect passive (transitive verbs)

These are the headings for these two subsections used in the textbook. Now, I didn't really understand these two subsections. One reason therefore is that my textbook showed me examples for

1) indirect passive (intransitive verbs)

with transitive verbs Oo Here's another example: 部長は私に仕事を頼んだ。-> 私は部長に仕事を頼まれた。 According to jisho, 頼む is a transtive verb. Did I understand something wrong about the category "intransitive" here? Do I need to understand this category in a different way here?

Furthermore, the translations for these examples didn't make any effort to transport the semantical difference between a "normal" sentence and these passive constructions. So, while I can definitely still translate and understand such sentences, the nuances brought in by these constructions are lost to me :(

Answer

This 受けられる is the potential form of 受ける. Passive voice is not used in this sentence.

れる/られる has four major meanings. See: Passive usage of 「済まされない」 in sentences

Regarding the usage of を, 研修が受けられる and 研修を受けられる are interchangeable here. See: The difference between が and を with the potential form of a verb.

(This potential れる/られる is a more basic topic than ～おかげで, so I personally don't think the textbook is bad.)

organic chemistry - Preparation of picric acid

My textbook says "Nowadays picric acid is prepared by treating phenol first with concentrated sulfuric acid which converts it to phenol-2,4-disulfuric acid and then with concentrated nitric acid to get 2,4,6-trinitrophenol"

My question: Why is this a better method than simply treating phenol with concentrated nitric acid to obtain picric acid?

What I think is that maybe because $\ce{NO2}$ is a stronger deactivating group than $\ce{SO3H}$, it is more difficult to add 3 $\ce{NO2}$ to a benzene ring directly.

But we would also have to substitute $\ce{SO3H}$ for $\ce{NO2}$. Wouldn't that be hard?

Answer

The benzene ring in phenol is highly activated toward electrophilic substitution and hence attempts to directly nitrate it result in charring and copious evolution of oxides of nitrogen. The reaction is highly exothermic and difficult to control.

To reduce the reactivity, the phenol is first mono-sulfonated ( some of the product which is substituted may also be used). The products are ortho and para isomers. The para isomer is separated and then nitrated. The nitration is comparatively far smoother (easier to handle). Ipso substitution of -SO3 groups occur.

---

Links: Wikipedia, Research paper

Link: (For ipso substitution): PDF, Wikipedia (see subsection: Ipso substitution)

discrete signals - What kind of filter I should use to remove the oscillations in this autocorrelation function?

I have an autocorrelation function which is shown as following,

I do believe these trailing oscillating wigs are spurious and should be removed by some kind of filter. But I am not familiar with any types of filters. Any help will be appreciated! The original data can be downloaded at

~

Update @ Apr 14 2017:

This result was processed with the filter suggested by Dan.

Answer

The following comb filter structure can be used to filter your data prior to computing the autocorrelation which will significantly reduce or eliminate the ringing due to the repetition in your signal. However be aware that it may also be filtering signal content based on the spectral occupation of your signal. The structure of the signal and the frequency response is shown below (for N=101).

Where N is the number of samples to be a 0.5 second delay. Implemented in Matlab using

out = filter([1 ones(1,N-1) -1],1,in)

This filter above as given is a highpass response so depending on the spectral content in the signal of interest, this may not be ideal.

Another option that will equally create nulls at the repetition rate and harmonics of the repetition rate is a moving average filter:

Implemented in Matlab using

out = filter([ones(1,N)],N,in)

Note in this case the filter was normalized, a 2 could be used as the second parameter in the first case to normalize that filter as well.

Note that the first filter removes DC and is a highpass function which may not be desirable, while the second filter has a 1/f frequency roll-off as a sinc function in addition to removing the harmonic nulls which will likely widen the autocorrelation response. If the high frequency content is desired, a third filter option that would be interesting to see can be done based on this post

Transfer function of second order notch filter

This filter also removes DC but can have a much tighter nulling response and flat elsewhere based on a nulling parameter $\alpha$

By inserting zeros in the frequency response we can get the base spectrum to repeat, so in your case you would insert N-1 zeros to perform the following filter function (see referenced post for structure of this filter):

out = filter((1+alpha)/2*[1 zeros(1,N-1) -1], [1 zeros(1,N-1) -alpha], in)

The frequency response for $N = 10$ and $\alpha = 0.9$ is shown below. (Your N of course is much larger). If you vary $\alpha$ less than or more than what I used up to but not equal to 1 (I recommend using $\alpha$ in the range between 0.7 and 0.999); the nulls get sharper as $\alpha$ approaches 1.

If you do implement these on your data, please post the results for all these cases so we can see the difference it has caused.

history - Is this the Chofetz Chaim?

What is the story behind this picture being known as the image of the great Rabbi Yisrael Meir Kagan(Chafetz Chaim)? Is it him in fact? Where does it come from?

The only true picture (that I am aware of) shows him sitting down, at a distance and is rather unclear:

This website mentions that the first image could be that of his Shamas or wagon driver. It describes the above photo (the second picture) as follows: "The above picture was taken in Radin and the Chofetz Chaim is seen speaking to his oldest son, Rabbi Leib Poupko who eventually became Rav in Radin and President of the Mizrachi in Poland. He passed away in 1939. The CC’s last Rebbetzin, also who was an adopted daughter, is standing behind him. There’s also a young woman in the picture standing next to the Chofetz Chaim. Rav Chaim Wolkin, the Mashgiach in Ateres, is seen posing (in a walking pose, which was apparently a popular thing in those days) in the street between the photographer and the house".

Are there any other photos of the Chofetz Chaim to which we can compare the first picture for authentication?

Answer

Yes, the painting is based on a popular picture of the Chofetz Chaim, which can be seen in The Schwadron Collection of the National Library of Israel (Jerusalem). The archive lists the picture as following:

A photo portrait of Rabbi Israel Meir Cohen ("Chafetz Chaim"): printed silver, black and white, 7X12 cm. Portfolio also includes a copy of this portrait, which contains on the back a summarized biography of the subject, written by Dr. Abraham Schwadron in Yiddish.

It is undated but clearly identified by the archive as the Chofetz Chaim. Here is that picture:

Although it is undated, we know it was circulated in his lifetime. The collection also includes clippings from Polish newspapers and other sources that reprint the picture, and label his name with shlita (may he live for many good days amen). For example:

The collection also contains another picture of the Chofetz Chaim from a newspaper, which alleges that he is 90 in the picture (notice the cane, also seen in the sitting picture you included):

matlab - Is it possible to do deconvolution with two data sets that have different sampling rates?

I have some terahertz spectroscopy time series data, a reference set with 2048 data points taken every 0.0521 picoseconds, and the sample data set with 544 data points taken every 0.0781 picoseconds. I'm using Matlab to take a FFT of both sets, with zero padding on the sample set up to 2048 points, and I'm supposed to do a deconvolution on those transformed sets on a specific range of frequencies, but because the sampling rate is different for the two, these correspond to different starting points in the data sets, and the lengths won't match up either. When plotting the two DFTs, they only line up properly when accounting for the different sampling rates when spacing the frequency axis.

Is there something I'm doing wrong, or how should I do the deconvolution?

grammar - になるほど in this sentence

I can not understand very well the meaning of "になるほど" in this sentence, from a Japanese Dragon Ball Guide Book.

(合体によるパワーアップは、)２人の戦闘力の合計ではなく、掛け算になるほどの凄まじさ！

Is it maybe something similar to the construction "The more... the more..."? Or does it have relations with "I see" なるほど?

And, in your opinion, can this translation be correct?

"The Power-Up of the fusion (context: by means of Potara earring), it's something impressive closer/more similar to multiplication rather than a sum of the power forces of the two warriors."

Answer

XほどY (or XくらいY) in this context means "Y to the point where X" or "so Y that X". The "the more ～, the more ～" construction is not relevant. This なるほど is not relevant, either. You have to add の if the following modified word is a noun. 凄まじい is an adjective, and 凄まじさ is its nominalized form ("-ness").

• 目で見えないほど小さい so small that it's invisible; too small to see


• 驚くほどの値段 the price so expensive/inexpensive that it's surprising; surprising price


• 問題になるほどの大きさ the size big enough to be problematic

So かけ算になるほどの凄まじさ is "intensiveness to the point where it becomes a multiplication" or "the tremendous power going up in a multiplicative manner (rather than the usual additive manner)". The sentence says the fused power increases in a way the fusion of 100 and 100 becomes 10000 rather than 200, for example.

gentiles - Publicly Teaching Halacha

I was looking at the Daily Halachah Yahoo group and was a little bit surprised at this statement:

It is therefore forbidden for Jews to teach non-Jews Halachah

Is this true?

(See a related question on the Meta).

parshanut torah comment - Was Adam placed in Gan Eden to keep the service of HaShem?

Bereshit 2:15 reads: לְעָבְדָהּ וּלְשָׁמְרָהּ, but it seems that this can't be referring to: 'to work the earth and keep it', because after the chet (the sin) it reads in Bereshit 3:23 that Adam was taken out of Eden to work the earth, לַעֲבֹד אֶת-הָאֲדָמָה

The Targums read: and set him to do service in the law, and to keep it('s commandments), while the Zohar points out that l'avdah refers to the positive commandments and uleshomrah to the negative commandments (the do's and don'ts).

But I wondered if there are any scriptures or comments referring to keeping the service, what kind of service was this?

Could Gan Eden be a part of The Heavenly Temple where Adam served and worshiped HaShem?

Hoping for some insight.

Answer

• Avot D'Rabbi Nattan (ch. 11) quotes R. Shimon Ben Elazar who is pretty explicit that this refers to physical labor; cited by Meiri to Avot (ch. 1). This is clear from the Hovot HaLevavot (Shaar HaBitahon ch. 3) and is the implication of Recanti as well.


• Sifrei (Ekev 41) writes that it refers to service of God Specifically, "to work it" refers to [Torah] study, and "to guard it" refers to mitzvot. This is paraphrased by the Maor HaAfelah (Parashat B'reishit p. 37) and the Midrash HaBeiur (Parashat B'reishit p. 34).


• B'reishit Rabbah (Vilna: ch. 16) indeed suggest that both verbs refer to offering sacrifices; cited by Ramban's commentary to the verse.



• The Zohar too writes that this refers specifically to sacrifices (Parshat: Toldot, Miketz, T'rumah, and Vaetchanan) If I understood it correctly.


• Tikkunei Zohar (70) interprets it more broadly to refer to Torah and mitzvot.

molecular orbital theory - Hyperlithiated Carbon Species

Below is replicated Question 1 from the Final Qualifying Exam of the Australian Chemistry Olympiad, 2004B, here.

Question:

The theory of promotion-hybridization is quite successful at explaining why covalent molecules with a central atom from period 3 or higher (e.g. $\ce{SF6}$, $\ce{PF5}$, $\ce{I3}$) are stable even though they are hypervalent (i.e. they appear to "exceed the octet rule"). However, hybridization is inadequate in explaining the remarkable stability of molecules such as the so-called "hyperlithiated" carbon species such as $\ce{CLi6}$. Molecular orbital theory can be applied to $\ce{CLi6}$ (Reed and Weinhold, 1985) for an alternate explanation to when and why hypervalent molecules with central atom of period 2 can be stable.

Of all the different types of atomic orbitals (e.g. $s$, $p_x$, $d_{yz}$, etc) of a general central atom, six of these will participate in the forming of sigma bonds with the molecular orbitals of six ligands in an octahedral geometry. The valence $s$ orbital is one of them. Note that the ligands approach the central atom along the three coordinate axes.

(a) Which other five atomic orbitals can also do this?

Now, we shall consider the molecular orbitals (MO) of the 6 $\ce{Li}$ $2s$ orbitals (in an octahedral geometry). The lowest energy MO is given below in figure 1:

According to quantitative calculations, the MO in figure 1 is the only MO lower in energy than the uncombined lithium $2s$ orbital; all other MOs are slightly higher in energy than uncombined $\ce{Li}$ $2s$ and can thus be considered as antibonding. (Note that this is generally true for 2 to 8 Li atoms arranged symmetrically around the central atom).

The $s$ orbital will overlap with the MO given in figure 1. Exactly one of the other five MOs will have the right symmetry to overlap (and form a sigma bond) with each of the other five atomic orbitals of the central atom.

(b) Hence draw the other five MOs obtained by linear combinations of the 6 $\ce{Li}$ $2s$ orbitals. Indicate their relative energies on an energy diagram.

When the central atom is carbon, three of the six MOs (of the 6 $\ce{Li}$ $2s$ orbitals) will not combine with the corresponding atomic orbital of carbon. One of these is the one drawn in figure 1. Because carbon is much more electronegative than lithium, the $2s$ orbital of carbon is too low in energy to mix with the MO in figure 1.

(c) Which other two MOs will not combine when carbon is the central atom? Provide a rationale as to why this is so.

(d) Hence, draw the molecular orbital energy diagram of $\ce{CLi6}$. Keep in mind the relative energies of carbon AOs and the six Li MOs (as carbon is more electronegative than lithium).

(e) Using the energy diagram in part (d), explain the stability of $\ce{CLi6}$.

(f) Propose an alternative to the "octet rule" when predicting stability of hyperlithiated species of period 2 atoms. Using this rule, identify possible neutral (i.e., not cations or anions) hyperlithiated species of nitrogen and oxygen that can exist due to the same stability reason as that of $\ce{CLi6}$.

(g) Can $\ce{CH6}$ exist? If so, why so? If not, why not?

My response:

(a) These are the $p_x$, $p_y$, $p_z$, $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals.

(b) Assuming analogy with octahedral complexes, Googling provides this, pg. 10 of which appears to answer the question. Given that $E_d$ < $E_s$ < $E_p$, the $d$ orbitals represented would be found at the bottom and $p$ at the top of the energy diagram.

• Is an octahedral complex analogous to $\ce{CLi6}$?


• What is a linear combination of atomic orbitals?

(c) By logical reasoning, these would be the two similar $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals. I am uncertain regarding the rationale.

• Is it because it is unfeasible for carbon's electrons to be in $3d$ orbitals?

(d) I really have no idea.

• If three of the six possible MOs will not combine, does that mean that they are not present in the MO diagram?


• Since carbon is more electronegative than lithium, it should contribute more to the bonding orbitals, while lithium should contribute more to antibonding orbitals.


• How are the 10 electrons in $\ce{CLi6}$ distributed? In order to impart stability, there must be more electrons in bonding than in antibonding orbitals. Can we simply have three bonding MO orbitals with 2 electrons and three antibonding MO orbitals with 4 electrons between them?

(e) By logical reasoning, the larger number of electrons in bonding rather than in antibonding orbitals imparts stability to the molecule since bonding electrons are not entirely counteracted by antibonding electrons.

(f) I really have no idea.

• By guesswork, I would propose the decuplet rule, in which hyperlithiated species from period 2 are stable with 10 valence electrons, producing the species $\ce{NLi5}$ and $\ce{OLi4}$.

(g) By logical reasoning, yes: it satisfies the conditions of (f).

• My gut says no.

Given that my background in chemistry is limited to high school AP Chemistry, please curtail the material provided in your responses. Thanks for all your help in advance!

Answer

(a) Correct.

(b) Let me start with Linear Combination of Atomic Orbitals.

That's a quantum calculation method that assumes that all molecular orbitals can be derived from (basic) atomic orbitals by a method called linear combination. You might have heard of linear combination from vector geometry. It's basically just adding things up with different preceding factors. So 2x - 3y is a linear combination of x and y.

The thing about linearly combining atomic orbitals is that you must end up with the same amount of orbitals as you started out with (so you combine 6 atomic orbitals to make 6 molecular orbitals). There's also additional stuff you need to take care of (the squares of all the preceding factors of one AO need to sum up to give one), but that's only of marginal importance here.

An example of linear combination would be your average H–H sigma bond. You have two atomic orbitals (1s of each of the hydrogens) which you combine. I'll call the 1s orbital of the left hydrogen l and that of the right hydrogen r, for simplicity. The two simplest linear combinations are l + r and l - r. In the first one, you get positive interference, i.e. you add a positive phase to a positive phase – think of it as the positive half of a sine wave added onto another positive half of a sine wave: it'll get bigger. This type of interaction is favourable, and will thus give you a molecular orbital energy that is less than that of a separated atomic orbital.

But you also need to remember to create a second MO, the l - r. Here, you would be adding plus to minus, which gives zero. So you have a plane in between the two protons where the resulting wave function has zero value, is non-existant (a nodal plane). Because the nodal plane is between the atoms, the resulting molecular orbital is antibonding, or higher in energy. The basic rule is that the more nodal planes an orbital contains, the higher its energy level will be.

The question asks you to find the further five linear combinations that will lead to those group orbitals capable of overlapping withthe central carbon's orbitals. So you basically know what to look for; and yes: the CLi₆ is analogous enough to an octahedral complex, that you can just copy that image. If you didn't have either the picture or the hint, you would need to use group theory and reduction formulae, which I won't cover here (for those interested: the group orbitals will transform as a_1g, e_g and t_1u).

I don't understand your reasoning for putting the d-type orbitals lower in energy (i.e. more stable) than s- or p-type ones. Usually, it's s < p < d. But the better way to answer the question would be using nodal plane counting: The orbital that will mix with the carbon'S s orbital has no nodal plane, and will therefore be the lowest (a_1g, for those interested). The three that will mix with the carbon's p orbitals are second, because they contain one nodal plane each (t_1u). Finally, those that will mix with d orbitals are highest, because they contain two nodal planes (either parallel or perpendicular to each other. Their energy level will be the same, though, says group theory) (you guessed it, those are e_g).

(c) Yes, that can be regarded as the easiest answer. Carbon, being an element of the second period, only 'has' (term is incorrect, but you get the picture) 1s, 2s and 2p orbitals. 3s and 3p might be accessible by irradiation. 3d is not, as it is even higher in energy than 4s.

The same rationale applies to the higher order main group elements. Even though silicon, being an element of the third period, could be considered to access d orbitals to create anions like $\mathrm{SiF}_6^{2-}$, the d orbitals' energy is too far removed from that of the s and p orbitals, that it cannot be considered to take part in bonding.

Note that transition metals – the elements usually associated with octahedral surroundings and d-orbitals – usually use 4s and 3d (and sometimes 4p) orbitals. 3d is a lot closer to 4s and 4p than to 3p.

(d) Start off by thinking what the relative energy levels of the orbitals in question are. Carbon, as was pointed out, is more electronegative than lithium, so all of its orbitals will be lower in energy than the lithium orbitals (the difference is significant here).

That should give you a picture of carbon orbitals on one side (2s, rather far-ish down, three 2p's, still pretty far down) and the three sets of lithium group orbitals on the other side (the one that I called a_1g being the lowest, but probably higher than carbon's 2p's; the three I called t_1u next and the two e_g last; those inside a group at the same height, of course).

Next, you need to find out which orbitals can recombine favourably. If you attempted to combine the orbital shown in figure 1 with a carbon 2p orbital, that intereference would be non-bonding or zero (you have bonding interactions on one side cancelled out by exactly the same amount of antibonding interactions on the opposite side). Note that they helped you by saying that the figure 1 orbital (and thus also carbon's 2s) and the two lithium orbitals that would bond with d-type orbitals do not take part in favourable interactions; so only the p-type ones remain, each of them creating a bonding-antibonding pair of molecular orbitals.

Moving from the edges to the centre of the diagram, you would move the carbon's 2s plus the three lithium group orbitals not taking part horizontally into the middle. Carbon's 2p and the other three lithium group orbitals are then mixed, lowering the 2p's energy while raising the lithium orbitals' energy.

You should end up with an energy diagram of the following type:
C(2s) - 3 C(2p) -- 1st Li -- 3 2nd Li - 2 3rd Li. It's not clear (nor can it be deducted a priori) whether the 2p orbitals will end up lower than the 2s (or the 2nd Lithium group higher than the third), but it is to be assumed that they shouldn't change their places too much.

Finally, fill in the electrons in the way you're used to. Two go into C(2s), six into the three C(2p), and two into the first Li orbital (the one shown in figure 1). To answer your questions:

• non-combining MOs can or can not be present in the MO diagram. They should, if they are part of the valence orbitals. They should not if they are core orbitals (note we always omitted the seven 1s orbitals). Not combining however means, that they do not change height from where you put them initially.



• Yes, that is correct, but that is more of a consequence we can think of after mixing has been done.


• If we include the nonparticipating orbitals, the question is no longer a question. If they weren't present, though, you would do exactly what you proposed. You would fill in three electrons spin up into three antibonding MOs and one further electron spin down. The resulting system would be degenerated. It would not be stable, and would reduce symmetry to force the three orbitals apart, creating a discrete energy level for the single paired electron.

(e) Yeah, that was a pretty good guess into the wild, that would always be correct ;)

To use the molecular orbital diagram as a reasoning, though: Consider the state before bonding. The four carbon orbitals have four electrons between them (2 in 2s, 2 degenerated in 2p). The six lithium orbitals have six electrons between them (2 in the figure 1 orbital, 4 in the three orbitals that will take part in bonding; this state is also degenerated). After the bonding, not only have we moved four electrons out of relatively unstable orbitals into a lot more stable ones (those from the lithium orbitals into the carbon-centered ones), but we have also removed two degenerated states and turned them into clearly defined ones. Both are good for the system's stability.

(f) I think your guesswork is pretty much spot on. (The orbital diagrams would be totally different ones, though; as NLi₅ would be a trigonal bipyramid (symmetry group D_3h) and OLi₄ a tetrahedron (symmetry group T_d) as opposed to the carbon octahedron (O_h).

(g) I'm pretty sure this is a yes-but case. Yes, just looking at what we have for Li would suggest that H would do the same thing. But, hydrogen's atomic orbitals are lower than lithium's, as it is a lot more electronegative. That in turn means that there is significant mixing to be observed between C(2s) and the figure 1 type orbital. Therefore, the C(2s) orbital would again be stabilised, while the hydrogen group orbital would be destabilised – most likely so far that it reaches the energy level of the three higher group orbitals. While they were clearly empty in the lithium case, they would most likely be populated in the hydrogen case, destabilising the system.

Another thing to consider is the relative stability of CH₄ versus CLi₄. Again, the all equal phase group orbital of the lithiums would be too high to significantly mix with the C(2s) orbital in a tetrahedric case. The all equal phase group orbital of hydrogens does mix quite significantly, creating a much more stable CH₄ molecule than CLi₄. Thus, CH₆ would eliminate hydrogen to create CH₄, while CLi₆ is stable enough (and CLi₄ unstable enough) to not eliminate Li₂

hashkafah philosophy - Is tearing the hymen something that should be avoided

I have always heard that tearing the hymen by beilas mitzva (first intercourse with a virgin after marriage) is a good thing and it should not be avoided (maybe to tear it is the mitzvah as it seems to me from shulchan aruch harav 280.3). (Though to be fair I did not know that it was possible to avoid it.)

@Aaron is claiming that tearing the hymen can and should be avoided. (As any other damages to a Jewish body)

Is @Aaron right?

I'm looking for sourced Jewish views on the subject.

inorganic chemistry - Why does tetrachlorocuprate(II) form in aqueous solution even though water should be the stronger ligand?

In aqueous solution, $\ce{Cu^{2+}}$ forms the $\ce{[Cu(H2O)6]^2+}$ complex. Given that water is a stronger ligand than $\ce{Cl-}$, though, why does why does the $\ce{[CuCl4]^2-}$ complex form upon addition of chloride ions?

Answer

[Water ligands] being stronger than $\ce{Cl-}$ ligands will form a $\ce{[Cu(H2O)6]^2+}$ complex.

This is incorrect. Water ligands may be stronger field ligands in the spectrochemical series but that does not in any way affect the strength of the ligand–metal bond or the kinetics and thermodynamics of certain complexes, their formation and their breakdown.

Which ligand forms bonds of which strength to which metal is only very generally determinable; and in general you would always denote binding affinities in a $1:1$ manner: ligand $\ce{X}$ forms strong bonds to metal $\ce{Y}$. So sulphide ions will form strong dative bonds to mercury(II) ions but not-so-strong bonds to titanium(IV) ions. Likewise, palladium(II) if presented with both phosphane and amino ligands will preferentially bind to phosphanes while copper(I) will preferentially bind to amino ligands.

The thermodynamic energy differences between different complexes are often relatively small and typically the kinetic barrier for ligand exchange is low, too. Thus, in many cases which complex is formed depends strongly on the concentrations of individual ions. The mere fact that complexes without aqua ligands can and are formed shows that water is not that good a ligand after all. In fact, due to the positive charge of the central metal having an anionic ligand bind is often preferred due to simple electrostatic interactions.

To answer the actual question you posed: $\ce{[CuCl4]^2-}$ will form because it is the predominant species under a given chloride ion concentration. This can be backrationalised by looking at formation constants $K_1$ to $K_4$. However, aside from quantum chemical calculations there is no way to properly rationalise it a priori.

organic chemistry - Why can't Pd/C and H2 reduce both the alkene and carbonyl portions of α,β-unsaturated carbonyls?

Why is it that the major product of the reduction of chalcones the ketone and not the monoalcohol? In other words, Why isn't the major product a benzyl alcohol?

From what I understand, catalytic hydrogenation can be used to reduce carbonyls as well as alkenes.

My TA told me that nucleophilic hydrides are preferred for reducing carbonyls. Why wouldn't hydrogenation work as well? Does it have to do with resonance involving the carbonyl since in a chalcone, the carbonyl is adjacent to an aromatic ring and is also conjugated with the alkene? Could the reason that catalytic hydrogenation can't effectively touch the α,β-unsaturated carbonyl be the same reason catalytic hydrogenation can't effectively reduce carboxylic acids, esters, and amides — all of which are also resonance stabilized?

My first thought had to do with heats of hydrogenation and how resonance-stabilization found in carboxylic acids and its derivatives reduce the heat of hydrogenation. However, later I found a resource online implying that all pi bonds - even the delocalized, resonance-stabilized ones found in benzene — could be reduced through catalytic hydrogenation with enough time.

In addition, it was noted elsewhere that chalcones could be completely reduced to the benzylic alcohol, although not with ease.

This leads me to believe that there is an activation energy barrier that's impeding the hydrogenation of certain substrates within the confines of a 3 hour undergraduate lab period … am I on the right path? Sterics, perhaps? It was noted that the more highly substituted an alkene is, the slower it is reduced because of the difficulty in getting a highly substituted alkene to approach the catalyst's surface with the $\ce{M-H}$ bonds in an appropriate manner.

halacha - if a woman went to mikveh after 7 days is it kosher bedieved?

There is a tradition in Orthodox (not Conservative) to keep niddah like a zava- meaning 4-5 bleeding days and then 7 additional clean days. In the Torah, niddda is 7 days and then mikveh. This was the way it was until the expulsion of the Jewish people when it was felt that people would get confused with the counting for niddah. My question is if the woman went to mikveh after the 7 days, is it ok bedieved? Thanks.

Answer

If the woman did a "hefsek taharah" (meaning she checked herself and was clean) before immersing the Mikva, then M'Doriasa (according to Biblical law), she is okay. But according to rabbinic law, she is still a niddah until she counts seven clean days and then immerses.

downsampling - Frequency Representation of Downsampled Signal

I'm trying to follow the steps from Oppenheim for the derivation of the frequency representation of a signal which is to be down sampled by a factor of M.

It makes sense to me that:

But then the book says to make a change of variables from $r$ to $i$ and $k$ where $i$ = $0, 1, ... M-1$, and $k$ = (-$\infty,\infty$). However, I don't really understand how this change of variables comes logically. After making this change of variables it is then easy to see how the downsampled frequency representation is just a sum of scaled and shifted representations of the original sampled sequence. But I would like to be able to make the bridge better between steps and how such a change of variable was concluded.

Answer

In the final result, you want to express the spectrum $X_d(e^{j\omega})$ in terms of $X(e^{j\omega})$, the spectrum of $x[n]=x_c(nT)$. Since $X(e^{j\omega})$ is already periodic, it must be possible to represent $X_d(e^{j\omega})$ as a sum of a finite number ($M$) of shifted versions of $X(e^{j\omega})$. This is why the original infinite sum is split up into a finite sum of infinite sums, the latter being shifted versions of $X(e^{j\omega})$.

I think you shouldn't worry if you think you wouldn't have come up with that change of variables yourself. What is important is that you understand what's going on.

Furthermore, I think it's instructive to understand the derivation of the expression for $X_d(e^{j\omega})$ without introducing an auxiliary continuous-time signal $x_c(t)$:

$$\begin{align}X_d(e^{j\omega})&=\sum_{n=-\infty}^{\infty}x[Mn]e^{-jn\omega}\\&=\sum_{n=kM}x[n]e^{-jn\omega /M}\end{align}\tag{1}$$

where in the second sum we only sum over indices $n$ that are integer multiples of $M$. If we introduce a sequence $d[n]$ which equals $1$ for $n=kM$ ($k\in\mathbb{Z}$), and zero otherwise, we can rewrite $(1)$ as

$$X_d(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]d[n]e^{-jn\omega /M}\tag{2}$$

An expression for the sequence $d[n]$ satisfying our requirements is

$$d[n]=\frac{1}{M}\sum_{l=0}^{M-1}e^{j2\pi ln/M}\tag{3}$$

Inserting $(3)$ into $(2)$ gives the final result:

$$\begin{align}X_d(e^{j\omega})&=\frac{1}{M}\sum_{n=-\infty}^{\infty}x[n]\sum_{l=0}^{M-1}e^{j2\pi ln/M}e^{-jn\omega /M}\\ &=\frac{1}{M}\sum_{l=0}^{M-1}\sum_{n=-\infty}^{\infty}x[n]e^{-jn(\omega-2\pi l)/M}\\ &=\frac{1}{M}\sum_{l=0}^{M-1}X\left(e^{j(\omega-2\pi l)/M}\right)\tag{4}\end{align}$$

organic chemistry - What is the priority of a phenyl group in a compound such as this?

I assumed this: http://www.chemspider.com/Chemical-Structure.12491.html?rid=222625d4-811c-4b07-b263-8930fafa831a

to be 1-phenylbut-4-ene Where is the priority here? No source I know of tells me of where the priority of the phenyl group lies.

It is wrong. The book states it is 4-PhenylBut-1ene.

What is the priority of the phenyl group?

signal detection - Noise reduction from very noisy audio

I am trying to write an algorithm that would automatically segment a piece of audio with bird calls recordings. My input data are 1 minute-long wave files and on the output I would like to get separate calls for further analysis. Problem is that signal-to-noise ratio is quite terrible due to environmental conditions and poor quality of a microphone (mono, 8 kHz sampling).

I would be most grateful for any advice on how to proceed further with noise reduction.

Here is an example of my input, one minute audio recording in wave format: http://goo.gl/16fG8P

This is how the signal looks like:

Band-pass filtering, in which I am keeping only anything in between 1500 - 2500 Hz, does improve situation, but still it is far from expectations. In this spectrum still a lot of noise is present.

I have also plotted long-term (over 32-sample intervals) average energy and removed some clicks from it. Here is the result:

With all the remaining noise I have to set a very low threshold to the onset detection algorithm to pick last 10 seconds of bird calls. Problem is if I tweak it in such a way then in next recording I can get load of false positives.

Moving average filter helps a bit with wind noise. Any other ideas? I was thinking of "Spectral Subtraction", but here it seems to me I have sort of chicken and egg problem - to find noise-only area I have to segment the audio and to segment the audio I need to remove the noise. Do you know of any libraries that have this algorithm or some implementations in pseudo-code? Methinks Audacity uses such a method to remove noise. It is very effective, but it is left to the user to mark noise-only area.

I am writing in Python and it is a free, open-source project.

Thanks for reading!

matlab - Need help cleaning noise

I have been recording sound sample from mouse vocalisations and noticed that a background noise appeared in the recording room. the noise is clearly visible in the spectrograms of the recordings as horizontal lines at frequencies 20, 40, 60, 80, and 100 kHz (ultrasonic range). I have tried to record in an acoustically isolated box, lined with sound absorbing material, and it helps, however, the box retains unwanted odours which effect the behaviour of the mice, rendering it unusable.

My question is if there is a method to isolate and remove the noise without affecting the rest of the sounds. I want to study the vocalisations, with an emphasis on the frequency in which the mice communicate under different social encounters, and having that constant noise in the background may affect the results of the study.

Vocal communications were recorded using a 1/4 inch microphone, connected to a preamplifier and an amplifier (Bruel & Kjaer) from mice. Vocalisations were sampled at 250 kHz with a CED Micro 1401-3 (Cambridge Electronic Design Limited, Sunnyvale, CA).

The samples were recorded using spike2, a "DC removal" filter was added to the recording. the files were then transferred to Matlab for splicing further analyses.

I know that there is a strict policy here against attaching files to questions, therefor I'm adding a link to the audio sample and the spectrogram

link to an audio sample and a spectrogram via google drive.

Answer

I disagree somewhat with Hilmar.

Technically, what you are calling noise is really an interfering source. Since the fundamental and harmonics are so clear in your spectrum, I think that the hypothesis that you have a steady tone interfering is a good bet.

Yes, it would be better to remove it from the source, but if it is a steady tone it is not that difficult to remove. All you have to do is get a good estimate of it and subtract it from your signal. If it is a steady hum, and you can get a good estimate (at least across several DFT frames), when you remove it, your signal's values in the affected bins should remain. You want to make sure that your DFT frames have a width of a whole number of cycles of your lowest harmonic. Your higher harmonics will also then be whole cylcles in your frame and have leakless bins.

This problem is very similar to removing unwanted powerline noise out of regular audio recordings. Do a search on "remove power main hum from audio recording" for more ideas, and possible hardware solutions for the hum removal.

experimental chemistry - Why add water first then acid?

From school, I remember a very important rule: first you need to pour the water and then the acid (when you need to mix them) not vice-versa. This is because otherwise the aсid becomes very hot and splashing may happen.

So, why does it get hotter when water is poured into it? What reaction takes place?

Answer

This is mostly the case for sulfuric acid. Commercially available sulfuric acid is dense (~1.8 g/ml) and when water is added, it may not mix. In this case a layer of hot weak acid solution is formed, which boils and sprays around. When acid is poured into water, it flows down the flask and mixes much better, so no boiling occurs.

The reason this occurs is due to the large amount of energy released in the hydration reaction of sulfuric acid ions. Do not believe that heat comes from dissociation, as the dissociation of acids, bases, and salts always consumes energy. The energy is released from subsequent hydration, and the release may be high, especially if $\ce{H+}$ or $\ce{OH-}$ ions are hydrated.

ki tavo - For purposes of how a Torah reader reads tochacha, where does it start and end?

Both parshat Bechukotai and parshat Ki Tavo contain "the tochacha", a section of the parsha with curses to befall B'nei Yisrael if they sin really badly. Traditionally, this section is read faster and in a low voice.

Where exactly does the tochacha start and end, in both parshiot?

Answer

See Kitzur Shulchan ARuch 78:4. When I read this paragraph, he lists other places in the Torah where we should read certain verses quietly. (BTW, he does not mention "quickly" anywhwere, so I'm uncertain when / how speed became a factor.) Excerpting the parts relevant to your question:

וגם הקללות שבפרשת בחקתי ופרשת כי תבא קורין בקול נמוך ואת הפסוק וזכרתי את בריתי יעקב קורין בקול רם ואחר כך הפסוק והארץ תעזב וגו' נמוך ואף גם זאת בקול רם עד הסוף ובפרשת כי תבא ליראה את השם הנכבד עד סוף הפסוק בקול רם ואחר כך נמוך עד ואין קונה

Also the curses in the portions ''Bechukotai'' and ''Ki Tavo'' we read quietly. The verse ''I will remember My covenant with Jacob...''13 we read in a normal voice, afterwards the verse ''the land also shall be forsaken by them...''14 quietly, ''And yet for all that...''15 in a normal voice upto the end. In the portion ''Ki Tavo'' ''that you may fear this glorious name...''16 until the end of the verse in a normal voice, and afterwards quietly until ''and no man shall buy you''17

So, technically, in terms of laining style, if you're referring about "normal" vs. "quiet" voice, there is a definite "start" in both situations where "start" means "low voice". In both situations, it starts at the beginning of the parsha, i.e. the start of the new Torah "paragraph". (I'm having trouble, now w/ my browser, WHen fixed, I'll try to edit in the exact verse.)

In terms of an "end", you can see that we switch voices back and forth a few times, so, you can calculate the "end" by the last silent verse, based on my understanding of the "laining" requirement in your question.

Otherwise, logically, in both cases, Behukotai and Ki Tavo tochahca are each in a single parsha (paragraph).

organic chemistry - What's the H-C-H bond angle in ethene?

The carbon is $\mathrm{sp^2}$ hybridised and is therefore planar and should also, theoretically be $120^\circ$. However, VSEPR theory suggests that the π bond would "need more space" due to greater electron repulsion. As a consequence the $\ce{H-C-H}$ bond angle would be smaller. However, since the π bond is out of the plane of the molecule does this actually happen?

Answer

The H-C-H bond angle in ethene is ca. 117 degrees and the H-C-C angle is ca. 121.5 degrees. There are two reasons that combine to explain this angular deformation in ethene.

First, from these bond angles and Coulson's Theorem (ref_1, ref_2) we can determine that the C-H sigma bonds are $\ce{sp^{2.2}}$ hybridized and the C-C sigma bond is $\ce{sp^{1.7}}$ hybridized.

From these hybridization indices (the index is the exponent "n" in the $\ce{sp^{n}}$ expression) we see that the C-C sigma bond has higher s-character content (1 part s to 1.7 parts p - 37% s) than the C-H bonds (1 part s to 2.2 parts p - 31% s). Since there is more s character in the C-C bond, it is lower in energy and the carbon sigma electrons will tend to flow towards this lower energy C-C bond. Consequently, the C-C sigma bond will contain more electron density than the C-H bonds. Therefore, the electron repulsion between the C-C sigma bond and C-H sigma bonds will be greater than the electron repulsion between the two C-H bonds. Hence the H-C-C bond angle will open up slightly from the $\ce{sp^{2}}$ ideal of 120 degrees and the H-C-H angle will close down slightly in order to minimize the bond-bond electrostatic repulsions.

Second, steric factors (which are also really just another way of describing electron-electron repulsion) may also come into play. To whatever extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, it will also serve to increase the C-C-H bond angle and shrink the H-C-H bond angle.