inorganic chemistry - Do all salts taste salty?

Recently, I am learning the production of soluble and insoluble salts. My friend and I have done this experiment at the school lab.

We wanted to taste them to see whether they are salty are not. The teacher luckily stopped us from doing that.

So without tasting them, I would really like to know whether all salts are salty.

Answer

No. There are sweet, bitter, and various other salts. (Likely, there are tasteless salts too). Pure salty taste is as far as I know exclusive for table salt, though I wouldn't bet on it.

Lead and Beryllium salts are said to be sweet, though toxic. Epsom salt, $\ce{MgSO4}$, is bitter. $\ce{CuSO4}$ has an incomprehensible, persistent metallic taste. (Based on personal experience. Copper salts are slightly toxic, but not extremely, so I survived with no consequences.)

Salts with hydrolysing cation (various alums) are acidic in addition to other notes.

Verifying Linear Time Invariance

I have a system of the form:

$$T(x(n))=x(n)+3x(n-2)-5x(n-3)x(2n)$$

I claim that

$$T(x(n-k))=x(n-k)+3x(n-k-2)-5x(n-k-3)x(2n-2k),$$ $$y(x(n-k)= x(n-k)+3x(n-k-2)-5x(n-k-3)x(2n-2k)$$

and

$$T(ax_1(n)+bx_2(n))=ax_1(n)+bx_2(n)+3\left( ax_1(n-2)+bx_2(n-2)\right) - 5 \left(ax_1(n-3)+bx_2(n-3) \right)\left( ax_1(2n)+bx_2(2n)\right),$$ $$aT(x_1(n))+bT(x_2(n))) = ax_1(n)+3x_1(n-2)-5x_1(n-3)x_1(2n) + ax_2(n)+3x_2(n-2)-5x_2(n-3)x_2(2n)$$

Hence, the system is not linear, but is time-invariant. I'm I applying the formulas correctly?

Answer

The term $x(2n)$ is not time invariant: it gives the even samples of $x$. Shift the input by 1 sample and you'll get the odd terms, which are different from the even terms shifted 1 place.

Your reasoning above is not correct. You should compare $T(x(n-k))$ to $T(x(n))|_{n-k}$ (that is, the original output, shifted by $k$ samples).

Edit: These type of exercises are a common source of confusion because of the following abuse of notation: $x(n)$ is not a signal, but the nth sample of the signal, and having $n$ inside of the expressions is formally wrong. Even Oppenheim's book makes this mistake when working with LTI systems and impulse response.

The correct way for expressing transforms between signals is as follows:

Let $x$ and $y$ be signals; $T\{\cdot\}$ be the transformation (system) under scrutiny; $R_k\{\cdot\}$ be a delay of $k$ samples.

So, in general you would write the relation between input and output signals as $y = T\{ x \}$, where the whole signals are involved. Using $n$ inside these expressions collapses the signal to a single sample and this is wrong. And having the same variable $n$ on both sides of the equation only leads to confusion.

Then, a system is time-invariant if $R_k\{T\{ \cdot \}\} = T\{ R_k\{ \cdot \}\}$ for all integer $k$.

This is why I gave the example above with even-odd samples, in order to avoid the usual confusion with $n$ inside the expressions.

Now, going back to the system $T$ that evaluates to $y[n] = x[2n]$ for each sample $n$:

Let $x_k = R_k\{x\}$ and $y_k = R_k\{y\} = R_k\{T\{x\}\}$.

Evaluating all these signals at sample $n$: $x_k[n] = x[n-k]$; $y_k[n] = x[2n - 2k]$.

Now, $T\{x_k\}$ at sample $n$ is $x_k[2n] = x[2n - k]$ which is different from $y_k$.

particles - ～にしては vs. ～としては

I know one implies the thing I'm comparing actually is the thing I'm comparing it to, and the other doesn't.

Can someone suggest an easy example or trick for remembering which is which?

combustion - What is the oxidation mechanism of gunpowder?

I've been recently looking into fireworks and the chemistry of explosives, and I found out that, amongst others, one of the most important reactions in gunpowder ignition is the oxidation of charcoal and sulfur by $\mathrm{KNO_3(s)}$. I am curious as to whether it is known if this reaction occurs between charcoal/sulfur and solid state potassium nitrate or if the potassium nitrate decomposes first and the reaction occurs via gaseous oxygen as an intermediate.

Is there a general mechanism of oxidation between solid oxidizing salts and organic matter/sulfur? Does the same thing happen with $\mathrm{KClO_3(s)}$ or $\mathrm{KMnO_4(s)}$ for example?

inorganic chemistry - What is formed when you leave iron(II) sulfate in plain air?

What is formed when you leave iron(II) sulfate in plain air?

Knowing that iron(II) is easily oxidised to iron(III), and assuming that the reactive component of air is oxygen, I solved it this way:

$$\ce{FeSO4 + O2 -> Fe2O3 + SO2}$$

But in my textbook it's given that iron(III) sulfate, $\ce{Fe2(SO4)3}$, is formed. Is this because $\ce{Fe2O3 + SO2}$ react to form $\ce{Fe2(SO4)3}$?

Answer

Usually, the original iron(II) sulfate is present as green $\ce{FeSO4.7H2O}$.

Oxygen can oxidize $\ce{Fe(II)}$ salts to $\ce{Fe(III)}$ salts; for example, at $\mathrm{pH}=0$:

$$\begin{alignat}{2} \ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\ \ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V} \end{alignat}$$

$\ce{Fe(II)}$ is even easier oxidized under alkaline conditions; for example, at $\mathrm{pH}=14$:

$$\begin{alignat}{2} \ce{FeO(OH) + H2O + e- \;&<=> Fe(OH)2 + OH-}\quad &&E^\circ = -0.69\ \mathrm{V}\\ \ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad &&E^\circ = +0.401\ \mathrm{V} \end{alignat}$$

However, you cannot simply oxidize iron(II) sulfate $\left(\ce{FeSO4}\right)$ to iron(III) sulfate $\left(\ce{Fe2(SO4)3}\right)$ in dry air since you would need additional sulfate to balance the equation:

$$\ce{2FeSO4 + SO4^2- -> Fe2(SO4)3 + 2e-}$$

Nevertheless, $\ce{Fe(II)}$ can be oxidized to $\ce{Fe(III)}$.

The resulting $\ce{Fe(III)}$ is subject to hydrolysis. By way of comparison, the ion $\ce{[Fe(H2O)6]^3+}$ is only stable under strong acidic conditions. Already at $\mathrm{pH}=0{-}2$, it turns into yellow $\ce{[Fe(OH)(H2O)5]^2+}$ and begins to form $\ce{[Fe(OH)2(H2O)4]+}$:

$$\begin{align} \ce{[Fe(H2O)6]^3+ \;&<=> [Fe(OH)(H2O)5]^2+ + H+}\\ \ce{[Fe(OH)(H2O)5]^2+ \;&<=> [Fe(OH)2(H2O)5]+ + H+}\\ \end{align}$$

Further addition of base causes precipitation of amorphous iron(III) hydroxide.

Accordingly, the likely product when iron(II) sulfate is oxidized is basic iron(III) sulfate, i.e. approximately:

$$\ce{4FeSO4 + O2 + 2H2O -> 4Fe(OH)SO4}$$

However, the real weathering and aging of iron(II) sulfate in dry air actually yields a mixture of various compounds, including iron(III) sulfate and iron(III) oxide-hydroxide.

You may observe this reaction in some iron fertilizers for lawns. The fresh product typically contains green iron(II) sulfate, which gradually becomes yellow.

physical chemistry - How does the inverted Marcus region explain chemiluminescence?

The invert Marcus region offers one explanation of chemiluminescence, the process by which light is one of the products of a chemical reaction.

I tried reading one of Marcus's original papers on the subject [1], but I don't really get his argument. I get confused about which of the potential energy surfaces in his Figure (copied here, hopefully that's okay) that refer to the ground state reactant, ground state product, excited reactant, and excited product.

[1] R.J. Marcus (1965), "On the Theory of Chemiluminescent Electron-Transfer Reactions", The Journal of Chemical Physics 43(8):2654-2657

tisha bav - Why not recite Havdalah without wine on Saturday Night Tishah BeAv?

When Tishah BeAv falls on Shabbat and the fast is postponed to the next day, or when Tishah BeAv falls on Sunday, the problem arises how to deal with the requirement of making havdalah over a cup of wine (or another drink). The positions generally discussed are: (1) to perform havdalah immediately after Shabbat but to give the wine to a minor; (2) to omit havdalah; (3) to perform havdalah the following night. The actual halachah has been determined by the Sages according to opinion (3).

I expected a fourth opinion to be part of this discussion, which is reciting the havdalah formula immediately after Shabbat but without a cup of wine (or another drink). According to this option only the blessing over the flame would be recited, followed by the havdalah formula (the blessings over wine and flagrances being omitted). In my opinion the reason for this fourth possibility lies in the fact that the havdalah formula doesn’t itself mention a cup of wine or in any way refer to it. Thus there seems to be no inherent necessity to recite this formula exclusively over a filled cup. So why can it not be recited “dry”?