8

If you dissolve $\ce{NH4Cl}$ in water, your solution contains $\ce{NH4+}$ and $\ce{Cl-}$ ions. If you dissolve $\ce{NaOH}$ in water, your solution contains $\ce{Na+}$ and $\ce{OH-}$ ions. Now if you mix those two solutions, nearly all $\ce{NH4+}$ and $\ce{OH-}$ will react and produce $\ce{NH3}$ and $\ce{H2O}.$ This shows that $\ce{NH4OH}$ cannot exist. It is ...


7

I think that the cheapest and fastest way is that of working on natural products, like actual tea leaves, or ground coffee (I just tried it with coffee, and it can be done): If you want only particles with neutral buoyancy, or, if I understand well, floating in the solvent, you can use said natural products, which contain particles of different shapes and ...


3

Measuring out volumes can sometimes be easier than dealing with mass, but in this case, converting the calculations to mass-centric can make the process easier. (I'll use your symbols, but they will now be in mass units (kg). If you desire percentages of ethanol by volume, you still need to convert to mass units because of the volume contraction!)) First, ...


2

Actually, I think the question has an error in the EMF of the cell given. It should be $\pu{-0.413 V}$ rather than $\pu{0.413 V}.$ So, actually you are right in saying that the left half cell would be cathode if the EMF of the cell has to be positive. The mistake that you are doing is when you wrote the reduction potential of anode, you should have written ...


2

Iron(III) chloride indeed hydrolyses in water, or to be precise, $\ce{Fe^3+}$ does assuring acidic medium. Hydrolysis isn't completely suppressed even in strongly acidic solutions. In the absence of other factors, what happens exactly depends on the concentration of chloride and temperature. In diluted cold solution a typical formation of metal aqua ions ...


2

A simple way to think about this is as follows (assuming we are talking about an aqueous solution). The dissolution of $\ce{CO2}$ in water is followed by three chemical reactions: $$ \begin{align} \ce{CO2 + H2O &<=> H2CO3}\label{rxn:R1}\tag{R1}\\ \ce{H2CO3 &<=> HCO3- + H+}\label{rxn:R2}\tag{R2}\\ \ce{HCO3- &<=> CO3^2- + H+}\...


1

Based on the comment to an answer elsewhere by OP, I'd try to solve this problem, purely based on volumes. However, these calculations has ignored the volume contraction of 96% ethanol may have showed initially (that should be minimal since it is only ~4% of water by volume in there). I'd say, your equation, regardless of how OP has derived it, is erroneous....


1

Of course there are such solutions. Take, as an example, Sigma-Alrich's offer of methanolic solution of HCl (3 mol/L, e.g. here), or the dry ones in diethyl ether (e.g., 2 mol/L here), in cyclopentyl methyl ether (e.g., here), or in 1,4-dioxane (e.g. 4 mol/L, here).


Only top voted, non community-wiki answers of a minimum length are eligible