New answers tagged

2

There are a few reasons. Some of the most important are: -Oxalic acid dihydrate isn't hygroscopic (it doesn't absorb water from the atmosphere, whereas sulfuric acid does). If a substance can absorb water from the atmosphere, then when you weigh it out, you won't be able to accurately calculate the number of moles used because you don't know how much is ...


1

Strong acid dissociate, because they are stronger acids then $\ce{H3O+}$, and react with its conjugated base $\ce{H2O}$: $$\ce{HA + H2O -> A- + H3O+}$$ Additionally, the $\ce{H3O+}$ activity coefficient steeply grows in concentrated acids, and so does the acid reactivity. In concentrated $\ce{H2SO4}$ and oleum, the role of $\ce{H3O+}$ is overtaken by $...


0

You must realize that one proton free, lonely, has not the same mass as when included in a nucleus. A proton looses a small part of its mass when entering a nucleus. This mass loss is transformed into energy through the Einstein law : E = mc2.


3

This is a question worth debating and discussion. First of let us realize the fact that we are used to looking at relative atomic weights or masses (if someone wants to be too accurate). No Table lists the absolute atomic masses of atoms in the periodic table, i.e., through out the history we have to assumed either the weight of hydrogen atom is exactly 1....


0

As others have answered, yes the potential of the cell will still change. It sounds like you're looking for a more "physics-y" answer than the ones provided so here goes. The Nernst equation (and equilibrium equations in general) work based on concentration and not total moles. To understand why, start with a homogeneous reaction, like $\ce{HCl + NaOH <-&...


0

In an electrochemical cell, increasing the concentration of reactants will increase the voltage difference, as you have indicated. A higher concentration of reactant allows more reactions in the forward direction so it reacts faster, and the result is observed as a higher voltage. If you have adjusted the cell volume to keep the total amount of reactants ...


2

"Parts per …" are dimensionless entities: to derive $\pu{ppm}$ from mass concentration $γ$ (e.g. $\pu{1 mg L-1})$ one uses density of water $\rho(\ce{H2O}) = \pu{1E6 mg L-1}$ for normalization: $$\frac{γ}{ρ(\ce{H2O})} = \frac{\pu{1 mg L-1}}{\pu{1E6 mg L-1}} = \pu{1E-6} = \pu{1 ppm}$$ Once the solution is too concentrated, you obviously are not allowed to ...


2

To summarize, and to answer the question, it is correct to say that the rate of a chemical reaction can be expressed in % per second if and only if the reaction is first order. This percentage is the first order rate constant $k$.


6

According to IUPAC Recommendations for chemical kinetics [1], rates $\nu$ are defined for reactant $\ce{B}$ and product $\ce{Y}$ as a time $(t)$ derivative of the amount $n$ (in general) or per unit volume $V$ (for the closed system): $$ \begin{array}{lll} \hline \text{System} & \text{Consumption} & \text{Formation} \\ \hline \text{Open} & \...


0

For a reaction like this: $\ce{aA +bB⟶cC +dD}$, the rate is given by Rate=$−a\frac{d[A]}{dt}=−b\frac{d[B]}{dt}=c\frac{d[C]}{dt}=d\frac{d[D]}{dt}$ where the unit is usually $\ce{mol L^{-1}{s^{-1}}}$. The rate of a reaction is different for different orders and has different units of the same. It's unit usually has the terms $\ce{mol, L, and s}$. For a zero ...


Top 50 recent answers are included