New answers tagged concentration
1
As I mentioned, I agree with Nisarg Bhavsar's conclusion.
For the justification, I think a more formal one can be obtained starting from this Wikipedia article on mass balance:
https://en.wikipedia.org/wiki/Mass_balance#Ideal_batch_reactor
In this case the system is indeed closed, so the mass (not concentration) balance is:
$r_A \cdot V = \frac {dn_A}{dt}$
...
2
There are a few things you can look at. First, we can write out the reaction as a whole: $\ce{Zn + Cu^2+ \rightarrow Zn^2+ + Cu}$
the nernst-equation is given by:
$\ce{E = E^\circ} - \ce{\frac{RT}{nF}lnQ}$, where $\ce{Q = \frac{[Zn^2+]}{[Cu^2+]}}$. We can see that if we increase $\ce{[Cu^2+]}$, we decreases the log value which then increases $\ce{E_{non-...
3
Initially in the solution the following equilibria exist
$$\ce{HA <=> H+ +A-}$$
and
$$\ce{H2O <=> H+ +OH-}$$
Let initial concentration of $\ce{HA}$ be $a_1$ and that of $\ce{A-}$ be $a_2$. Now to a litre of this buffer solution, let's say you have added $\mathrm dn$ moles of strong acid to it. The acid dissociation constant of $\ce{HA}$ be $K_a$ ...
3
Suppose initially you had 1 mole of $\ce{N2}$ and 3 moles of $\ce{H2}$, and that $n$ moles of $\ce{N2}$ had reacted. Then you would have, at equilibrium, $1-n$ moles of $\ce{N2}$, $3(1-n)$ moles of $\ce{H2}$, and $2n$ moles of $\ce{NH_3}$. So the total moles at equilibrium would be $4-2n$, and the mole fractions (equal to the partial pressures in atm) ...
3
The answer and the solution given in the book are wrong
I have searched and found the following in Chemical Kinetics by Laidler which talks about the volume change during reaction. It states:
....Equation (1.11) gives a general definition of the rate of reaction, but Eqs. (1.13)-(1.14) apply only if the volume remains constant during reaction. To remove the ...
1
When the reactants and products are both gaseous and at same pressure and temperature you might want to prefer using volume as a unit of measurement instead of concentrations as it simplifies the equations tremendously. Using volume as
a unit of measurement is possible because the moles of each gas $\ce A$ and $\ce B$ will be proportional to their volumes.
...
7
Buttonwood has given an excellent answer with a nice demonstrational video. However, there is one point I'd like to emphasize. In ideal gases, the molar mass is directly proportional to the density:
$$PV = nRT \ \Rightarrow \ \frac{n}{V} = \frac{P}{RT} \tag1$$
If the molar mass of a gas is $M$ and its mass is $m$, $n = \frac{m}{M}$ and $\frac{n}{V} = \frac{m}...
9
If your sample is solid and crystalline, knowledge of the unit cell's dimension and symmetry (e.g., fcc) and density allows you to determine the molecular mass of your compound.
In fact, in crystallography, you may determine the macroscopic density of your sample (e.g., find a liquid which i) wets your crystals and ii) lets your crystals float) in first ...
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