New answers tagged concentration
2
Option C is the right answer
Solubility of AgCN will be equal to sum of concentrations of Ag+ and [Ag(CN)2]-
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As molarity is defined as the ratio of molar amount of the dissolved solute and solution volume:
$$c=\frac nV$$
molarity increases when solvent ( like water ) evaporates.
Molarity can be considered constant only in context of later refilling the evaporated solvent and mixing the solution. Then molarity returns to the original value. It assumes the solute ( ...
0
In water at 25 ºC $[H+]=[OH-]=10^{-7}M$, so a concentration of $[H+]=10^{-6}M$ is very low. With a strong acid like HCl you have to dissolve only around 35 micrograms in a liter of water to get this concentration.
Sea water on the other hand has a high conductivity and, in average, 35 grams per liter of salts dissolved in it. If you dissolve 35 grams of HCl ...
0
The symbol “w/w” is more often used in chemistry, whereas its equivalent “m/m” is favored in pharmaceutical industry.
For instance, European Pharmacopoeia in section 1.2. Other provisions applying to general
chapters and monographs suggests the following methods for expressing concentration [1, p. 4]:
Expression of content. In defining content, the ...
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“[M]ass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity.” (Wikipedia quoting [de Silva, G.M.S. (2002), Basic Metrology for ISO 9000 Certification, Butterworth-Heinemann, 214p.]
Well-known relationship:
$$
W = m \times g
$$
where W is weight, m ...
1
During titration of small amounts of acids, the molar amount of the indicator in 1-2 drops of $\pu{1 \%}$ indicator solution may not be negligible compared to the acid molar amount, affecting the result.
So for that cases, $\pu{0.1 \%}$ solution is used, to be able to dose smaller indicator amounts.
As the phenolphalein molar mass is about $M=\pu{318 g/mol}$,...
1
Indeed, the only thing you'll need is solving the first order differential equation (but before that, let me define $R_0 := \ce{[RSH]_0}$ and $C :=\ce{[Ker-S-S-Ker]}$.
Now, let's solve it assuming (that's what is given) that $R_0$ can be considered a constant.
Then, we have : $$\frac{\operatorname{d}C}{\operatorname{d}t} = -kR_0C \implies \frac{\operatorname{...
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