New answers tagged concentration
2
Your problem seems to have no solution without further information. You may be interested in the the general method for any acid /base calculation is summarised below.
When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution ...
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Your qualitative understanding is correct. As you increase the temperature, you have two effects:
The dissociation reaction, which is endothermic, shifts to the right. This effect increases the concentration of hydrogen ions.
The concentration of hydrogen ions is expressed as molarity, which is number per unit volume. As you increase $T$, water expands, ...
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For pure water, $\mathrm{pH}=- \frac 12 \cdot \log{K_\mathrm{w}}$. As $K_\mathrm{w}$ grows with temperature, $\mathrm{pH}$ decreases.
The influence of water thermal dilation is much smaller than the temperature dependency of water autodissociation constant.
For solutions with acids, bases, acidic or alkaline salts, buffering systems it is much more ...
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Since the volume is given and it's a constant, the initial amount of ammonia $n_0(\ce{NH3})$ can be found from its initial concentration $c_0(\ce{NH3}):$
$$n_0(\ce{NH3}) = c_0(\ce{NH3})\times V\tag{1}$$
To find $c_0(\ce{NH3}),$ an ICE table might indeed come in handy; however, yours needs corrections.
First, I suggest to rewrite it according to the process ...
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These are the absorbance spectra of copper sulfate in water (open markers) and heavy water (filled markers) (1):
The absorbance in water is maximum at 780 nm, as stated in the reference. From the spectrum, the molal absorption coefficient, at 780 nm, appears to be approximately 12.5 $\mathrm{(mol/55.51 mol \ of \ water)^{-1} cm^{-1}}$ with aquamolality ...
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The reaction $\ce{Cu^2+ + 4 Cl- <=> CuCl4^2-}$ may lead to slowing down reaction with aluminium, decreasing the equilibrium concentration of $\ce{Cu^2+}$ in $\pu{0.8 M}$ $\ce{CuCl2}$. That all would lead to lower peak temperature due bigger heat dissipation before reaching maximal temperature.
Contributing factors may be
decreasing of chemical ...
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Your title and the main body of the post are two different questions. Anyway, it is important to understand the difference in each case.
Let us talk about the analyte. Your analyte can come in two forms: (i) either as a solid or (ii) as a solution.
In the case of a solid analyte you would quantitatively transfer a known exact weight of the sample to a ...
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