Clearly volume will INCREASE almost sevenfold, as we would expect if we reduce the pressure. ... Assuming this is an ideal gas, and that the number of molecules *n* of this gas remains constant, and given that the temperature is also ... That is, in these conditions, pressure times volume is constant :
Answer:
Explanation:
Hello,
In this case, since the pH defines the concentration of hydrogen:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-3.4%7D%3D3.98x10%5E%7B-4%7D)
And the percent ionization is:
![\% \ ionization=\frac{[H^+]}{[HA]}*100\%](https://tex.z-dn.net/?f=%5C%25%20%5C%20ionization%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D%2A100%5C%25)
We compute the concentration of the acid, HA:
![[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\](https://tex.z-dn.net/?f=%5BHA%5D%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5C%25%20%5C%20ionization%7D%2A100%5C%25%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%7D%7B66%5C%25%7D%20%20%2A100%5C%25%5C%5C%5C%5C)
![[HA]=6.03x10^{-4}](https://tex.z-dn.net/?f=%5BHA%5D%3D6.03x10%5E%7B-4%7D)
Thus, the Ka is:
![Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%2A3.98x10%5E%7B-4%7D%7D%7B6.03x10%5E%7B-4%7D%7D%5C%5C%20%20%5C%5CKa%3D2.63x10%5E%7B-4%7D)
So the pKa is:
Regards.
Answer:
the salt dissolves in water because of covalent bonds.
Explanation:
mark me brainliest.
ur mom ur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur momur mom
Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
