Answer:
35.9 ml
Explanation:
Start with the balanced equation:
3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)
This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-
∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4
We know that concentration = moles/volume i.e:
c= n/v
∴n=c×v
∴nCuCl2=0.107×91.01000=9.737×10−3
I divided by 1000 to convert ml to L
∴nNa3PO4=9.737×10−3×23=6.491×10−3
v=nc=6.491×10−30.181=35.86×10−3L
∴v=35.86ml
Answer:

Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
Answer:
400 miles because it is going 50 mph
They will become unequivalent