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.
add up the mass of protons and neutrons
Answer:
Evaporation
Explanation:
Evaporation is the certain process that requires water to gain heat energy from the environment.
1. 2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. 2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. 2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
<h3>Further explanation</h3>
There are several reactions that can occur in a chemical reaction: single replacement, double replacement, synthesis, decomposition or combustion, etc.
1.Al(s)+HCl(aq)⇒AlCl₃(aq)+H₂(g)
type : single replacement
balance :
2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. AgNO₃ (aq) + Cu (s) ⇒ Cu(NO₃)₂ (aq) + Ag (s)
type : single replacement
balance :
2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. C₃H₈O + O₂ ⇒ CO₂ + H₂O
type : combustion of alcohol
balance :
2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
