Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
Answer:
Energy is released when bonds are broken, and energy is absorbed when bonds are formed.
I believe the answer is B. PO4-3
Answer:
- 278.85 J
Explanation:
Given that:
Pressure = 1.1 atm
The initial volume V₁ = 0.0 L
The final volume V₂ = 2.5 L
The work that takes place in a reaction at constant pressure can be expressed by using the equation:
W = P(V₂ - V₁ )
Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:
W = - P(V₂ - V₁ )
W = -1.1 atm ( 2.5 - 0.0) L
W = -1.1 atm (2.5 L)
W = -2.75 atm L
Recall that:
1 atm L = 101.4 J
Therefore;
-2.75 atm L = ( -2.75 × 101.4 )J
= -278.85 J
Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm = - 278.85 J
