Answer:
C
Explanation:
75 mile shallow flat area just off coastlines
Answer:
Recycling and reuse of materials
Explanation:
One of the greatest problems facing the human population is the problem of solid waste disposal. The menace of solid waste disposal has led to the idea of landfills. Land fills are depressions on the earth surface prepared for the purpose of solid waste disposal.
The most important approach towards solid waste disposal is the idea of recycling of materials. A material can be collected after use and processed into the same material or serve as a precursor in another manufacturing process. This means that no waste is generated as the materials which are supposed to be disposed of as solid waste are processed into other useful materials. This will reduce the volume of solid wastes generated that may need to be disposed in a landfill.
Answer:
A) pH of Buffer solution = 4.59
B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original buffer solution = 4.65
Explanation:
This is the Henderson-Hasselbalch Equation:
![pH = pKa + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
to calculate the pH of the following Buffer solutions.
Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
