<span>2Fe2O3(s) + 3C(s) →4 Fe(s) + 3CO2(g)
3 mol 3mol
4 mol x mol
x=4*3/3= 4.0 mol
</span>2Fe2O3(s) + 3C(s) →4 Fe(s) + 3CO2(g)<span>
2 mol 3 mol
14 mol x mol
x=14*3/2= 21.0 mol</span>
<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
Answer : The mass of of water present in the jar is, 298.79 g
Solution : Given,
Mass of barium nitrate = 27 g
The solubility of barium nitrate at 
is 9.02 gram per 100 ml of water.
As, 9.02 gram of barium nitrate present in 100 ml of water
So, 27 gram of barium nitrate present in 
of water
The volume of water is 299.33 ml.
As we know that the density of water at is 0.9982 g/ml
Now we have to calculate the mass of water.
Therefore, the mass of of water present in the jar is, 298.79 g
