Answer:
hope this image is helpful for you
Answer:
The answer is "Option B".
Explanation:
![\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\](https://tex.z-dn.net/?f=%5Cto%20C%20%5C%20CH_3COONa%20%3D%20%20%5Cfrac%7B%280.01%5C%20%20mol%20%2B%205%20%5C%20E-4%5C%20%20mol%29%7D%7B%280.105%5C%20L%20%29%7D%5C%5C%5C%5C%5Cto%20C%20%5C%20CH_3COONa%20%3D%200.1%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20Ka%20%3D%20%28%5BH_3O%5E%7B%2B%7D%5D%5Ctimes%20%5Cfrac%7B%280.1%20%2B%20%5BH_3O%5E%2B%5D%29%29%7D%7B%280.0905%20-%20%5BH_3O%5E%2B%5D%29%7D%20%3D%201.75%5C%20E-5%5C%5C%5C%5C%5Cto%200.1%5BH_3O%5E%2B%5D%20%2B%20%5BH_3O%5E%2B%5D%5E2%20%3D%20%281.75%20E-5%29%5Ctimes%20%280.0905%20-%20%5BH_3O%5E%2B%5D%29%5C%5C%5C%5C)
![\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7](https://tex.z-dn.net/?f=%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%5C%200.1%5BH_3O%5E%2B%5D%20%3D%201.584%5C%20%20E-6%20-%201.75%5C%20%20E-5%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%2B%200.1000175%5BH_3O%5E%2B%5D%20-%201.584%20%5C%20E-6%20%3D%200%5C%5C%5C%5C%5Cto%20%20%5BH_3O%5E%2B%5D%20%3D%201.5835%5C%20%20E-5%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20pH%20%3D%20-%20%5Clog%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%20pH%20%3D%20-%20%5Clog%20%281.5835%20%5C%20E-5%29%5C%5C%5C%5C%20%5Cto%20pH%20%3D%204.8004%20%3E%204.7)
Answer:
88.50 %
The balance chemical equation is as follow,
2 H₂ + O₂ → 2 H₂O
Step 1: Find the limiting reactant;
According to eq.
4.032 g (2 mole) H₂ reacts with = 32 g (1 mole) of O₂
So,
11 g of H₂ will react with = X g of O₂
Solving for X,
X = (11 g × 32 g) ÷ 4.032 g
X = 87.301 g of O₂
Therefore, H₂ is the limiting reactant as O₂ is present in excess.
Step 2: Calculating %age Yield;
According to eq.
4.032 g (2 mole) H₂ produces = 36.032 g (1 mole) of H₂O
So,
11 g of H₂ will react with = X g of H₂O
Solving for X,
X = (11 g × 36.032 g) ÷ 4.032 g
X = 98.301 g of H₂O
So,
Actual Yield = 87 g
Theoretical Yield = 98.301 g
Using formula = Actual Yield / Theoretical Yield × 100
= 87 g / 98.301 × 100
= 88.50 %
Answer: 24 gram is a final Answer.
Explanation:1 mol of carbon weight is 12 gram then
12.044×1023 contain 2 mole carbon so
