it has to me b because it shows what they did in this part
The percentage yield is 72.8 %.
<em>Step 1</em>. Calculate the <em>mass of Br₂</em>
Mass of Br₂ = 20.0 mL Br₂ × (3.10 g Br₂/1 mL Br₂) = 62.00 g Br₂
<em>Step 2</em>. Calculate the <em>theoretical yield</em>
M_r: 159.81 266.69
2Al + 3Br₂ → 2AlBr₃
Moles of Br₂ = 62.00 g Br₂ × (1 mol Br₂/(159.81 g Br₂) = 0.3880 mol Br₂
Moles of AlBr₃ = 0.3880 mol Br₂ × (2 mol AlBr₃/(3 mol Br₂) = 0.2586 mol AlBr₃
Theor. yield of AlBr₃ = 0.2586 mol AlBr₃ × 266.99 g AlBr₃)/(1 mol AlBr₃)
= 69.05 g AlCl₃
<em>Step 3</em>. Calculate the <em>percentage yield
</em>
% yield = (actual yield/theoretical yield) × 100 % = (50.3 g/69.05 g) × 100 %
= 72.8 %
Answer: b. the volume of water vapour increases
Explanation:
when there's an increase in concentration, the equilibrium favours the forward reaction, more product is produced, froward reaction is faster
The rate constants, K, can be expressed in many different terms. In this case, Kp is the equilibrium constant expressed in terms of gas partial pressure. The formula for this is:
Kp = [P(product C) × P(product D)] / [P(reactant A) × P(reactant B)]
As there is only one product, we will use only its pressure in the numerator.
Kp = [P(COCl2)] / [P(CO) × P(Cl2)]
P(COCl2) = 1.49 × 10⁸ × 2.22 × 10⁻⁴ × 2.22 × 10⁻⁴
P(COCl2) = 7.34 atm
Answer:
the partial pressure of Xe is 452.4 mmHg
Explanation:
Dalton's law of partial pressures says that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
The partial pressures can be calculated with the molar fraction of the gas, in this case, Xe.
Molar fraction of Xe is calculated as follows:
Then, 0.29 is the molar fraction of Xe in the mixture of gases given.
To know the parcial pressure of Xe, we have to multiply the molar fraction by the total pressure:
Partial Pressure of Xe=1560mmHg*0.29
Partial Pressure of Xe=452.4mmHg
