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Sergio [31]
3 years ago
8

How is the Hvap used to calculate the mass of liquid boiled by 1 kJ of energy?

Chemistry
1 answer:
coldgirl [10]3 years ago
6 0
The heat used in phase changes is calculated by multiplying the mass of the substance by the energy of the phase change. In this case, for liquid to boil, we would find total heat by multiply the mass of liquid by the latent heat of vaporization (Hvap). If we are instead given the Hvap and the total heat of 1 kJ, we would divide 1 kJ by the Hvap (which is usually in kJ/kg) to get the mass of liquid boiled (in kg).
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Solid potassium chlorate (kclo3) decomposes into potassium chloride and oxygen gas when heated. how many moles of oxygen form wh
IrinaVladis [17]

Answer:

  • <u>0.665 mol of O₂.</u>

Explanation:

<u>1. Molecular chemical equation:</u>

  • 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

<u>2. Mole ratios:</u>

  • 2 mol KClO₃ : 2 mol KCl : 3 mol O₂

<u>3. Number of moles of KClO₃</u>

  • Number of moles = mass in grams / molar mass

  • Molar mass of KClO₃ = 122.55 g/mol

  • Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol

<u>3. Number of moles of O₂</u>

As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many  moles of O₂ will be produced from 0.44038 mol of KClO₃.

  • 3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃

  • x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂

Round to 3 significant figures: 0.665 mol of O₂ ← answer

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You have 2.2 mol Xe and 2.0 mol F₂, but when you carry out the reaction you end up with only 0.25 mol XeF₄. What is the percent
alexira [117]

Answer:

The correct answer is 25 %

Explanation:

According to the chemical reaction:

Xe(g) + 2 F₂ (g) → XeF₄ (g)

1 mol of Xe(g) reacts with 2 mol of F₂(g), so the stoichiometric ratio os reactants is 2 mol F₂/mol Xe.

We have 2.2 mol Xe and 2.0 mol F₂, so the ratio is:

2.0 mol F₂/2.2 mol Xe = 0.909 mol F₂/mol Xe

The molar ratio of reactant we have is lower than the required, so F₂ is the limiting reactant.

By using the limiting reactant, we calculate the theoretical amount of product (XeF₄). For this, we know that 1 mol of XeF₄ is formed from 2 mol of F₂ (1 mol XeF₄/ 2 mol F₂), and we have 2.0 mol F₂:

2.0 mol F₂ x (1 mol XeF₄/ 2 mol F₂)= 1 mol XeF₄

If we only obtained 0.25 mol XeF₄, the percent yield of the experiment is:

Yield = experimental amount/theoretical amount x 100%

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