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gayaneshka [121]
3 years ago
13

What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (Kf = 1.86°C/m) –1.86°C –7.44°C –5.58°C –3.72°C

Chemistry
1 answer:
tresset_1 [31]3 years ago
6 0

Answer:

  • Last choice: <em><u>- 3.72°C</u></em>

Explanation:

The freezing point depression in a solvent is a colligative property: it depends on the number of solute particles.

The equation to predict the freezing point depression in a solvent is:

  • ΔTf = Kf × m × i

Where,

  • ΔTf is the freezing point depression of the solvent,
  • m is the molality,
  • Kf is the cryoscopic molal constant of the solvent, and i is the Van'f Hoff factor, which is the number of ions produced by each unit formula of the ionic compound.

The calcualtions are in the attached pdf file. Please, open it by clicking on the image of the file.

Download pdf
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Answer:

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3 years ago
Find the molecules in 32 grams of CH4
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════════ ∘◦❁◦∘ ════════

<h3>Answer = 2</h3>

════════════════════

<h3>Known</h3>

Mass = 32grams

Name of atom = Methane (CH4)

Molar mass C = 12

Molar mass H = 1

════════════════════

<h3>Question</h3>

molecules (mol?)

════════════════════

<h3>Way to do</h3>

mass = mol × molar mass

32g = mol × (CH4)

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32g = mol × 16

mol = 32 : 16

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════════════════════

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3 years ago
Zn + 2 HCl → ZnCl2 + H2
MrRissso [65]

Answer:

10.1g of H₂ are produced

Explanation:

To solve this question we need, first, to convert the mass of each reactant to moles and, using the chemical reaction, find limiting reactant. With limiting reactant we can find the moles of H2 and its mass:

<em>Moles Zn -Molar mass: 65.38g/mol-:</em>

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<em>Moles HCl -Molar mass: 36.46g/mol-:</em>

381g HCl * (1mol / 36.46g) = 10.45 moles

For a complete reaction of 10.45 moles of HCl are required:

10.45 moles HCl * (1mol Zn / 2mol HCl) = 5.22 moles Zn

As there are 4.696 moles of Zn, <em>Zn is the limiting reactant</em>

<em />

The moles of H₂ produced = Moles of Zn added = 4.696 moles. The mass is-Molar mass H₂ = 2.16g/mol-:

4.696 moles * (2.16g / mol) =

<h3>10.1g of H₂ are produced</h3>
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