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3 years ago

How many liters of 0.37 M solution can be made with 29.53 grams of lithium fluoride. (LiF)?

Chemistry

1 answer:

dsp733 years ago

4 0

Answer:

V = 3.1 L

Explanation:

Given data:

Molarity of solution = 0.37 M

Mass of LiF = 29.53 g

Volume of solution = ?

Solution:

Number of moles of LiF:

Number of moles = mass/molar mass

Number of moles = 29.53 g/ 25.94g/mol

Number of moles = 1.14 mol

Volume:

Molarity = number of moles of solute / Volume in L

0.37 M = 1.14 mol / V

V = 1.14 mol / 0.37 M

V = 3.1 L (M = mol/L)

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The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi

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Answer:

See explanation

Explanation:

Number of moles of each gas is

Nitrogen = 15/28 = 0.536 kmoles

Helium = 5/4 = 1.25 kmoles

Methane = 60/16 = 3.75 kmoles

Ethane = 20/30 = 0.67 kmoles

Total number of moles = 0.536 kmoles + 1.25 kmoles + 3.75 kmoles + 0.67 kmoles = 6.206 kmoles

Mole fraction of each gas;

Nitrogen = 0.536 kmoles/6.206 kmoles = 0.086

Helium = 1.25 kmoles/6.206 kmoles = 0.201

Methane = 3.75 kmoles/6.206 kmoles =0.604

Ethane = 0.67 kmoles/6.206 kmoles =0.108

Partial pressure of each gas;

Nitrogen = 0.086 * 1200 kPa = 103.2 kPa

Helium = 0.201 * 1200 kPa = 241.2 kPa

Methane = 0.604 * 1200 kPa = 724.8 kPa

Ethane = 0.108 * 1200 kPa = 129.6 kPa

Apparent specific heat at constant pressure;

Cp = (0.15 * 1.039) + (0.05 * 5.1926) + ( 0.6 * 2.2537) + (0.2 * 1.7662)

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2 years ago

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3 years ago

Which has the highest boiling point .33 m NH3 or .10 m Na2SO4

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Answer:

0.33 mol/kg NH₃

Explanation:

Data:

b(NH₃) = 0.33 mol/kg

b(Na₂SO₄) = 0.10 mol/ kg

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from a solute.

(a) For NH₃,

The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.

1 mol NH₃ ⟶ 1 mol particles

i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water

(b) For Na₂SO₄,

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)

1 mol Na₂SO₄ ⟶ 3 mol particles

i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water

The NH₃ has more moles of particles, so it has the higher boiling point.

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3 years ago