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igomit [66]
3 years ago
11

Week 8 of Quarter 2

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
4 0

Answer:

blurred an pic mo paki ayos

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Suppose you begin with 1.50~g of the hydrate copper(II)sulfate · x-hydrate (CuSO4· x H2O), where x is an integer. After dehydrat
chubhunter [2.5K]

These are two questions and two answers.

Question 1.

Answer: x = 5

Explanation:

1) Data:

m₁ = 1.50 g

compound₁: CuSO₄· x H₂O

m₂ = 0.96g

compound₂ = CuSO₄

x = ? (round to the nearest integer)

2) Solution:

i) molar mass of CuSO₄ = 63.546g/mol + 32.065g/mol + 4×15.999g/mol = 159.607g/mol

ii) number of moles of CuSO₄

number of moles = mass in grams / molar mass = 1.50 g/ 159.607 g/mol = 0.006265 mol

iii) molar mass of H₂O = 18.015 g/mol

iv) mass of H₂O = 1.50g - 0.96g = 0.54g

v) number of moles of H₂O = mass in grams / molar mass = 0.54 g / 18.015 g/mol = 0.0300 mol

vi) Ratio moles H₂O / moles CuSO₄ = 0.0300 / 0.0062625 ≈ 5

∴ x = 5.

Question 2.

Answer: 5.5 g

1) Data:

compound₁ = KAl(SO₄)₂ · 12H₂O.

compound₂ = KAl(SO₄)₂

m₂ = 3.0 g KAl(SO₄)₂

m₁ = ? (two significant figures)

2) Solution:

i) molar mass of KAl(SO₄)₂ = 39.098g/mol + 26.982g/mol + 2×32.065g/mol + 8×15.999g/mol = 258.202

ii) number of moles of KAl(SO₄)₂ = mass in grams / molar mass = 3.0g / 258.202 = 0.011619 mol

iii) number of moles of H₂O = 12 × number of moles of KAl(SO₄)₂ = 12 × 0.011619*12 mol = 0.1394 moles

iv) mass of H₂O = number of moles × molar mass = 0.1394 moles × 18.015 g/mol = 2.5 g (rounded to two significant figures)

v) mass of the original compound = mass of KAl(SO₄)₂ + mass of H₂O = 5.5g

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Historians find proof in the past and scientists keep questioning and look at the future
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Calculate the enthalpy change, ΔH, for the process in which 30.1 g of water is converted from liquid at 10.1 ∘C to vapor at 25.0
gayaneshka [121]
You can split the process in two parts:

1) heating the liquid water from 10.1 °C to 25.0 °C , and

2) vaporization of liquid water at constant temperature of 25.0 °C.


For the first part, you use the formula ΔH = m*Cs*ΔT

ΔH = 30.1g * 4.18 j/(g°C)*(25.0°C - 10.1°C) = 1,874 J

For the second part, you use the formula ΔH = n*ΔHvap

Where n is the number of moles, which is calculated using the mass and the molar mass of the water:

n = mass / [molar mass] = 30.1 g / 18.0 g/mol = 1.67 mol

=> ΔH = 1.67 mol * 44,000 J / mol = 73,480 J

3) The enthalpy change of the process is the sum of both changes:

ΔH total =  1,874 J + 73,480 J = 75,354 J

Answer: 75,354 J
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