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timama [110]
3 years ago
7

How many grams of ice at -12.1 ∘C can be completely converted to liquid at 15.2 ∘C if the available heat for this process is 4.4

9×103 kJ ? For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus=6.01kJ/mol .
Chemistry
1 answer:
ra1l [238]3 years ago
8 0

Answer:

The mass of ice is 1,06 kg

Explanation:

Formula for calorimetry:  Q = m . C . ΔT

Formula for phase transition, with latent heat: Q = L . m

m . C . ΔT + L . m + m . C . ΔT = 4490 kJ

m . 2,01 J/g°C (0° - (-12,1°C) + 334 J/g . m + m . 4,186 J/g°C . (15,2°C - 0°C)

Our unknown value is mass

<em><u>You should look for latent heat of Ice and of course, specific heat of water.</u></em>

Those are constant and they are known.

<u>As the units are in J, and the heat for the process is in Kj we have to convert the final number into J</u>

4490kJ . 1000 = 449000 J

m. 2,01 J/g°C (0°-(-12,1°C) + 334 J/g .m +m .4,186 J/g°C .(15,2°C - 0°C) = 449000 J

m. 24,321 J/g + 334 J/g .m + 63,62 J/g . m = 449000 J

421,94 J/g . m =  449000 J

m =  449000 J /421,94 g/J = 1064,1 g ---->1,06 kg

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5 0
3 years ago
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
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Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

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Answer:

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A student was trying to obtain lithium by electrolysis of aqueous lithium chloride. Was he successful?
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The student was not successful.

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However, it takes only 0.83 V to reduce water to hydrogen.

Thus, the student will get H₂ instead of Li.

6 0
3 years ago
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