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maxonik [38]
3 years ago
11

20 points

Chemistry
1 answer:
9966 [12]3 years ago
8 0

Answer:

I`m pretty sure it`s electrical tape.

Explanation:

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= 1 x 10^-4 m

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1. If the water is not completely removed from the salt during the heating process will the reported water-to-salt mole ratio (X
Vitek1552 [10]

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The water in the hydrate (referred to as "water of hydration") can be removed by heating the hydrate. When all hydrating water is removed, the material is said to be anhydrous and is referred to as an anhydrate.

Explanation:

4 0
2 years ago
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

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

7 0
3 years ago
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