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xz_007 [3.2K]
2 years ago
9

Solve pls brainliest

Chemistry
1 answer:
sammy [17]2 years ago
4 0

Answer:

Even though the two substances possess many similarities, they have some unique properties. In turn, since they have the same properties, if they were the same substance, it would make matters worse, if the same chemical was in two different places, there would not be a difference between them since they are the same, just as it is with are two different chemicals would have differing properties since they are two properties would vary from one another since they are 2 totally different things!

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<span>362.51 Kelvin ln (p1/p2) =( dH / R) (1/T2 - 1/T1) ln (760 Torr /520Torr) =( 40,700 Joules / 8.314 J molâ’1K-1)(1/T2 - 1/373K) ln (1.4615) =( 4895.35)(1/T2 - 0.002681) 0.37946 = 4895.35/T2 - (0.002681)(4895.35) 0.37946 = 4895.35/T2 - (13.124) 0.37946 + 13.124 = 4895.35/T2 13.5039 = 4895.35/T2 T2 = 4895.35 / 13.5039 T2 = 362.51 answer is 362.51 Kelvin - 273 answer is also 89.5 Celsius</span>
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Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
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