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Lelechka [254]
3 years ago
13

Based on the chart, determine the identity of a substance that increases in temperature by 11.4oC when 1250J of energy are added

to a 55g sample.
Substance Specific Heat (J/g oC)
Lithium 3.56
Vegetable Oil 1.99
Air 1.02
Iron 0.444
Sand 0.290
Gold 0.129

Salt

Air

Lithium

Vegetable Oil
Chemistry
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

LIthium vegetable oil

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7 0
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Examine the given reaction. NH4NO3(s) → NH4+(aq) + NO3–(aq) ΔH° = 25.45 kJ/mol ΔS° = 108.7 J/mol·K Which of the given is correct
kobusy [5.1K]

Answer:

B)−6,942 J /mol

Explanation:

At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:

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Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):

ΔG = ΔH - TΔS

ΔG = 25450J/mol - 298K×108.7J/molK

ΔG = -6942.6J/mol

Right solution is:

<h3>B)−6,942 J /mol</h3>

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