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Lera25 [3.4K]
3 years ago
9

5. The heat of fusion of ice is 333.5J/g. The entropy change for the water when freezing 5.0 g of water at 0°C

Chemistry
1 answer:
lara [203]3 years ago
7 0

Answer:

(e) -6.1 J/K

Explanation:

Step 1: Given data

  • Heat of fusion of ice (ΔH°fus): 333.5 J/g
  • Mass of water (m): 5.0 g

Step 2: Calculate the heat (Qfreezing) required to freeze 5.0 g of water

We will use the following expression.

Qfreezing = -ΔH°fus × m

Qfreezing = -333.5 J/g × 5.0 g = -1.7 × 10³ J

Step 3: Calculate the entropy change (ΔS°) at 0 °C (273.15 K) and 1 atm

We will use the following expression.

ΔS° = Qfreezing/T

ΔS° = -1.7 × 10³ J/273.15 K = -6.1 J/K

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The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

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Answer:

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Explanation:3

Initial concentration if HF = c = 0.280 M

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Initially

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(c-x)      x             x

The expression of disassociation constant is given as:

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K_a=\frac{x\times x}{(c-x)}

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