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den301095 [7]
3 years ago
9

Calculate the heat released when 25.0 grams of water freezes at 0 degree Celsius.

Chemistry
1 answer:
mafiozo [28]3 years ago
5 0

Answer:

= 8350 joules

Solution and explanation:

  • The heat of fusion refers to the quantity of heat released when a given amount of water freezes.
  • For example, 1 g of water releases 334 J when it freezes at 0°C.

Therefore; For 25.0 g of water.

Heat released = Mass of water × heat of fusion

                        = 25 g × 334 J/g

                        = 8350 Joules

Hence, the amount of heat released when 25.0 g of water freezes at 0°C is 8350 J.

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8_murik_8 [283]
<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>
  • According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
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At varying pressure and volume;

P1V1=P2V2

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Initial volume, V1 = 2.0 L

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Replacing the known variables with the values;

P2=\frac{(101.5)(2.0L)}{0.25L}

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