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atroni [7]
3 years ago
6

How much heat energy would be released if 78.1 g of water at 0.00 °c were converted to ice at −57.1 °c. give your answer as a po

sitive number in kilojoules (kj)?
Chemistry
1 answer:
kolezko [41]3 years ago
3 0
To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g; 
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C 
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348 
 = 26250.0348 J or 26.250 kJ
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21.10g of H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

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Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:

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Therefore, 1.172 mole of H2O is produced from the reaction.

Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:

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