The mole of ammonia in this chemical equation is 0.36933 moles of NH3... I hope it helped? :)
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Answer:</h3>
7.502 kilo calories
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Explanation:</h3>
<u>We are given;</u>
- Quantity of heat, Q as 31.39 kJ
We are required to convert it to kilo calories
First, we convert it to Joules
1 kJ = 1000 J
Therefore;
31.39 kJ will be equivalent to 31390 Joules
Second we convert Joules to calories using a suitable conversion factor
The suitable conversion factor is 4.184 J/Cal
Thus;
31390 Joules will be equivalent to;
= 31390 J ÷ 4.184 J/Cal
= 7502.39 Cal
Third, we convert calories to kilo calories
1 kCal = 1000 cal
Therefore;
7502. 39 Cal will be equivalent to;
= 7502.39 ÷ 1000
= 7.50239
=7.502 kilo calories
Hence, 31.39 kJ will correspond to 7.502 kilo calories
Lower as KCl has ionic bonding which is less stable than CaCl2s covalent bond which requires less energy.
I’m guessing A or C because I can’t really see clearly on both questions.