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Answer:
There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C
Explanation:
To decrease the temperature of the solution there are necessaries:
4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y
8368J + 83,68J/gX = Y <em>(1)</em>
Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.
Also, the energy Y will be:
Y = 25700J/mol×X
Y = 321J/g X <em>(2)</em>
Replacing (2) in (1)
8368J + 83,68J/g X = 321J/g X
8363J = 237,32J/gX
<em>X = 35,2g</em>
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Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C
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The masses of the components are obtained as;
- Sodium hydrogen carbonate = 3.51 g
- Sodium carbonate = 8.708 g
The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.
Now putting down the equation of the reaction, we have;
Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g
Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol
= 3.51 g
Learn more about anhydrous sodium carbonate :brainly.com/question/20479996
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