A 50.0 50.0 mL solution of 0.127 0.127 M KOH KOH is titrated with 0.254 0.254 M HCl HCl. Calculate the pH of the solution after
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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
I hope it helps!
Answer:
Explanation:
The quantity of heat transfered from the jellybean to the water is: