What is the percentage of water in the following compound? Answer using three significant figures. Sodium carbonate decahydrate,
2 answers:
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Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction .
where,
We are given:
Putting values in above equation, we get:
Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Answer: V= 3.13 L
Explanation: solution attached:
Use combine gas law equation:
P1 V1 / T1 = P2 V2/ T2
Derive to find V2
V2 = P1 V1 T2 / T1 P2
Convert temperatures in K
T1= 13.0°C + 273 = 286 K
T2= 22.5°C + 273 = 295.5 K
Substitute the values.
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!