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Leya [2.2K]
2 years ago
15

When a pure sample of KNO3(s) spontaneously dissolves in water at room temperature, the solution becomes quite cold to the touch

. For this reaction the enthalpy of the solution is _____ and the entropy is _____.
Chemistry
1 answer:
jekas [21]2 years ago
4 0

The enthalpy of the solution is <u>positive </u>and the entropy is <u>positive</u>.

Potassium trioxonitrate (V) KNO₃(s) is a strong oxidizing solid substance that when dissolved in water changes to aqueous solution.

In its aqueous solution state, the randomness of molecules increases as a result of that the entropy will also increase leading to the positive state of the entropy.

Similarly, provided that the solution becomes quite cold to the touch, the enthalpy is also in it positive state.

Therefore, we can conclude that the enthalpy of the solution is <u>positive </u>and the entropy is <u>positive</u>.

Learn more about Potassium trioxonitrate (V) KNO₃(s) here:

brainly.com/question/25303112

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A mixture of0.161 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO
labwork [276]

Answer:

number of moles of CO2 is 0.054

number of moles of CO is 0.107

number of moles of O2 remaining is 0.01 mole

mole fraction of CO is 0.63

Explanation:

Firstly, we write the equation of reaction;

3C(s) +2O2(g) → CO2(g) +2CO(g)

Now, we proceed.

From the written equation, we can deduce that

3 mol C = 2 mol O2 = 1 mol CO2 = 2 mol CO

No of mol of C reacted = 0.161 mol

limiting reactant according to the question is Carbon

a. no of mol of CO2 formed = 0.161*1/3 = 0.054 moles ( no of moles of CO2 formed is one-third of no of moles of carbon reacted. This is obtainable from their mole ratio 1:3)

b. no of mol of CO formed = 0.161*2/3 = 0.107 mol

c. no of mol of O2 remaining = 0.117 - (0.151*2/3) = 0.117-0.107 = 0.01 mole

d. mole fraction of CO = no of mol of CO/Total number of moles

= 0.107/(0.107+0.054+0.01)

= 0.625730994152 which is approximately 0.63

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