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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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Which statement best describes the development of theories that connected microscopic and macroscopic phenomena?
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Most scientific theories involving microscopic and macroscopic phenomenon have taken several years to be developed; however, this theories are still under revision till date.

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What happens if you try to move the atoms very close to each other?
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2 years ago
A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon
Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0
3 years ago
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