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Sati [7]
3 years ago
11

Sunidhi made a study chart about changes in states of matter. Which headings best complete the chart? X: Solid Directly to Liqui

d Y: Liquid Directly to Solid X: Liquid Directly to Solid Y: Solid Directly to Liquid X: Heat Is Released Y: Heat Is Absorbed X: Heat Is Absorbed Y: Heat Is Released
Chemistry
1 answer:
mixas84 [53]3 years ago
3 0

Answer:

The answer is "freezing and melting sublimating x y".

Explanation:

In the given question, the given table name is used as the matter condition, which is freezing or melting sublimation x y, that must be the best headings throughout order to cancel a column of state of matter.

The process goes from the solid-state to a gaseous state, that is achieved by bringing heat into another system.

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Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol..[3]

NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?..[4]

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Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol

2\times \Delta H^o_{4}=470.3 kJ/mol

\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

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