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mash [69]
2 years ago
15

Hen solid kbr is dissoved in water, the solution gets colder. this is an example of a(n) ________ reaction.

Chemistry
1 answer:
Sonja [21]2 years ago
7 0
Endothermic reactions. These are reactionsthat take in energy from the surroundings. The energy is usually transferred as heat energy, causing the reaction mixture and its surroundings to get colder
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Oxygen and nitrogen are both gases at room temperature. Explain why oxygen has a greater force of attraction between its particl
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Explanation:

Oxygen molecules and Nitrogen molecules forms in a very similar way. The attraction between particles of oxygen is great due to its very high electronegativity value. Oxygen has a higher electronegative value compared to nitrogen.

Electronegativity of an atom is the relative tendency with which atoms of an element attracts valence electrons in a chemical bond. Valence electrons are used in forming chemical bonds. They can be transferred from one atom to the other or they can be shared.

Oxygen is the second most electronegative atom on the periodic table. To form a bond, it shares the valence electrons in order for its octet to be complete. Pull for the valence electrons between the contributing atoms is very strong due to their large electronegative values. This pull is stronger compared to that between nitrogen atoms.

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If mass increases, force increases.

Explanation:

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Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of
bulgar [2K]

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]

Using Hess's law:

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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