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guajiro [1.7K]
3 years ago
9

For each of the environments below: (1) identify your system and surroundings and (2) predict whether it would be endothermic or

exothermic.a.Wood burningSystem:Surroundings:Enthalpy Change:b.Water freezingSystem:Surroundings:Enthalpy Change:c.Sweat evaporatingSystem:Surroundings:Enthalpy Change:d.Chemical hand-warmerSystem:Surroundings:Enthalpy Change:
Chemistry
1 answer:
SCORPION-xisa [38]3 years ago
6 0

Answer:

<u>A) Wood burning </u>

system : Wood.

surroundings : atmosphere

Enthalpy : Exothermic

<u>B) Water Freezing system </u>

System : Refrigerator

surroundings :  water in the refrigerator

enthalpy change : Endothermic

<u>C) Sweat evaporating </u>

System :  Human being

surroundings : Air nearby

Enthalpy change : exothermic

<u>D) Chemical Hand-warmer </u>

system : Hand warmer pack

Surroundings : human palms

Enthalpy Change : Exothermic

Explanation:

<u>A) Wood burning </u>

system : Wood

surroundings : atmosphere

Enthalpy : Exothermic

This system give away energy to its surroundings hence its enthalpy change is exothermic

<u>B) Water Freezing system </u>

System : Refrigerator

surroundings :  water in the refrigerator

enthalpy change : Endothermic

The system absorbs heat from what is put inside(surroundings ) of it hence this is na endothermic system

<u>C) Sweat evaporating </u>

System :  Human being

surroundings : Air nearby

Enthalpy change : exothermic

This is an exothermic reaction ( enthalpy change ) because the system gives out heat to the surrounding

<u>D) Chemical Hand-warmer </u>

system : Hand warmer pack

Surroundings : human palms

Enthalpy Change : Exothermic

There is movement of heat from the system to the surrounding hence it is an exothermic reaction

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<u><em>calculation</em></u>

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5 0
3 years ago
Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
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Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0
3 years ago
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