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Alecsey [184]
3 years ago
12

Now imagine that you're waiting for your soup to cool down, and you leave it on the counter while you go do your homework. Your

homework is so interesting that you forget all about your soup, and now it's cold! Actually, if you were to measure the temperature, you'd find that your soup is the same temperature as the air in the room. What's going on?
Chemistry
1 answer:
Vlad [161]3 years ago
4 0

Explanation:

Generally, heat flows from a hot environment to a cold (lesser temperature) environment. In this case, the soup is the hot environment and the air is the cold temperature.

Heat would continue to flow from one environment to another until thermal equilibrium is reached. At this thermal equilibrium, both environments would have the same temperature.

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Answer:

1.1grams

Explanation:

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4.5x10^22/(6.02x10^23)=0.07mol

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Finger [1]

Answer:

X 154

Check solution in explanation

Explanation:

Average atomic mass = ( mass 1× abudance) + ( mass 2× abudance)+ ( mass 3× abudance) / 100

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8 0
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Consider the reaction Ca(OH)2(s)→CaO(s)+H2O(l) with enthalpy of reaction ΔHrxn∘=65.2kJ/mol What is the enthalpy of formation of
Ganezh [65]

Answer:

ΔH °fCaO= -655.09 KJ/mol =  ΔH °fCaO

Explanation:

ΔH °(reaction)a=ΔH °f(product)-ΔH °f(reactant)

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ΔH °f(product)=is the standard heat of formation of products

-ΔH °f(reactant)=is the standard heat of formation of reactants

The standard heat of formation of  H 2 O ( l )  is  − 285.8 k J / m o l .

The standard heat of formation of  Ca(OH )2  is  − 986.09 k J / m o l .

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

ΔH °(reaction)a=ΔH °f(CaO+H2O)-ΔH °f(Ca(OH)₂

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65.2                        =  ΔH °fCaO+ 720.29

65.2-720.29      =  ΔH °fCaO

-655.09 KJ/mol ==  ΔH °fCaO

3 0
3 years ago
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