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Rina8888 [55]
3 years ago
12

What is the change in heat energy when 114.32g of water at 14.85oC is raised to 18.00oC?

Chemistry
1 answer:
iris [78.8K]3 years ago
8 0

Given:

114.32g of water

14.85oC is raised to 18.00oC

Required:

Change in heat energy

Solution:

This can be solved through the equation H = mCpT where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.

The specific heat of the water is 4.18 J/g-K

Plugging in the values into the equation

H = mCpT

H = (114.32g) (4.18 J/g °C) (18 – 14.85)

H = 1,505.3 J

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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
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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

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

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

65.2                =  ΔH °fCaO+( -265.8) -(-986.09)

                       = ΔH °fCaO-265.8+986.09

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65.2-720.29      =  ΔH °fCaO

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