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egoroff_w [7]
3 years ago
8

Calculate the amount of heat needed to boil of benzene (), beginning from a temperature of . Be sure your answer has a unit symb

ol and the correct number of significant digit
Chemistry
1 answer:
Pavlova-9 [17]3 years ago
8 0

Answer:

Check explanation

Explanation:

From the question, the parameters given are 64.7g of benzene,C6H6; a starting temperature of 41.9°C and bringing it to 33.2°C.

Molar mass of benzene,C6H6= 78.11236 g/mol.

Things to know: heat capacity of benzene, C6H6= 1.63 J/g.K, the heat of fusion = 9.87 kj/mol.

STEP ONE(1): ENERGY USED IN MELTING BENZENE SOLID.

Using the formula below;

Energy used in melting the solid(in JOULES) = (mass of benzene/molar mass of benzene) × heat of Fusion.

=(64.7 g of C6H6/ 78.11236(g per mol) of C6H6) × 9.87 kJ per mol.

= 8.175 J.

= 0.008175 kJ.

STEP TWO (2): ENERGY OF HEATING THE LIQUID.

It can be calculated from the formula below;

Energy= heat capacity (J/g.K) × mass of benzene× (∆T).

= 1.63 J/g.K × 64.7 × (41.9-33.2).

= 917.5J.

= 0.9175 kJ.

Energy required to boil benzene= Energy required to melt the bezene + energy required for boiling.

= 0.008175+ 0.9175.

= 0.93kJ

Approximately, 1 kJ

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<h3>What is the contribution of each  step?</h3>

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I        0.12                                     0                    0

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Ka1= [H3O^+] [HA^-]/[ H2A]

Ka1= x^2/  0.12 - x  

1.0×10^−4 = x^2/  0.12 - x  

1.0×10^−4(0.12 - x ) = x^2

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          HA^-(aq) + H20(l)    -------> A^-(aq)   + H3O^+

I       0.0034                                  0               0

C       -x                                          + x            +x

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5.0×10^−5 = x^2/ 0.0034 - x  

5.0×10^−5 (0.0034 - x ) = x^2

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x^2 + 5.0×10^−5x - 1.7 * 10^-7 = 0

x=0.00039 M

Learn more about the dissociation of a polyprotic acid:brainly.com/question/14481763

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