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adelina 88 [10]
3 years ago
14

The specific heat capacity of gold is 0.128 J/g C. How much would be needed to warm 250.0 grams of gold from 25.0 C to 100.0 C.

Chemistry
1 answer:
madam [21]3 years ago
7 0

Answer:

3,200 joules

Explanation:

q = mcΔT = (250.0 g)(0.128 j/g°C)(100°C -25°C) = 3,200 joules

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The reaction of dissociation of the benzoic acid in water is given by the following equation:

C₆H₅-COOH + H₂O  ⇄  C₆H₅-COO⁻  +  H₃O⁺    (1)

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HA ⇄ A⁻ + H⁺        (2)

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<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA].     </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}

I hope it helps you!          

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