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2 years ago

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consider the balanced chemical equation below. when the chemical reaction was carried out calculated theoretical was yield for s

odium bromide 162 grams but the measured yield was 150 grams what is the percent yield?

1 answer:

8_murik_8 [283]2 years ago

7 0

Answer:

Explanation:

% yield = (actual yield / theoretical yield) X 100

For this question,

% yield = (150g/ 162 g) X 100 = 92.6%

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Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

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$\Delta G^o=-RT\ln K$

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$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

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Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

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