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2 years ago

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How many silver (Ag) atoms are in a silver bar that is 20.0 g? A. 1.3 3 1027 Ag atoms B. 6.02 3 1023 Ag atoms C. 1.1 3 1023 Ag a

toms D. 1.8 3 1023 Ag atoms

2 answers:

ad-work [718]2 years ago

6 0

Answer: The correct option is, (C) $1.13\times 10^{23}$

Explanation: Given,

Mass of Ag = 20 g

Molar mass of Ag = 108 g/mole

First we have to calculate the moles of Ag.

$\text{Moles of }Ag=\frac{\text{Mass of }Ag}{\text{Molar mass of }Ag}=\frac{20g}{108g/mole}=0.185moles$

Now we have to calculate the number of atoms in Ag.

As, 1 mole of Ag contains $6.022\times 10^{23}$ number of Ag atoms

So, 0.185 mole of Ag contains $0.185\times 6.022\times 10^{23}=1.13\times 10^{23}$ number of Ag atoms

Therefore, the number of atoms present in Ag are, $1.13\times 10^{23}$

postnew [5]2 years ago

4 0

C. 1.1 3 1023 Ag atoms

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$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

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$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

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Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

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$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

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