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3 years ago

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What is the volume of 45.6g of silver if the density of silver is 10.5g/mL? A. 4.34mL B. 479mL C. 0.23mL

1 answer:

nirvana33 [79]3 years ago

8 0

Explanation:

first you get moles of silver

n=m/M

hence you add no of moles to this equation

c=nv

v=n/c

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Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which

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Answer:

Explanation:

From the information given:

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However; the workdone = -PdV

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A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch

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Answer:

The reactions free energy $\Delta G = -49.36 kJ$

Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

The pressure of (Cl) gas is $P_{Cl} = 9.15 \ atm$

The pressure of nitrosly chloride (NOCl) is $P_{(NOCl)} = 7.70 \ atm$

The reaction is

$2NO_{(g)} + Cl_2 (g)$ ⇆ $2 NOCl_{(g)}$

From the reaction we can mathematically evaluate the $\Delta G^o$ (Standard state free energy ) as

$\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}$

The Standard state free energy for NO is constant with a value

$\Delta G^o _{NO} = 86.55 kJ/mol$

The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

The Standard state free energy for $NOCl$ is constant with a value

$\Delta G^o _{NOCl} =66.1kJ/mol$

Now substituting this into the equation

$\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6$

$= -43 kJ/mol$

The pressure constant is evaluated as

$Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }$

Substituting values

$Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}$

$= 0.0765$

The free energy for this reaction is evaluated as

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

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

T is temperature in K with a given value of $T = 25+273 = 298 K$

Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

$= -43-6.36$

$\Delta G = -49.36 kJ$

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