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Chemistry

1 answer:

stiv31 [10]2 years ago

7 0

The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the following Charle's law equation:

V1/T1 = V2/T2

Where;

T1 = initial temperature
T2 = final temperature
V1 = initial volume
V2 = final volume

V1 = 9L
V2 = ?
T1 = 325K
T2 = 450K

9/325 = V2/450

325V2 = 4050

V2 = 4050/325

V2 = 12.46L

Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

Learn more about volume at: brainly.com/question/2817451

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Moreover, waves are further classified into:

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(2): $H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)$

(3): $H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)$

(4): $H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)$

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$q=m\times \Delta H_{(f , v)}$ ......(i)

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q is the amount of heat absorbed, m is the mass of sample and $\Delta H_{(f , v)}$ is the enthalpy of fusion or vaporization

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$C_{s,l}$ = specific heat of solid or liquid

$T_2\text{ and }T_1$ are final and initial temperatures respectively

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$m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC$

Putting values in equation (i), we get:

$q_1=46g\times 2.108J/g^oC\times (0-(-25))\\\\q_1=2424.2J$

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We are given:

$m=46g\\\Delta H_{fusion}=334J/g$

Putting values in equation (i), we get:

$q_2=46g\times 334J/g\\\\q_2=15364J$

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We are given:

$m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC$

Putting values in equation (i), we get:

$q_3=46g\times 4.186J/g^oC\times (100-0)\\\\q_3=19255.6J$

For process 4:

We are given:

$m=46g\\\Delta H_{vap}=2260J/g$

Putting values in equation (i), we get:

$q_4=46g\times 2260J/g\\\\q_4=103960J$

Calculating the total amount of heat released:

$Q=q_1+q_2+q_3+q_4$

$Q=[(2424.2)+(15364)+(19255.6)+(103960)]$

$Q=141003.8J=141.004kJ$ (Conversion factor: 1 kJ = 1000J)

Hence, the amount of heat absorbed is 141.004 kJ.

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