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2 years ago

6

I will give brainliest!!! What is the density of a sample of chlorine gas that exerts a pressure of 4.5 atm in a 12.6 L containe

r at 36 c (Answer in 4 digits)
Please help I'm so confused!

1 answer:

labwork [276]2 years ago

7 0

Answer : The density of a sample of chlorine gas is, 12.59 g/L

Explanation :

To calculate the density of chlorine gas we are using ideal gas equation.

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\frac{RT}{M}\\\\P=\frac{\rho RT}{M}$

where,

P = pressure of chlorine gas = 4.5 atm

V = volume of chlorine gas = 12.6 L

n = number of moles of chlorine gas

w = mass of chlorine gas

R = gas constant = 0.0821 L.atm/mol.K

T = temperature of chlorine gas = $36^oC=273+36=309K$

M = molar mass of chlorine gas = 71 g/mol

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

$P=\frac{\rho RT}{M}$

$4.5atm=\frac{\rho \times (0.0821L.atm/mol.K)\times (309K)}{71g/mol}$

$\rho=12.59g/L$

Therefore, the density of a sample of chlorine gas is, 12.59 g/L

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An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form

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In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

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1 year ago

Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, G

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Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat required = ?

m = mass of water = 45 g

$c_{p,s}$ = specific heat of solid water = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$c_{p,g}$ = specific heat of liquid water = $1.84J/g^oC$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]$

$\Delta H=139287.75J=139.28775KJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ

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