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Yuki888 [10]
3 years ago
12

A 1.44-g sample of an unknown pure gas occupies a volume of 0.335 L at a pressure of and a temperature of 100.0°C. The unknown g

as is ________.
Chemistry
1 answer:
Zepler [3.9K]3 years ago
8 0

Answer:

Xenon

Explanation:

Step 1: Given data

  • Mass (m): 1.44 g
  • Volume (V): 0.335 L
  • Pressure (P): 1.00 atm (I looked it up)
  • Temperature (T): 100.0°C

Step 2: Convert the temperature to Kelvin

K = °C + 273.15 = 100.0°C + 273.15 = 373.2 K

Step 3: Calculate the number of moles (n)

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.00 atm × 0.335 L / (0.0821 atm.L/mol.K) × 373.2 K

n = 0.0109 mol

Step 4: Calculate the molar mass of the gas

M = 1.44 g / 0.0109 mol = 132 g/mol

Step 5: Identify the gas

The gas with a molar mass of about 132 g/mol is xenon.

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aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

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Answer:

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Explanation:

Recall that

n= CV where n=m/M

Hence:

m/M= CV

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