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The volume of a sample of gas (4.32 g) was 752 mL at 1.98 ATM and 62°C. The gas is​

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Answer:

Bromine

Step-by-step explanation:

To identify the gas, you first need to find the moles of the gas. You can do this using the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol *K)

-----> T = temperature (K)

After converting mL to L and Celsius to Kelvin, you can plug the values into the equation and simplify to find the moles.

P = 1.98 atm R = 0.08206 atm*L/mol *K

V = 752 mL / 1,000 = 0.752 L T = 62 °C + 273.15 = 335.15 K

n = ? moles

PV = nRT

(1.98 atm)(0.752 L) = n(0.08206 atm*L/mol *K)(335.15 K)

1.48896 = n(27.5024)

0.0541 = n

You can identify the gas by determining the molar mass of the gas, which is specific to each element. The molar mass exists as a ratio that compares the mass per 1 mole.

Molar Mass = mass / moles

Molar Mass = 4.32 g / 0.0541 moles

Molar Mass = 79.8 g/mol

This molar mass is closest to the molar mass of bromine (79.904 g/mol).

User Michael Goldshteyn
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