Part A Calculate the density of oxygen, O 2 , under each of the following condit

Question

Part A Calculate the density of oxygen, O2, under each of the following conditions:

• STP
• 1.00 atm and 25.0 ?C

Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma. g/L SubmitHintsMy AnswersGive UpReview Part Part B To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 0.65-L bulb, then filled it with the gas at 2.00atm and 24.0 ?C and weighed it again. The difference in mass was 1.5g. Identify the gas. Express your answer as a chemical formula. SubmitHintsMy AnswersGive UpReview Part
Help me solve?? Part A Calculate the density of oxygen, O2, under each of the following conditions:

• STP
• 1.00 atm and 25.0 ?C

Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma. g/L SubmitHintsMy AnswersGive UpReview Part Part A Calculate the density of oxygen, O2, under each of the following conditions:

• STP
• 1.00 atm and 25.0 ?C

Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma. g/L SubmitHintsMy AnswersGive UpReview Part g/L g/L SubmitHintsMy AnswersGive UpReview Part Part B To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 0.65-L bulb, then filled it with the gas at 2.00atm and 24.0 ?C and weighed it again. The difference in mass was 1.5g. Identify the gas. Express your answer as a chemical formula. SubmitHintsMy AnswersGive UpReview Part
Help me solve?? Part B To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 0.65-L bulb, then filled it with the gas at 2.00atm and 24.0 ?C and weighed it again. The difference in mass was 1.5g. Identify the gas. Express your answer as a chemical formula. SubmitHintsMy AnswersGive UpReview Part
Help me solve?? SubmitHintsMy AnswersGive UpReview Part
Help me solve?? g/L

Explanation / Answer

PART A

1)

At STP,

Molar mass of oxygen=32 g=0.032 kg

T=273.15 K

P=100kPa

From ideal gas equation,

PM=(ro)* RT

ro=PM/RT=(10^5*0.032)/(8.314*273.15)=1.409

density=1.409 kg/m^3= 1.409 g/L

2)

Molar mass of oxygen=32 g=0.032 kg

T=273.15 +25= 298.15 K

P=101.325 kPa

From ideal gas equation,

PM=(ro)* RT

ro=PM/RT=(101325*0.032)/(8.314*298.15)=1.308

density=1.308 kg/m^3 =1.308 g/L

density= 1.409 g/L, 1.308 g/L

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PART B

Volume of the bulb= 0.65 L= 0.65 *10^-3 m^3

P= 2 atm=2*101.325 = 202650 Pa

Temperature=273.15 +24= 297.15 k

Mass of gas= Difference in mass= 1.5 gm

From ideal gas equation,

PV=n RT

PV=(m/M)*RT

M=mRT/PV=(1.58*10^-3)*8.314*297.15/(202650*0.65 *10^-3)=0.02963 kg

Molecular mass of X2=Molar mass of gas(X2)=29.63 g

Atomic mass of X=29.63/2 =14.82 g

Since, nitrogen has a atomic mass of nearly 14 amu

X2 is N2

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