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Answer:
Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488
Answer:
Answer:
Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,
n = 96/32 = 3 Moles
Mole fraction = Mole conc./Total moles
Mole fraction of CO
= 3.14/(3.14 + 3)
= 3.14/6.14
= 0.511
Mole Fraction of O2
= 3/(3.14 + 3)
= 3/6.14
= 0.488Given mixture contains:
88g of CO gas
96g of O2 gas
To find:
Mole fraction of the two gases.
Solution:
No. of moles (n) = Given mass/Molar Mass
For CO,
n = 88/28 = 3.14 Moles
For O2,