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oxygen gas occupies 2dm3 at a pressure of 0.5 atm and a temperature of 27°...

oxygen gas occupies 2dm3 at a pressure of 0.5 atm and a temperature of 27° calculate the mass of oxygen (r=0.0082 atm dm3 k-3)

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Answers (4)

ESTHER
1 week ago
OLUOMA is correct😎😎😎😎😎😎😎
EmX
1 week ago
Oluoma
1 week ago
Pv=nrt
P=0.5
V=2dm3
R=0.082
T=27+273=300
n=m/mm
Mm of oxygen=16
m=?

Pv=(m/mm)*rt
0.5*2=(m/16)*0.082*300
1=24.6m/16
16=24.6m
m=16/24.6
m=0.65
Therefore, mole(n)=m/mm= 0.65/16
=0.041
  • Oluoma: Correction.. R=0.0082
    Pv=nrt
    P=0.5
    V=2dm3
    R=0.0082
    T=27+273=300
    n=m/mm
    Mm of oxygen=16
    m=?

    Pv=(m/mm)*rt
    0.5*2=(m/16)*0.0082*300
    1=2.46m/16
    16=2.46m
    m=16/2.46
    m=6.5
    Therefore, mole(n)=m/mm= 6.5/16
    =0.41
    Like 0    Dislike 0   1 week ago
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