The decomposition of 11.0 g of fe2o3 results in
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We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.
The final temperature is: 75.11 °C.
The work done at constant pressure, W=nR(T₂-T₁)
n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).
W=2.4 × 10³ Joule (Given)
From the expression,
(T₂-T₁)=
(T₂-T₁)=
(T₂-T₁)= 48.11
T₂=300+48.11=348.11 K= 75.11 °C
Final temperature is 75.11 °C.
Circumference of carbon and sulphur is 483.56 & 646.84 respectively
Answer : The partial pressure of X and Y gases are, 0.525 and 1.575 atm respectively.
Explanation : Given,
Moles of X = 2.0 mole
Moles of Y = 6.0 mole
Total pressure = 2.1 atm
Now we have to calculate the mole fraction of X and Y.
and,
Now we have to calculate the partial pressure of X and Y.
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas = 2.1 atm
= mole fraction of gas
and,
Thus, the partial pressure of X and Y gases are, 0.525 and 1.575 atm respectively.