A gas goes from 1L at 273 K to 2 L when heated. What is the new temp?
1 answer:
You might be interested in
The chemical reaction equation for this is
XeF6 + 3H2 ---> Xe + 6HF
Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.
we get n = 0.1433 moles H2
to get the mass of XeF6,
we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.
0.1433 moles H2 x
x 
= 11.71 g XeF6
Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
XeF6 + 3H2 ---> Xe + 6HF
Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.
we get n = 0.1433 moles H2
to get the mass of XeF6,
we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.
0.1433 moles H2 x
Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
Answer:
Oxide of M is and sulfate of
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:
Moles of hydrogen gas produced = 0.01225 mol
Moles of metal =
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.
x = 2.9 ≈ 3
Formulas for the oxide and sulfate of M will be:
Oxide of M is and sulfate of
.