A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.
We can calculate the total number of moles using the ideal gas equation.
At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.
We can calculate the moles of nitrogen using the ideal gas equation.
The mole fraction of nitrogen in the mixture is:
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
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<span>26.833 liters
Aluminum oxide has a formula of Al</span>₂O₃,<span> which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).
Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.
Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.
Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815
Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,
1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
</span>
The average atomic mass of your mixture is 1.03 u
.
The average atomic mass of H is the weighted average of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).
Thus,
0.99 × 1.01 u = 0.998 u
0.002 × 2.01 u = 0.004 u
0.008 × 3.02 u = <u>0.024 u</u>
TOTAL = 1.03 u
Answer:
1.387 moles
Explanation:
Step 1:
The balanced equation for the reaction. This is illustrated below:
4Fe + 3O2 —> 2Fe2O3
Step 2:
Determination of the number of mole of Fe in 155.321g of Fe. This can be achieved by doing the following:
Mass of Fe = 155.321g
Molar Mass of Fe = 56g/mol
Number of mole of Fe =?
Number of mole = Mass/Molar Mass
Number of mole of Fe = 155.321/56
Number of mole of Fe = 2.774 mol
Step 3:
Determination of the number of mole of rust (Fe2O3) produced. This is illustrated below:
From the balanced equation above,
4 moles of Fe produced 2 moles of Fe2O3.
Therefore, 2.774 moles of Fe will produce = (2.774 x 2)/4 = 1.387 moles of Fe2O3.
Therefore, 1.387 moles of rust (Fe2O3) is produced from the reaction
Answer:
The concentration of the solution is 5.8168 × 
mol.
Explanation:
Here, we want to calculate the concentration of the solution.
The unit of this is mol/dm^3
So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3
The number of moles = mass/molar mass
molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol
Number of moles = 33.4/87 = 0.384 moles
This 0.384 moles is present in 660 L
x moles will be present in 1 dm^3
Recall 1 dm^3 = 1L
x * 660 = 0.384 * 1
x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3
