Answer:
The gas that Dr. Brightguy added was O₂
Explanation:
Ideal Gases Law to solve this:
P . V = n . R . T
Firstly, let's convert 736 Torr in atm
736 Torr is atmospheric pressure = 1 atm
20°C = 273 + 20 = 293 T°K
125 mL = 0.125L
0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K
(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n
5.20x10⁻³ mol = n
mass / mol = molar mass
0.1727 g / 5.20x10⁻³ mol = 33.2 g/m
This molar mass corresponds nearly to O₂
40.1g of nitrogen gas is produced.
The equation given is
2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O
This equation is already balanced.
When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.
We get 1 mole of nitrogen from 3 moles of copper oxide.
We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.
4.3/3 x 1 = 1.433 mols
- 1.433 mols of nitrogen gas are produced
- The molar mass of nitrogen gas is 14+14 = 28g
- The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g
40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.
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Answer:
It would be to the fourth power
Explanation:
We’re going to use the mass percent formula shown below:
For the percent by mass N, we’re going to rewrite the equation as:
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Problem Details
Calculate the mass percent composition of nitrogen in each nitrogen-containing compound:
c. NO2
