Answer:
I think that the answer is N20:2(14.01)+1(16.06)=44.02g/mol(c) fluorine Di nitrogen monoxide contains how many mom 6.022 x
1023 mol N20
Answer:
0.800 mol of O2
Explanation:
<em>Calculate the moles of oxygen produced by the reaction of 0.800mol of carbon dioxide.</em>
The balanced equation for the reaction is given as;
6CO2 + 6H2O → C6H12O6 + 6O2
From the reaction;
6 mol of CO2 produces 6 mol of O2
0.0800 mol of CO2 would produce x mol of O2
6 = 6
0.0800 = x
Solving for x;
x = 6 * 0.800 / 6
x = 0.800 mol
Answer:
4.13×10²⁷ molecules of N₂ are in the room
Explanation:
ideal gases Law → P . V = n . R . T
Pressure . volume = moles . Ideal Gases Constant . T° K
T°K = T°C + 273 → 20°C + 273 = 293K
Let's determine the volume of the room:
18 ft . 18 ft . 18ft = 5832 ft³
We convert the ft³ to L → 5832 ft³ . 28.3L / 1 ft³ = 165045.6 L
1 atm . 165045.6 L = n . 0.082 L.atm/mol.K . 293K
(1 atm . 165045.6 L) / 0.082 L.atm/mol.K . 293K = n
6869.4 moles of N₂ are in the room
If we want to find out the number of molecules we multiply the moles by NA
6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules
