<span>T = 22.0 + 273.15 =295.15 K
Molar mass Cl2 = 70.906 g/mol
d = molar mass x p / RT = 70.906 x 1.00 / 0.0821 x 295.15 = 2.93 g/L
Hope this isn't to much? If it doesn't help you know where to find me :)</span>
Answer:
0.15 L
Explanation:
You need to first find the volume of the container. You can do this by dividing the mass by the density. This will give you the mass in mL.
5.00 kg = 5,000 g
(5,000 g)/(1.00 g/mL) = 5,000 mL
5,000 mL = 5 L
Now, find the volume the seawater will take up.
(5,000 g)(1.03 g/mL) = 4854.4 mL
4854.4 mL = 4.85 L
Subtract the two volumes to find the volume that left unfilled.
5 L - 4.85 L = 0.15 L
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
