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
Explanation:
Oxidation state of Nitrogen in N2O5 is +5
Answer: the answer is C oxidizing
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g