69,57% + 30,43% = 100%
100% ----------- 92,02g
69,57% oxygen ----- x
x = 64,004 ≈ 64,00g oxygen
100% ---------- 92,02g
30,43% nitrogen ----x
x = 27,99 ≈ 28g nitrogen
Mass of oxygen = 16g
Mass of nitrogen = 14g
moles of oxygen = 64g : 16g = 4
moless of oxygen = 28g : 14g = 2
N : O
4 : 2 ||:2
2 : 1
empirical formulas = N₄O₂
molecular formulas = N₂O
N₂ + ¹/₂O₂ ---> N2O
3, 9, 16 are correct! 17 should be b because buffered means that it resists change in pH
Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol