Answer:
10.70grams
Explanation:
Density of a substance = mass/volume
At STP of a gas (standard temperature and pressure), the density of oxygen gas is 1.429 g/L
Hence, according to this question, in 7.49 L of oxygen gas, there would be:
Using D = m/V
1.429 = m/7.49
m = 1.429g/L × 7.49L
m = 10.70g
Molecular weight of AgBr = 187.7
moles of Ag =
moles of Br = moles of Ag = 2.96 x 10⁻³ mol
concentration of HBr (Molarity) = conc. of Br (strong acid) =
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
