Answer:
The answer is 37amu
Explanation:
We are given:
Chlorine-35: atomic mass =35 amu and percent abundance =75.5% ≡ 0.755
Chlorine-37: atomic mass = x amu and percent abundance =24.5% ≡ 0.25
From table, relative atomic mass of Chlorine is 35.5
⇒(0.755 × 35) + (24.5 × x) = 35.5
26.402 + 24.5x = 35.5
∴ x = 9.075 ÷ 24.5 = 0.37 ≡ 37.
∴ mass of Chlorine 37 = 37amu
Answer:
[ N₂(g) ] = 0.016 M
Explanation:
N₂(g) + 3 H₂(g) ↔ 2 NH₃(g)
The equilibrium constant for the above reaction , can be written as the product of the concentration of product raised to the power of stoichiometric coefficients in a balanced equation of dissociation divided by the product of the concentration of reactant raised to the power of stoichiometric coefficients in the balanced equation of dissociation .
Hence ,
Kc = [ NH₃ (g) ]² / [ N₂(g) ] [ H₂(g) ]³
From the question ,
[ NH₃ (g) ] = 0.5 M
[ N₂(g) ] = ?
[ H₂(g) ] = 2.0 M
Kc = 2
Now, putting it in the above equation ,
Kc = [ NH₃ (g) ]² / [ N₂(g) ] [ H₂(g) ]³
2 = [ 0.5 M ]² / [ N₂(g) ] [ 2.0 M ]³
[ N₂(g) ] = 0.016 M .
I believe it’s A, they form whole-number ratios in the compound
D
B
those are the correct answers
