Answer:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
E decreseas 3/2 as fast as G increases = 0.30 M/s
Explanation:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:
Given data = d[D]/dt = 0.10 M/s
-d[D] / 2dt = d[H]/dt
d[H]/dt = 0.05 M/s
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:
d[G] / 2dt = -d[H]/3dt
E decreseas 3/2 as fast as G increases = 0.30 M/s
False, a gas that produces a pop is hydrogen
Hope this helped :)
Answer:
all of the above. they all are chemical reactions
Answer:
0.123 moles of ammonia, can be produced
Explanation:
First of all, we need to determine the reaction:
Ammonia is produced by the reaction of hydrogen and nitrogen.
3H₂(g) + N₂(g) → 2NH₃(g)
Ratio is 3:2. Let's solve the question with a rule of three:
If 3 moles of hydrogen can produce 2 moles of ammonia
Then, 0.37 moles will produce (0.37 . 2) /3 = 0.123 moles
