<span>d.2HNO3 (aq) + Sr(OH)2 (aq) → 2H2O (l) + Sr(NO3)2(aq)
4H </span>4H
8O 8O
2N 2N
1Sr 1Sr<span>
</span>
I believe it would be 1.660539040 × 10−24 gram.
I think it's 2 I tried looking it up because I was not sure.
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
<h3><u>Answer;</u></h3>
= 3032.15 kPa
<h3><u>Explanation;</u></h3>
Using the equation;
PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.
Volume = 7.5 L, T = 274 +273 = 547 K, N = 5 moles
Therefore;
Pressure = nRT/V
= (5 × 0.08206 × 547)/7.5 L
= 29.925 atm
But; 1 atm = 101325 pascals
Hence; Pressure = 3032150.63 pascals
<u>= 3032.15 kPa</u>
