NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
V
1
/T
1
=V
2
/T
2
(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.
Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.
Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
Answer: Add more information please! Thanks!
Explanation:
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
