Gravitational pull coming from the sun
Answer:
It's nitrogen
Explanation:
cuz it has valence 3 and a diatomic gas at room temperature
The balanced equation
for the reaction is
CO(g) + 2H₂(g) ⇄ CH₃OH(g)
The given
concentrations are at equilibrium state. Hence we can use them directly in
calculation with the expression for the equilibrium constant, k.
expression for k can be written as
k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²
</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M
From substitution,
k = 0.030
M / 0.020 M x (0.072 M)²
k =
289.35 M⁻²
<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².
<span> </span>
43 degrees
angle of reflection is equal to the angle of incidence
Answer: E
=
1.55
⋅
10
−
19
J
Explanation:
The energy transition will be equal to 1.55
⋅
10
−
1
J
.
So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition
1
λ =
R
⋅
(
1
n
2
final −
1
n
2
initial )
, where
λ
- the wavelength of the emitted photon;
R
- Rydberg's constant - 1.0974
⋅
10
7
m
−
1
;
n
final
- the final energy level - in your case equal to 3;
n
initial
- the initial energy level - in your case equal to 5.
So, you've got all you need to solve for λ
, so
1
λ =
1.0974
⋅10 7
m
−
1
⋅
(....
−152
)
1
λ
=
0.07804
⋅
10
7
m
−
1
⇒
λ
=
1.28
⋅
10
−
6
m
Since
E
=
h
c
λ
, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by
h
⋅
c
, where
h
- Planck's constant -
6.626
⋅
10
−
34
J
⋅
s
c
- the speed of light -
299,792,458 m/s
So, the transition energy for your particular transition (which is part of the Paschen Series) is
E
=
6.626
⋅
10
−
34
J
⋅
s
⋅
299,792,458
m/s
1.28
⋅
10
−
6
m
E
=
1.55
⋅
10
−
19
J
