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
A two carbon chain attached to a benzene ring: ethylbenzene
I think that the answer is A because if you think about water for an example. When water turns into ice, it's still technically water, just the molecules are frozen. The water is physically changed. It could also see why you think it is B but I don't think so completely. It doesn't really make sense to me. So I'd choose A. It's definitely not C or D.
Answer:
K=CHANGE IN CONCENTRATION/TIME TAKEN
Explanation:
Answer:
See Explanation
Explanation:
Let us consider the ionization of these compounds;
H3PO4 ⇔3H^+ + PO4^3-
H3BO3 ⇔3H^+ + BO3^3-
The next to consider is the type of electrolyte the both solution are; the both solutions are weak electrolytes and weak electrolytes do not ionize to a large extent.
The implication of this is that, not so much number of ions is added to the solution due to the poor ionization of these weak electrolytes. Hence, in spite of the subscript of 3, the conductivity of the solution does not significantly improve for the reason stated here quite unlike when strong electrolytes are used.
