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
Answer:
Molality = 1.46 molal
The freezing point of the solution = -2.72 °C
Explanation:
Step 1: Data given
The freezing point of water H2O is 0.00°C at 1 atmosphere
urea = nonelectrolyte = van't Hoff factor = 1
Mass urea = 13.40 grams
Molar mass urea = 60.1 g/mol
Mass of water = 153.2 grams
Molar mass H2O = 18.02 g/mol
Kf = 1.86 °C/m
Step 2: Calculate moles urea
Moles urea = mass urea /molar mass urea
Moles urea = 13.40 grams / 60.1 g/mol
Moles urea = 0.223 moles
Step 3: Calculate the molality
Molality = moles urea / mass water
Molality = 0.223 moles / 0.1532 kg
Molality = 1.46 molal
Step 4: Calculate the freezing point of the solution
ΔT = i * Kf * m
ΔT = 1* 1.86 °C/m * 1.46 m
ΔT = 2.72 °C
The freezing point = -2.72 °C
Answer:
SO3
Explanation:
Data obtained from the question include:
S = 40%
O = 59%
To obtain the empirical formula, do the following:
Divide the above by their molar mass as shown below:
S = 40/32 = 1.25
O = 59/16 = 3.69
Next, divide by the smallest as shown below:
S = 1.25/125 = 1
O = 3.69/1.25 = 3
Therefore, the empirical formula is SO3
Answer:
Two-third of the toxic metal, mercury comes from gold mining.
Explanation:
During gold mining small pieces of gold which are within the soil are separated with the help of mercury. Both gold and mercury are mixed and they forms amalgam. Further gold can be extracted by vaporizing the mercury, thus releasing it into air and contributing to two-third of the air pollution by human activities.
Therefore (d) gold mining is the correct answer.
