The answer is
6.8 * 10^-15
The explanation:
1- we have to convert all measurements to the same units:
Conversions:
when 1 m = 100 cm
and 1 m = 10^12 pm
So,
proton radius: 1.0*10^-13 cm * (1m / 100 cm) = 10^-15 m
proton volume: 4/3 * pi * r^3 = 4/3 * pi * (10^-15 m)^3 = 4.2 * 10^-45 cu. meters
and
H atom radius: 52.9 pm * (1m / 10^12 pm) = 5.29 * 10^-11 m
H atom volume: 4/3 * pi * r^3 = 4/3 * pi * (5.29 * 10^-11 m)^3 = 6.2 × 10^-31 cu. meters
So,
2- Fraction of space occupied by nucleus = proton volume / H atom volume
= (4.2 * 10^-45 cu. meters) / (6.2 × 10^-31 cu. meters)
= 6.8 * 10^-15
So, the "fraction" would be 6.8 * 10^-15 out of 1.
Explanation:
It is given that,
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,
The amount of energy change during the transition is given by :
![\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5CDelta%20E%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
And
![\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5Cdfrac%7Bhc%7D%7B%5Clambda%7D%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
Plugging all the values we get :
![\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5](https://tex.z-dn.net/?f=%5Cdfrac%7B6.63%5Ctimes%2010%5E%7B-34%7D%5Ctimes%203%5Ctimes%2010%5E8%7D%7B3745%5Ctimes%2010%5E%7B-9%7D%7D%3D2.179%5Ctimes%2010%5E%7B-18%7D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C%5Cdfrac%7B5.31%5Ctimes%2010%5E%7B-20%7D%7D%7B2.179%5Ctimes%2010%5E%7B-18%7D%7D%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C0.0243%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B64%7D%5D%5C%5C%5C%5C0.0243%2B%5Cdfrac%7B1%7D%7B64%7D%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5C0.039925%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5Cn_f%5E2%3D25%5C%5C%5C%5Cn_f%3D5)
So, the final level of the electron is 5.
Answer:
The majority of the weight in an atom is found in the nucleus.
Explanation:
The protons and neutrons that make up the nucleus of the atom may take up a tiny amount of space in comparison to the rest of the atom, but they are far more dense than the electrons that orbit the nucleus.
Explanation:
Moles of metal,
=
4.86
⋅
g
24.305
⋅
g
⋅
m
o
l
−
1
=
0.200
m
o
l
.
Moles of
H
C
l
=
100
⋅
c
m
−
3
×
2.00
⋅
m
o
l
⋅
d
m
−
3
=
0.200
m
o
l
Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.
So if
0.200
m
o
l
acid react, then (by the stoichiometry), 1/2 this quantity, i.e.
0.100
m
o
l
of dihydrogen will evolve.
So,
0.100
m
o
l
dihydrogen are evolved; this has a mass of
0.100
⋅
m
o
l
×
2.00
⋅
g
⋅
m
o
l
−
1
=
?
?
g
.
If 1 mol dihydrogen gas occupies
24.5
d
m
3
at room temperature and pressure, what will be the VOLUME of gas evolved?
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
