Answer: The bonds are intermediate between double and single bonds
Explanation:
A closer look at the diagram below shows that the bonds in sulphur IV oxide are intermediate between double and single bonds. Hence they do not have the exact bond angle of single bonds. This is why the bond angle is not exactly 120°. There are two resonance structures in the diagram that clearly show this point.
When an object enters the Earth's atmosphere, it experiences a few forces, including gravity and drag. Gravity will naturally pull an object back to earth.
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Answer:
75 kJ/mol
Explanation:
The reactions occur at a rate, which means that the concentration of the reagents decays at a time. The rate law is a function of the concentrations and of the rate constant (k) which depends on the temperature of the reaction.
The activation energy (Ea) is the minimum energy that the reagents must have so the reaction will happen. The rate constant is related to the activation energy by the Arrhenius equation:
ln(k) = ln(A) -Ea/RT
Where A is a constant of the reaction, which doesn't depend on the temperature, R is the gas constant (8.314 J/mol.K), and T is the temperature. So, for two different temperatures, if we make the difference between the two equations:
ln(k1) - ln(k2) = ln(A) - Ea/RT1 - ln(A) + Ea/RT2
ln (k1/k2) = (Ea/R)*(1/T2 - 1/T1)
k1 = 8.3x10⁸, T1 = 142.0°C = 415 K
k2 = 6.9x10⁶, T2 = 67.0°C = 340 K
ln(8.3x10⁸/6.9x10⁶) = (Ea/8.314)*(1/340 - 1/415)
4.8 = 6.39x10⁻⁵Ea
Ea = 75078 J/mol
Ea = 75 kJ/mol
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
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