Missing question:
(a) 1s2 2s2 2p6 3s1
(b) 1s2 2s2 2p6 3s2
(c) 1s2 2s2 2p6 3s2 3p1
(d) 1s2 2s2 2p6 3s2 3p4
(e) 1s2 2s2 2p6 3s2 3p5
Answer is: a) 1s²2s²2p⁶3s¹ (sodium).
Sodium have the largest second ionization energy, because when he lost one electron(first ionization energy), he have stable electron configuration of noble gas neon (1s²2s²2p⁶), so sodium do not need to lost second electron, because he will have unstable electron configuration.
The standard enthalpy of formation for chlorine is zero but the standard entropy is larger than 0 because it is the elemental state of chlorine.
The standard enthalpy of formation for chlorine is zero because cl2 is the elemental state of chlorine and it does not require any energy for the formation of the standard state of chlorine.
The entropy of any system cannot be negative. It can only be positive or zero.
The entropy of a system will become zero only at a absolute zero temperature.
That's why the entropy of chlorine in elemental state is more than zero because absolutely zero temperature can't be obtained.
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Answer:
6.52×10⁴ GHz
Explanation:
From the question given above, the following data were obtained:
Wavelength (λ) = 4.6 μm
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
Next we shall convert 4.6 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
4.6 μm = 4.6 μm × 1×10¯⁶ m / 1 μm
4.6 μm = 4.6×10¯⁶ m
Next, we shall determine frequency of the light. This can be obtained as follow:
Wavelength (λ) = 4.6×10¯⁶ m
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
v = λf
2.998×10⁸ = 4.6×10¯⁶ × f
Divide both side by 4.6×10¯⁶
f = 2.998×10⁸ / 4.6×10¯⁶
f = 6.52×10¹³ Hz
Finally, we shall convert 6.52×10¹³ Hz to gigahertz. This can be obtained as follow:
1 Hz = 1×10¯⁹ GHz
Therefore,
6.52×10¹³ Hz = 6.52×10¹³ Hz × 1×10¯⁹ GHz / 1Hz
6.52×10¹³ Hz = 6.52×10⁴ GHz
Thus, the frequency of the light is 6.52×10⁴ GHz
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Answer:
For finding frequency, we need to first find the period of the graph.
The period of a sinusoidal graph is the time interval in which it repeats its pattern.
In the graph, we can see, after 
time, it repeats its pattern.
Hence the period of the graph is 
.
Now we need to find its frequency 
The formula for frequency is 
This is the answer
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