Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa
∴ on rearranging above eq., we get V2 =

∴ V2 = 4 m3
Answer:
4.4×10² cm³
Explanation:
From the question given above, the following data were obtained:
Diameter (d) = 68.3 mm
Height (h) = 0.120 m
Volume (V) =?
Next, we shall convert the diameter (i.e 68.3 mm) to cm.
This can be obtained as follow:
10 mm = 1 cm
Therefore
68.3 mm = 68.3 mm / 10 mm × 1 cm
68.3 mm = 6.83 cm
Therefore, the diameter 68.3 mm is equivalent 6.83 cm.
Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:
1 m = 100 cm
Therefore,
0.120 m = 0.120 m/ 1 m × 100 cm
0.120 m = 12 cm
Therefore, the height 0.120 m is equivalent 12 cm.
Next, we shall determine the radius of the cylinder. This can be obtained as follow:
Radius (r) is simply half of a diameter i.e
Radius (r) = Diameter (d) /2
r = d/2
Diameter (d) = 6.83 cm
Radius (r) =?
r = d/2
r = 6.83/2
r = 3.415 cm
Finally, we shall determine the volume of the cylinder as follow:
Radius (r) = 3.415 cm
Height (h) = 12 cm
Volume (V) =?
Pi (π) = 3.14
V = πr²h
V = 3.14 × (3.415) ² × 12
V = 440 cm³
V = 4.4×10² cm³
Therefore, the volume of the cylinder is 4.4×10² cm³
Thermal energy is caused by the motion of atoms and molecules. That's what creates heat, or thermal energy.