If you have a rival employee trying to beat you to a higher paying job that will encourage competition.
We solve for the number of mols of CO2 given its volume, assuming that it is an ideal gas.
n = PV / RT
Substituting,
n = (1 atm)(15.67 L) / (0.082 L.atm/mol.K)(273.15 K) = 0.7 moles
From the balanced chemical reaction above, we determine the number of O2 moles needed to produce 0.7 moles CO2.
0.7 moles CO2 x (3 moles O2/ 2 moles CO2) = 1.05 moles O2
Then, we multiply this value by 22.4 L/mol
1.05 moles O2 x (22.4 L / mol) = 23.52 L O2
Thus, the answer is the fourth choice.
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
