!!LONG ANSWER !!
Let's start with the number of nickel atoms, since it's the easiest one to calculate.
Before determining how many atoms of nickel you have in 1 gram, you must first determine how many moles of nickel you have. This is done by using its molar mass, 58.7 g/mol.
#"1 g" * "1 mole"/"58.7 g" = "0.0170 moles Ni"#
Use Avodagro's number - #6.022 * 10^(23)# atoms in 1 mole - to determine how many atoms of nickel you have
#"0.0170 moles" * (6.022 * 10^(23)"atoms")/"1 mole" = 1.02 * 10^(22)"atoms of Ni"#
Now for the lithium aluminium hydride (LAH), or #LiAlH_4#. Notice that 1 molecule of LAH has 1 atom of lithium, 1 of aluminium, and 4 of hydrogen. Determine how many molecules of LAH you have first, then go to its component atoms.
#"0.10 g" * "1 mole"/"38.0 g" = "0.00263 moles LAH"#
#"0.00263 moles" * (6.022 * 10^(23)"molecules")/"1 mole" = 1.6 * 10^(21)"molecules of LAH"#
The number of atoms of Li, Al, and H will be
#1.6 * 10^(21)"molecules LAH" * "1 atom Li"/"1 molecule LAH" = 1.6 * 10^(21)"atoms Li"#
#1.6 * 10^(21)"molecules LAH" * "1 atom Al"/"1 molecule LAH" = 1.6 * 10^(21)"atoms Al"#
#1.6 * 10^(21)"molecules LAH" * "4 atoms H"/"1 molecule LAH" = 6.4 * 10^(21)"atoms H"#
Now for the oxygen. Gases are a little trickier because you need values for temperature and pressure to determine the number of moles. Since no mention of particular conditions was made, I'll assume STP - 273.15 K and 1 atm.
The great part about STP is that 1 mole of any ideal gas occupies exactly 22.4 L - this is called the molar volume of a gas. So if you know the volume, you can determine how many moles of oxygen you have.
Remember that air is only 20.9% oxygen, so the volume of oxygen gas will be
#"1 mL air" * "20.9 mL oxygen"/"100 mL air" = "0.209 mL "# #O_2#
Another thing to look out for - notice that you've got diatomic oxygen, or #O_2#. This means that the number of oxygen atoms will be twice the number of #O_2# molecules. So,
#n = V/V_("molar") = (0.209 * 10^(-3)"L")/"22.4 L" = 9.3 * 10^(-6)"moles "# #O_2#
#9.3 * 10^(-6)"moles" * (6.022 * 10^(23)"molecules" O_2)/("1 mole"O_2) = 5.6 * 10^(18)"molecules oxygen"#
Therefore, the number of oxygen atoms will be
#5.6 * 10^(18)"molecules oxygen" * "2 oxygen atoms"/"1 oxygen molecule" = 1.1 * 10^(19)"atoms of oxygen"#
You can use the same strategy to calculate how many atoms of nitrogen you have. Keep in mind that it too is a diatomic molecule, #N_2#, and that it's percentage in air's composition is 78.1%.
