This is how you solve it you use Avagoadro's constant which is 6.022 X1023
this number is not showing up correctly here. It is supposed to be 6.022 X 10 to the 23 power.
This is the number of atoms per mole of a molecule. Just like you have 12 eggs in a dozen. For example how many eggs would 2.4 dozen eggs have. 2.4 X 12 = 28.8 eggs.
So in the following problem the moles will cancel each other out and give you the number of atoms. I would calculate for you but I don't have my calculator on me, but this is how you set it up.
2.4 mole Zn X 6.022 X10 23 atoms/mole
2.4 mole hydrogen (6.022 X 10^23/1 mole H) x 2 atoms of hydrogen per mole of H2
= 2.8 X 10^24 atoms of hydrogen
--------------------------------------------------wrong!!!
Whoever " corrected this is wrong. A mole of H, or H2 is still a mole and designation of that in the conversions mean nothing.
2.4 mole hydrogen ( in gaseous form ) (6.022 X 10^23/1 mole H2)
= 1.4 X 10^24 atoms of hydrogen
------------------------------------------------correct!
--------------------------------------------( no 2 atoms of hydrogen per mole H2!! )
For this problem, the Atomic Mass is not required. Take the mass in moles and multiply it by Avogadro's constant, 6.02 × 1023. Divide by one mole for the units to cancel.
2.4 moles H2 × (6.02 × 1023 atoms) = 1.44 × 1024 atoms
1 mol is about 6.022 x 1023 atoms so 2.4 mol is 1.44528 x 1024 atoms.... I think! Its a lot anyway.
1 mole of Be weighs 9.012 g
So 2 moles will weigh 18.024 g
Since hydrogen exists normally as diatomic molecules, the answer is 2 X 2.4 X Avogadro's Number or 2.9 X 1024 atoms, to the justified number of significant digits.
2.4 moles hydrogen (6.022 X 10^23/1 mole H2)
= 1.4 X 10^24 atoms of hydrogen ( gas, or otherwise )
12 g of carbon has 6.023 x 1023 atoms or 1 mole of atom
So 2.4 g will have 1.205 x 1023 atoms or 0.2 mole of atom
1 mol Be = 9.01g = 6.02*10^23 atoms
8.97 mol Be = 80.8g = 5.4*10^24 atoms
The mass is 80,811 g.
1 mole of water is 18g. 36g of water is thereforeequivalent to 2 moles. 2 moles of water contains 4 moles of hydrogen and 2 moles of oxygen and so there are6 moles of atoms in 2 moles of water. 6 moles of atoms x (3.6x10^24) = 36.1x10^24 atoms
1,36 moles C5H12 (12 moles H/1 mole C5H12)(6.022 X 10^23/1 mole H) = 9.83 X 10^24 atoms of hydrogen
One Mole of C2H4 will containt 6.0221415×10^23 molecules of C2H4. Therefore 2.23 Moles of C2H4 will contain 1.39713683x10^24 molecules of C2H4. There are 4 Hydrogen atoms in C2H4, so 1.39713683x10^24 x 4 = 5.58854732X10^25 atoms of hydrogen.
10 moles of oxygen atoms or 5 moles of oxygen molecules.
To answer this kind of question, multiply the coefficient, 2, by the subscript after the atomic symbol for hydrogen, H, to obtain 24 hydrogen atoms.
1 mole of water is 18g. 36g of water is thereforeequivalent to 2 moles. 2 moles of water contains 4 moles of hydrogen and 2 moles of oxygen and so there are6 moles of atoms in 2 moles of water. 6 moles of atoms x (3.6x10^24) = 36.1x10^24 atoms
0.86 moles CH4 (4 mole H/1 mole CH4)(6.022 X 10^23/1 mole H) = 2.1 X 10^24 atoms of hydrogen --------------------------------------------
4,515.10e24 atoms of magnesium is equal to 7,5 moles.
1.81x10^24 atoms
5.75x10^24 atoms x 1 mole/6.02x10^23 atoms = 9.55 moles of Al
1 mol Cr/6.02x10^23 atoms x 3.6x10^24 atoms = 5.98 moles
3,00 moles of Li have 18,066422571.10e23 atoms.
You can figure this question out by looking up quinine to find its chemical formula. The number of moles of hydrogen will be the number of hydrogen in the chemical formula. Once you see the chemical formula is C20H24N2O2 then you know that there are 24 moles of hydrogen for every one mole of quinine.
2.80x10^24 atoms x 1 mol/6.02x10^23 atoms = 4.65 moles
The answer is 3,587 moles.
1,36 moles C5H12 (12 moles H/1 mole C5H12)(6.022 X 10^23/1 mole H) = 9.83 X 10^24 atoms of hydrogen
One Mole of C2H4 will containt 6.0221415×10^23 molecules of C2H4. Therefore 2.23 Moles of C2H4 will contain 1.39713683x10^24 molecules of C2H4. There are 4 Hydrogen atoms in C2H4, so 1.39713683x10^24 x 4 = 5.58854732X10^25 atoms of hydrogen.