5 moles H2O (1 mole O/1 mole H2O)(6.022 X 1023/1 mole O)(1 mole O atoms/6.022 X 1023)
= 5 moles oxygen atoms
=================( as you see this is a formal set up that you do not need to use fully as Avogadro's number is over itself as a form of one )
5 moles H2O (2 moles H/1 mole H2O)6.022 X 1023/1 mole H)(1 mole H atoms/6.022 X 1023)
= 10 moles hydrogen atoms
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4 moles of oxygen atoms are present in 4 moles of H2O
In one mole of hydrogen peroxide (H2O2), there are two moles of oxygen atoms.
The number of hydrogen atoms is 14,290540253661.10e23.
Well if one mole of water = 2 moles of hydrogen and 1 mole of oxygen, than 2moles of water = 4 moles of hydrogen and 2moles of oxygen.
Methane (CH4) has four atoms of hydrogen per molecule. If there are 3 moles of methane, then there are 12 moles of hydrogen.
Glucose is C6H12O6 and thus the mole ratio is 6 moles of carbon to 6 moles of Hydrogen Molecules (12 moles of Hydrogen atoms) and 3 moles of oxygen molecules (6 moles of oxygen atoms)
4 moles of oxygen atoms are present in 4 moles of H2O
2 moles of benzene gives 12 moles of hydrogen atoms since benzene is C6H6
I was wondering about this... but I think if you combined..The four oxygen gas O2, and the two of Hydrogen gas H2.. and predict was will happened I guess this is what it will or might be calculate, but Im not sure.KKKO2O2+H2H1O4+H2O2H
In one mole of hydrogen peroxide (H2O2), there are two moles of oxygen atoms.
The number of hydrogen atoms is 14,290540253661.10e23.
Well if one mole of water = 2 moles of hydrogen and 1 mole of oxygen, than 2moles of water = 4 moles of hydrogen and 2moles of oxygen.
Methane (CH4) has four atoms of hydrogen per molecule. If there are 3 moles of methane, then there are 12 moles of hydrogen.
200. The formula is for every 1 Oxygen atom, 2 Hydrogen atoms must be present in water. Otherwise you would produce H2O2 (you cannot make it HO because it is never found in molecules on it's own) which is bleach.
five
One molecule has four H atoms.So two moles have 8 moles
3KNO3, so 9 oxygen atoms.