No, due to Einstein''s wormhole theory, it equals approximately twelve point three and a half of a black man's left thumb.
1.5*10^23
The molecular mass of NH3 is the sum of the atomic mass of nitrogen and three times the atomic mass of hydrogen, or 14.007 + 3(1.008) = 17.031. Therefore, the number of moles of NH3 in 14.0 grams is 14.007/17.031 = 0.822. Since each molecule of N2 supplies two nitrogen atoms and each molecule of NH3 needs only one nitrogen atom, the number of moles of N2 needed is half the number of moles of NH3 formed = 0.411.
28g
The atomic number of nitrogen (N) is 7. The atomic weight of N is 14.00674 grams per mole. Nitrogen gas exists as a diatomic molecule: N2.
38 grams oxygen gas (1 mole O2/32 grams)(6.022 X 1023/1 mole O2) = 7.2 X 1023 molecules of oxygen gas -------------------------------------------------
molar weight of N = 14 grams/mole 35.7 grams/14 grams/mole = 2.55 moles
The atomic number of nitrogen (N) is 7. The atomic weight of N is 14.00674 grams per mole. Nitrogen gas exists as a diatomic molecule: N2.
1 mole of helium equal 4,002602(2) grams.
Oxygen has a molar mass of 16 grams per mole 1 mole = 16 grams 0.8834 mole = x grams x=14.1344 grams
Two atoms of nitrogen form the gaseous, natural state, of nitrogen. 10.62 grams N2 (1 mole N2/28.02 grams)(6.022 X 10^23/1 mole N2)(1 mole N2 atoms/6.022 X 10^23) 0.3790 mole of gaseous nitrogen atoms ------------------------------------------------------ * as you may see, Avogadro's number is over itself as a form of one and is a superfluous step put there for formality's sake
A mole of gas is equal to 6.02 x 1023 gas molecules.
1.5*10^23
The molecular mass of NH3 is the sum of the atomic mass of nitrogen and three times the atomic mass of hydrogen, or 14.007 + 3(1.008) = 17.031. Therefore, the number of moles of NH3 in 14.0 grams is 14.007/17.031 = 0.822. Since each molecule of N2 supplies two nitrogen atoms and each molecule of NH3 needs only one nitrogen atom, the number of moles of N2 needed is half the number of moles of NH3 formed = 0.411.
28g
The atomic number of nitrogen (N) is 7. The atomic weight of N is 14.00674 grams per mole. Nitrogen gas exists as a diatomic molecule: N2.
Balanced equation. 4Na + O2 ->2Na2O 14.6 grams Na (1 mole Na/22.99 grams)(1 mole O2/4 mole Na)(32.0 grams/1 mole O2) = 5.08 grams oxygen gas needed --------------------------------------------
38 grams oxygen gas (1 mole O2/32 grams)(6.022 X 1023/1 mole O2) = 7.2 X 1023 molecules of oxygen gas -------------------------------------------------