The molecular formula for ammonia is NH3, showing that each molecule contains four atoms. Therefore, the number of atoms in one mole of ammonia is 4 times Avogadro's Number or about 24.1 X 1024.
There are approximately 0.023 moles of ammonia in 1 g of ammonia (NH3).
Three moles of nitrogen are required to produce 2 moles of ammonia according to the balanced chemical reaction for ammonia synthesis. Therefore, 27 moles of nitrogen are required to produce 18 moles of ammonia.
4.25 grams. .050 M = .050 mol/1 L 5.0 L x .050 mol/L (cancel out L to get mol as a unit)= .25 mol Atomic mass of Ammonia (NH3)= 17 g/mol .25 mol x 17 g/mol (cancel out mol to get g as a unit)= 4.25 g
36.084g/mol is INcorrect1*14.01 + 4*1.008 = 18.042 g/mol for NH4, but this is not an exsiting compound.Ammonia is NH3 (17.034), ammonium is NH4+(18.042).
Ammonia (NH3) has one lone pair of nonbonding electrons on the nitrogen atom.
1 g of ammonia (NH3) is equal to 0,059 mol.
There are approximately 0.023 moles of ammonia in 1 g of ammonia (NH3).
To determine the amount of hydrogen in 150 g of ammonia (NH3), we first need to calculate the molar mass of ammonia. The molar mass of NH3 is approximately 17 g/mol (1 nitrogen atom with a molar mass of 14 g/mol and 3 hydrogen atoms with a molar mass of 1 g/mol each). Next, we find the molar ratio of hydrogen to ammonia, which is 3:1. Therefore, in 150 g of ammonia, there are approximately 33.53 grams of hydrogen (150 g / 17 g/mol * 3 mol H2 / 1 mol NH3).
Ammonia sulfide is (NH4)2SO4. And the molar mass is 116 g mol-1.
Three moles of nitrogen are required to produce 2 moles of ammonia according to the balanced chemical reaction for ammonia synthesis. Therefore, 27 moles of nitrogen are required to produce 18 moles of ammonia.
In ammonia (NH3), the molar mass is 17 g/mol. To find the mass of nitrogen in 125 g of ammonia, first, calculate the number of moles of ammonia in 125 g. Then, multiply the moles of ammonia by the molar ratio of nitrogen in ammonia (1 mol of N for every 1 mol of NH3), and finally, multiply by the molar mass of nitrogen (14 g/mol) to find the mass of nitrogen. This will give the mass of nitrogen in 125 g of ammonia.
The gram molecular mass of ammonia is 17.03. The formula shows that only one atom of nitrogen is required for each mole of ammonia; 4.12 mol of diatomic nitrogen contains 8.24 mol of nitrogen atoms, and with excess H, all of this nitrogen can be converted to ammonia. Therefore, 8.24 mol of ammonia can be produced, and multiplying this number by17.03 yields a total mass of 140.3 grams of ammonia, to the justified number of significant digits.
1 pair
4.25 grams. .050 M = .050 mol/1 L 5.0 L x .050 mol/L (cancel out L to get mol as a unit)= .25 mol Atomic mass of Ammonia (NH3)= 17 g/mol .25 mol x 17 g/mol (cancel out mol to get g as a unit)= 4.25 g
36.084g/mol is INcorrect1*14.01 + 4*1.008 = 18.042 g/mol for NH4, but this is not an exsiting compound.Ammonia is NH3 (17.034), ammonium is NH4+(18.042).
Ammonia (NH3) has one lone pair of nonbonding electrons on the nitrogen atom.
Ammonia is NH3 so 1 nitrogen and 3 hydrogen