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Ah, what a lovely question! To find the number of molecules in 1 gram of NH3, we first need to know its molar mass, which is approximately 17 grams per mole. Then, we use Avogadro's number, which tells us there are about 6.022 x 10^23 molecules in one mole of a substance. By dividing 1 gram by the molar mass of NH3 and then multiplying by Avogadro's number, we find there are roughly 3.55 x 10^22 molecules in 1 gram of NH3.
What else can I help you with?
How many molecules -not moles- of NH3 are produced from 3.07104 g of H2?
Balanced equation is N2 + 3H2 ==> 2NH33.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 presentmoles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formedmolecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)
How many molecules are in 75g of N2O3?
1 mole N2O3 = 76.0116g N2O3 (atomic weights x subscripts) 1 mole N2O3 = 6.022 x 1023 molecules N2O3 75g N2O3 x (1mol N2O3/76.0116g N2O3) = 0.99mol N2O3 0.99mol N2O3 x (6.022 x 1023 molecules N2O3/1mol N2O3) = 6.0 x 1023 molecules N2O3
What is the mass of 1.75 1024 molecules of NH3?
To find the mass of 1.75 x 10^24 molecules of NH3, you would first calculate the molar mass of NH3 (17.03 g/mol). Then, you would convert the number of molecules to moles by dividing by Avogadro's number (6.022 x 10^23 molecules/mol). Finally, multiply the number of moles by the molar mass to find the mass.
How many molecules (not moles) of NH3 are produced from 5.69104 g of H2?
The reaction would be H2 + 3N2 ==>2NH3moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 producedMolecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
Results for Calculate The Amount In Moles Of Molecules For Each Substance. A) A Sample Of Ammonia NH3(g) Containing 8.1x1020 Atoms Of Hydrogen?
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
How many molecules (not moles) of NH3 are produced from 1.57104 g of H2?
The answer is 1,57.10e27 molecules.
How many grams are their in 3.4x10molecules of NH3?
The mass is 9,6.10e-22 g for 34 molecules.
How many molecules -not moles- of NH3 are produced from 3.07104 g of H2?
Balanced equation is N2 + 3H2 ==> 2NH33.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 presentmoles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formedmolecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)
How many molecules are in 75g of N2O3?
1 mole N2O3 = 76.0116g N2O3 (atomic weights x subscripts) 1 mole N2O3 = 6.022 x 1023 molecules N2O3 75g N2O3 x (1mol N2O3/76.0116g N2O3) = 0.99mol N2O3 0.99mol N2O3 x (6.022 x 1023 molecules N2O3/1mol N2O3) = 6.0 x 1023 molecules N2O3
How many grams does 7.74 x 1025 molecules of ammonia (NH3) represent?
This value is 21,85 g.
What is the mass of 1.75 1024 molecules of NH3?
To find the mass of 1.75 x 10^24 molecules of NH3, you would first calculate the molar mass of NH3 (17.03 g/mol). Then, you would convert the number of molecules to moles by dividing by Avogadro's number (6.022 x 10^23 molecules/mol). Finally, multiply the number of moles by the molar mass to find the mass.
How many molecules (not moles) of NH3 are produced from 5.69104 g of H2?
The reaction would be H2 + 3N2 ==>2NH3moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 producedMolecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
If you have 500 grams of NH3 how many moles to you have Please help and explain how you got the answer?
The molar mass of ammonia is 17,031 g.17,031 g NH3-------------------1 mol500 g NH3-------------------------xx= 500/17,031 = 29,36 moles
How many moles s are present in 1 g of ammonia?
1 g of ammonia (NH3) is equal to 0,059 mol.
HOW many grams pf ammonia are in 387x1021 molecules?
Assuming that this ammonia gas is at STP, you can use Avogadro's number to gind the number of moles of gas:(387 x 1021 molecules) x (1 mol / 6.02x1023particles) x (17.03 g / 1 mol) =110 g NH3
Results for Calculate The Amount In Moles Of Molecules For Each Substance. A) A Sample Of Ammonia NH3(g) Containing 8.1x1020 Atoms Of Hydrogen?
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
How many grams of oxygen are needed to react completely with 200.0 g of ammonia, NH3?
The balanced chemical reaction is: 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O We use the reaction and the amount of the reactant , NH3, to determine the amount of oxygen needed. We do as follows: 200 g NH3 ( 1 mol / 17.04 g) (5 mol O2 / 4 mol NH3) (32 g / 1 mol) = 469.48 g O2 needed
