9.982 is the answer because you take the given mole and multiply it by the mass of N. So it would be 0.713 mol x 14.00 = 9.982 g
4.27562e22
5.418E23 molecules
There are 6.02 x 1023 NH3 molecules in 1 mole of NH3.Therefore 0.850mol of ammonia would have 0.85 x 6.02 x 1023 molecules of NH3.In each NH3 molecule there are 4 atoms (one N and three H atoms).Therefore the number of atoms in 0.850mol of NH3 is4 x 0.85 x 6.02 x 1023 = 2.05 x 1024
18 mol of NH3 (ammonia) contains 18 mol of N atoms. 1 mole of N2 (nitrogen gas) contains 2 mol N atoms. so 9 mol N2 is used to produce 18 mol NH3.
There are 6.022x10^23 molecules in 1.00 mol of anything.
1 mol of any substance contains 6.02 x 1023 constituent particles. This is the avogadro constant. So in 10 moles of NH3, there would be 10 x 6.02 x 1023 = 6.02 x 1024 NH3 molecules.
The number of molecules is 0,391439155705.10e23.
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
5.418E23 molecules
I actually have this problem in my chemistry book too. This is what is given in the "Solutions to Red Exercises" book that went along with the text: Analyze. Given: mol NH3, Find: N atoms Plan. mol NH3---> mol N atoms---> N atoms Solve. 0.410 mol NH3 * 1 mol N atoms/ 1 mol NH3 * 6.022*10^23 atoms/1mol = 2.47*10^23 N atoms Check. (0.4)(6*10^23)=2.4*10^23. I hope this helps.
Ok, so I'm assuming that the chemical formula is written as - 3H2 + N2 ----> 2NH3 2.80 = moles of N2 17.03052 g/mol = Molar mass of NH3 (2.80 mol N2) x (2 NH3) / (1 N2) = 5.6 mol NH3 x (17.03052 g) / (1 mol NH3) = 95.4 g NH3
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
There are 6.02 x 1023 NH3 molecules in 1 mole of NH3.Therefore 0.850mol of ammonia would have 0.85 x 6.02 x 1023 molecules of NH3.In each NH3 molecule there are 4 atoms (one N and three H atoms).Therefore the number of atoms in 0.850mol of NH3 is4 x 0.85 x 6.02 x 1023 = 2.05 x 1024
You need the balanced chemical equation for N2 and H2 combining to form ammonia, NH3.N2 (g) + 3 H2 (g) -----> 2 NH3 (g)Moles NH3 = ( 55.5 g NH3 ) / ( 17.03 g/mol NH3 ) = 3.259 moles of NH3n N2 required = ( 3.259 mol NH3 ) ( 1 N2 mol / 2 NH3 mol ) = 1.629 moles N2m N2 required = ( 1.629 mol N2 ) ( 28.103 g N2 / mol N2 ) = 45.67 g N2 needed
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
Molecular mass = sum of all atoms masses = 1(molN/mol NH3)*14.01(g/mol N) + 3(molH/mol NH3)*1.008(g/mol H) = 17.03 g/mol NH3
molar mass NH3 = 17 g/molmolar mass SF6 = 146 g/molmolecules in 0.55g SF6 = 0.55g x 1mol/146g x 6.02x10^23 molecules/mole = 2.27x10^21 moleculesgrams NH3 needed = 2.27x10^21 molecules x 1mol/6.02x10^23 molecules x 17g/mol = 0.064 grams
18 mol of NH3 (ammonia) contains 18 mol of N atoms. 1 mole of N2 (nitrogen gas) contains 2 mol N atoms. so 9 mol N2 is used to produce 18 mol NH3.