How many molecules of NH3 are there in 0.70 mol of NH3?
471277
in 1mol of NH3, there will be 6.023 E 23 molecules will be present\nIn 0.7 mols 6.023 X 0.7 =4.21 E23 molecules will be there\n\nIs it clear?
773122
n(H2O) = m/M = 8.02 / 18.02 = 0.4451 mols\n\nno. of molecules = n(H2O) x Avogadro's Number = 0.4451 x (6.02 x 10^23) = 2.68 x 10^23\n\n1 molecule of H20 = 2 atoms of H and 1 atom of 0\n\nno. of atoms of H = 2 x no. of molecules of H2O = 2 x (2.68 x 10^23) = 5.36 x 10^23
959309
#3:   113.6g ammonia is the maxium that can be produced, i think.\n\nBecause:   when you set up the problem you do:   5 moles N2 x 2 moles of NH3 (coming from the balanced equation) x 17.04g NH3 (formula mass) equaling.....170.4g NH3\n\nthen....\n\nyou have to do it again with the other number(10 moles of H2)\n\n10 moles H2 x 2 moles NH3 / 3 moles H2 x 17.04 equaling....113.6.   \n\nyou pick the lowest number becuase that is the maxium you can have.   so your answer is 113.6g NH3.\n\n\n\n#4:   15.33g NO2 is the maxium that can be produced \n\nbecause: 20g of O2 to grams of NO2.  To do that, you do...20g O2 / 32g O2 (Formula mass of O2) x 46.0g NO2 (formula mass of NO2)\nwhich equals......28.76\n\nthen do:\n\n10gNO / 30.0g NO (formula mass of NO) x 46.01(formula mass of NO2) and that equals....15.33\n\nso your answer is the lowest number which is 15.33.\n\n\ni hope that helps and i hope i'm right...
260523
well you convert using Avogadros number. so if you have 1.26mol, you have 1.26 * Na molecules. and Na is 6.02 x 10^23. so itz about 7.58*10^23. same thing for 2. but its about 3.37 * 10^23 molecules. \nfor part c and d, you have to convert the values to mols first. in order to do that, you have to divide the answers by their molar masses. CH4 molar mass is about 16 i think, so 16g/16g/mol = 1mol which is 6.02*10^23 atoms.\npart d 1000 grams divided by molar mass and multiply by avogadros number.\nhope this helps. let me know if u need more help or whatever. laterz
782126
The molecular mass of Iodomethane (CH3I) is 141.94 g/mol.\n\n1g/141.94g/mol = 0.007 moles.\n\n.007 x 6.02 x 10^23 = 4.21 x 10^21 molecules.
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
553883
27.8 g of ammonium nitrate = 0.348 moles (MW = 80)\n\nAssuming you mean molality and not molarity:\n\nkg solvent = 0.348 moles/0.452 m = 0.770 kg or 770 g
693257
From your reaction equation, 3 moles of hydrogen gas produces 2 moles of ammonia if the reaction was 100% efficient.  From the way you phrased your question, it's difficult to tell if you mean 450g of hydrogen (H), or 450 grams of hydrogen gas (H2).  I'll assume you mean hydrogen gas.\n\nFirst you have to calculate the number of moles hydrogen gas you started with:\n\nMoles H2 = grams H2 / Molecular weight H2\nMoles H2= 450 g H2 / 2.016 g per mole H2\nMoles H2 =223.21 moles\n\nIf the reaction was 100% efficient:\nMoles NH3 = 2/3 * moles H2\nMoles NH3 = 2/3 * 223.21\nMoles NH3 = 148.80 moles \n\nNow to calculate the grams NH3 expected from 100% efficiency:\ngrams NH3 = moles NH3 at 100% efficiency * mol wt NH3\ngrams NH3 = 148.80 moles * 17.024 grams/mole\ngrams NH3 = 2533.17 grams\n\nSo to calculate the percent yield for the reaction:\n\n% yield = (actual yield/ theoretical yield) *100\n% yield = (1575 grams/ 2533.17 grams) *100\n% yield = 62.17%\n\nNow if you meant H instead of H2 for the original grams hydrogen in your question, you'll have to recalculate as appropriate.
960538
90.12g of H2O are produced. you have to first convert to molar mass, then use the mole ratio of the balanced equation for NH3 and H20, then use molar mass again.
842680
2 KBrO3 + 3 N2H4 -> 2 KBr + 3 N2 + 6 H2O\n\nYou can check by finding out the total number of atoms of both sides:\n 2 Potassium\n 2 Bromine\n 6 Oxygen\n 6 Nitrogen\n12 Hydrogen
788534
N2 + 3H2 -> 2NH3\n\nN2 has a gfm of 28g/mol and H2 has a gfm of 2g/mol.  (you do the math to find the moles I will make a rough guess) you've got about .5 mols of N2 and about 1 mole of H2, so H2 is your limiting reagent.  your ratio of H2 to NH3 is 2molNH3/3moleH2 and you multiply that by the one mole of H2.\n\nNH3's gfm is 17g/mol so .667molNH3 x 17g/molNH3 ~ 12g\n\n*edit* make sure you do the math over because I was guessing and estimating where you should not *edit*
424662
the fumes of HCl and NH3 react and form NH4Cl
457002
That may be right, but the website just says that its 7 X 10^27-it doesnt explain how they figured it..\n\nSo lets see if we can figure how they got that number: There are 6 X 10^23 atoms per mole.  Lets say the average atom (most being hydrogen-1, followed by oxygen-16, and carbon-12) has a molar weight of 7.  So, 6 X 10^23 average atoms in the body would weigh 7 grams.  To get 70 kilograms, we have to multiply that by 10,000.  So, 2000 times 6 X 10^23 gets us about 6 X 10^27.  That's pretty close to 7 X 10^27.  I have a feeling that the real average molar weight per atom would be quite a bit smaller than 7, though.
851404
At room temperature and pressure one mole (6 x 10^23 particles/atoms/molecules) of a gas fills 24 dm^3 (Litres).\n\nThis means that in every cubic centimeter (one millilitre) of air there are 2.5 x 10^19 molecules.\n\nIf we forget about the rapid movement and think how many static molecules this means in each dimension we find that each 1-dimensional 1-cm long line contains 2,924,017 molecules (let's say 3 million to keep things simpler).\n\nOver 70% of air is Nitrogen and the length of an N2 molecule is about (keeping things simple) 100 pm (picometers; 10^-12 m) so 3 million molecules take up 300 million picometers or 300 micrometers (following conversion)\n\nThere are 10,000 micrometers in a centimeter so the molecules only 3% of the available distance.\n\nIf all 3 million molecules were equally spaced they would be approximately 3.3 nm (nanometers; 10^-9 m) apart.\n\nA division of the molecular length (100 pm) by the separation (3.3 nm) shows that they are about 33 molecules length apart.\n\nNow I've made plenty of assumptions here. I'm more than happy for people to point out where I've gone wrong (if I have). Please PM me.
805524
Hydrogen and nitrogen react to form ammonia in what is called the Haber process.\n\nThe reaction is as follows:\nN2 + 3 H2 <--> 2 NH3\n\nThis reaction will give about a 85% yield at 200°C (473 °K) and 200 atm of pressure.  This is typically what is done in industry.\n\nWith special considerations to safety, you can get about 100% yield at 200°C and 750 atm of pressure.\n\nThe only remaining question is, how much do you want to make?\nI will assume 100 kg per batch.  You can recalculate for different amounts.\n\nThe molecular weight of ammonia is 17 g/mol, so 100 kg will have 100 kg / 0.017 kg/mol = 5882 mol\n\nTo produce 5882 mol of NH3, you need 2941 mol of N2 and 17646 mol of H2\nSo, the maximum number of moles of gas will be at the beginning of the reaction: 2941 + 17646 mol = 20587 mol\n\nUsing the ideal gas law:\nPV = nRT, where R is the gas constant and n is the number of moles\nV = nRT/P = 20587 mol * .082057 L*atm / mol*K * 473°K / 750 atm = 1065 L = 1.065 m3
937526
If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)\nFill in the atomic masses of H = 1.00797 and N = 14.0067:\n[3 x 2 x (1.00797)] +[ 2 x (14.0067)]  = 2 x (14.0067) x 3 x (1.00797) = 34.06122\nThis is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:\n2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.
15541
Yes.  NH3 obeys the octet rule because nitrogen shares two electrons in each of three covalent bonds to hydrogen, and has a free electron pair. 2*3+2=8.