how many molecules are in 6.37 mol of cabon monoxide?
930413
6.37 moles of any molecular substance just contains 6.37 times Avogadro's number of molecules, because a mole of matter always contains a mole of particles, that is, Avogadro's number many.  6.37*6.022 x 10^23 = 3.84 x 10^24.
445028
Carbondioxide gas, Oxygen gas and Nitrogen gas
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
769316
6.02 x 10^23 of NO2 molecules make 1 mol.\nThus\n4.65 x10^24 of NO2 molecules\n= 1/(6.02 x10^23) x 4.65x10^24\n= 7.724252492 mol of NO2.
306971
Cr has 6 electrons\nEach Benzene contributes 6, so that's \n\n6 + 6(2) = 18. CrPh2 does\n\nFor Co(CO)4-\n\nCo provides 9 (uncharged)\nNegative charge provides 1\nCO (carbonyl) provides 2 each, so that's 8\n\n9 + 1 + 8 = 18 electrons\n\nBoth organometallics obey the 18 electron rule.
174643
The formula in the above answer is incorrect since it omits the aromatic ring and the 9 subsequent repeating units (it's not 9 repeating -OH groups, that makes no sense).\n\nThe empirical formula by my calculations seems to be: C33 H60 O10
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
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.
322620
This is simple, since O=C=O (Carbon dioxide) includes the molecule C==O (carbon monoxide) and 28 gms of CO equals 1 mole, you'd need 1 mole of CO2 to produce 1 mole of CO.\n\nNow, if you want to take this to the next level, you'd need 44 gms of CO2 to equal one mole.
844628
a) 53.4\n34gCO  x 1mol CO/28gCO x 2molCO2/2molCO x44gCO2/1molCO2 = 53.4gCO2\nThe 2 moles CO cancel the 2 moles CO2 (from the balanced equation).
911009
C6H6O + 7O2 yields 3H2O + 6CO2
621030
see if this page helps:\n\nhttp://www.chemsoc.org/visElements/pages/data/intro_groupi_data.html
865243
C3H6 +  9/2 O2 ---> 3 CO2 + 3 H2O\nassumed oxygen is ideal gas, so\nPV=nRT\nR = 82.057 cm3.atm/gmol.K\nV = 15 l = 15000 cm3 ; T = 37 C = 310 K\nn (mol) O2 = PV/RT = 2.4*15000/82.057*310 = 1.415 gmol\nfrom balanced reaction CO2 = 3/(9/2) * 1.415 gmol = 0.943 gmol\n                                         = 0.943 * 44 = 41.507 gram
768474
3.24 kmol= 3.24*10^3mol=3.24*10^3*261.992 [molecular wt of the compd.] g=848854.08 g
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)