C6H6 is the molecule of benzene and it's molecular weight is 78, calculated as
6 x 12 +6 x 1=78
and you will understand it if you know some basics of chemistry.
then,
You know that each mole of a molecule weighs exactly those number of grams as much as is its molecular weight. That means if a molecule has 'M' as its molecular weight, then one mole of it weighs exactly M grams when actually weighed using a weighing machine.
This means one mole of benzene weighs 78 grams. then 195 grams of C6H6 has (1/78)x195 moles or 195/78 moles in it.
Molar mass C6H6 = 12.0x6 + 1.00x6 = 72 + 6 = 78 g/mole12.7 moles x 78 g/mole = 990.6 grams = 991 g (to 3 significant figures)
Approx 3.29 moles.
what is ag2o
How many molecules are in 94 grams of sodium fluoride
For this you need the atomic mass of Na. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel.11.5 grams Na / (23.0 grams) = .500 moles Na
10.08 moles (there are six carbon atoms per molecule of C6H6, hence, six times the number of moles of carbon can be produced).
Molar mass C6H6 = 12.0x6 + 1.00x6 = 72 + 6 = 78 g/mole12.7 moles x 78 g/mole = 990.6 grams = 991 g (to 3 significant figures)
0.06667 moles
2 moles of benzene gives 12 moles of hydrogen atoms since benzene is C6H6
By using conversion of units: Benzene = C6H6 4.5 mol C x (1 mol Benzene / 6 mol C) = .75 mol Benzene
2C6H6 + 15O2 ==> 12CO2 + 6H2Omoles C6H6 combusted = 45.0 g x 1 mole/78 g = 0.577 molesmoles CO2 formed = 0.577 moles C6H6 x 12 moles CO2/2 moles C6H6 = 3.46 moles CO2
Atomic mass from my Periodic Table for Zinc is 65.38. This means 1 mole of naturally occurring zinc has a mass of 65.38 grams. If your zinc sample is pure zinc then: (mass of your sample)/(65.38 grams) = # moles of zinc.
Molality = moles of solute / kg of solvent solute = 44.9 grams of C10H8. There are approx 128 g/mole in C10H8. 44.9 grams x 1/128 moles/gram = 0.351 moles (approx) convert 175 grams C6H6 into 0.175 kilograms Molality = 0.351 moles solute (or C10H8) / 0.175 kg solvent (C6H6) Answer is 2.00 Molal
You first need to find the mass weight of NH3. wt. of N + (wt. of H)= 14.0067 + 3(1.0067)= 17.03052 Now that you have the mass weight, you divide 15 into 17.03052... 15.0/17.03052 = 0.8807 moles in 15. g of NH3
Balance this combustion reaction first! 2C4H10 + 13O2 -> 8CO2 + 10H2O 0.86 moles C4H10 (13 moles O2/2 moles C4H10) = 5.6 moles of oxygen required ----------------------------------------
17.2 (g) / 78.11 (g.mol−1) = 0.2202 = 0.220 mole C6H6
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