The answer is 144,007 g for the anhydrous aluminium chloride.
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For this you need the atomic (molecular) mass of C12H22O11. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. C12H22O11= 342 grams.105 moles C12H22O11 × (342 grams)= 35.9 grams C12H22O11
molar mass AlCN3 = 77 g/mole229g x 1 mole/77 g = 2.97 moles If you meant aluminum cyanide, Al(CN)3 then... 229 g x 1 mole/105 g = 2.18 moles
I assume you are talking about the Haber Process:3H2 + N2 ----> 2NH3If 3.71 mol of N2 is produced, there will be 7.42 mol of ammonia produced (as per mol ratios).Using the formula, n = g/mw ---> g = 7.42 x 17.034= 126.39228 grams of NH3
The mass of CO2 is 1 105 g.
Quite a few. 6.32 X 10^25 atoms sodium (1 mole Na/6.022 X 10^23) = 105 moles of sodium
1.54 x 105 mg
1.54 x 105 mg
Density is grams per cm3. 615/105 = 5.86 g/cm3
For this you need the atomic (molecular) mass of C12H22O11. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. C12H22O11= 342 grams.105 moles C12H22O11 × (342 grams)= 35.9 grams C12H22O11
9.375 lbs (on earth?) or mass is4,252.42847 grams
105 pounds is 47,627.2 grams.
The density is about 5.86 g/cm3
105 grams is about 0.231 pounds.
Assuming that the reaction is combustion, the balanced reaction equationwould be2 C4H10 +13 O2 ->8 CO2 +10 H2O. This shows that each mole of O2 would produce 8/13 mole of CO2. 105 grams of O2 corresponds to [105/(2 exact)(15.9994)] or about 3.281 moles of O2. or (8/13)(3.281) or 2.02 moles of carbon dioxide, to the justified number of significant digits.
105 grams = 3.7 oz, approx.
520 kg = 5.20 X 105 grams. The gram atomic mass of helium is 4.0026; therefore the number of moles of helium in 520 kg is (5.20/4.0026) X 105 or 1.30 X 105. Multiplying this number by Avogadro's Number, 6.022 X 1023, yields the number of atoms, which is about 7.82 X 1028, to the justified number of significant digits.
Since 1 dry ounce = approximately 28.4 grams, it would be approximately 298 grams.