The answer is 144,007 g for the anhydrous aluminium chloride.
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First, we calculate the molar mass of MgCl2, which is 95.21 g/mol. Then, we divide the given mass by the molar mass to determine the number of moles of MgCl2. So, 105 g of MgCl2 contains approximately 1.10 moles of MgCl2.
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
The total mass of the solution is 105 grams, which is the sum of the mass of the salt (5 grams) and the mass of the water (100 grams). The mass of the solute (salt) and the solvent (water) are additive in a solution.
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
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
105 ounces of water weigh approximately 6.56 pounds.
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.