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Q: What is the mass of 0.50 mol of aluminum foil?
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What is the mass of 7050 mol of uranium?

The mass of 7 050 moles of natural uranium is 238,02891 x 7 050.


How many grams of ammonia are presented in 5.0 l of a 0.050 m solution?

4.25 grams. .050 M = .050 mol/1 L 5.0 L x .050 mol/L (cancel out L to get mol as a unit)= .25 mol Atomic mass of Ammonia (NH3)= 17 g/mol .25 mol x 17 g/mol (cancel out mol to get g as a unit)= 4.25 g


Calculate the molecuar mass of alluminium sulphate?

The atomic mass of Aluminum Sulfide is 150.158 g/mol


What is the molar mass of aluminum nitride?

40.9882 g/mol


How do you calculate the mass of aluminum oxide produced when 3.75 moles of aluminum burn in oxygen?

Aluminum Oxide is Al2O3 and Al = 27 and oxygen =16 so the molar mass is 102 g/mol Al2O33.75 mol Al ~ 3.75/2 mol Al2O3 ~ (3.75/2)mol * 102 g/mol = 191.25 = 191 gram Al2O3


What is the chemical composition of aluminum?

Molar Mass of Al: 2(27.0g/mol) = 54.0g/mol Molar Mass of O: 3(16.0g/mol) = 48.0g/mol Molar Mass of compound: 102.0g.mol (54.0g/mol / 102.0g/mol) x 100% = 52.9%


What percentage composition of Al2O3 is aluminum?

Molar Mass of Al: 2(27.0g/mol) = 54.0g/mol Molar Mass of O: 3(16.0g/mol) = 48.0g/mol Molar Mass of compound: 102.0g.mol (54.0g/mol / 102.0g/mol) x 100% = 52.9%


What is the mass in grams for 4.5 moles of aluminum?

2.38 mol Al


What is atomic mass of aluminium sulphide?

The atomic mass of Aluminum Sulfide is 150.158 g/mol


What is the mass of four moles of aluminium?

For this you need the atomic mass of Al. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.2.00 moles Al × (27.0 grams) = 54.0 grams Al


Which best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?

Each compound has a different molar mass.


How many moles aluminum exist in 100.0 grams of aluminum?

The formular you need is M = n/m (molar mass = amount of substance/mass), or n = m/MAluminium has a molar mass of 26.982 g/mol. The given mass is 3 grams. Therefore:n(Al) = 3g / (26.982g/mol) = 0.11mol