3.5 x 10^-4 is the answer ! :)
Answer:
All of these oxides of manganese have a mass percent of metal greater than 50%
Explanation:
Step 1: Data given
Molar mass of Mn = 54.94 g/mol
Molar mass of O = 16 g/mol
<em><u>Step 2: Calculate mass % of Mn in MnO</u></em>
Molar mass of MnO = 54.94 + 16 = 70.94 g/mol
% Mn = (54.94/70.94)*100 % = 77.45 %
<u><em>Step 3: Calculate % of Mn in MnO2</em></u>
Molar mass of MnO2 = 54.94 + 2*16 = 86.94 g/mol
% Mn = (54.94/86.94) * 100% = 63.19 %
<u><em>Step 4: Calculate % of Mn in Mn2O3</em></u>
Molar mass of Mn2O3 = 2*54.94 + 3*16 = 157.88 g/mol
% Mn = ((2*54.94)/157.88)*100 % = 69.6 %
All of these oxides of manganese have a mass percent of metal greater than 50%
Answer:
None of the options are correct. The correct answer is:
56.67g
Explanation:
N2 + 3H2 —> 2NH3
Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
From the equation,
6g of H2 produced 34g of NH3.
Therefore, 10g of H2 will produce = (10 x 34)/6 = 56.67g of NH3
Therefore, 56.67g of NH3 are produced
We are given with 7 moles of a compound with a molecular formula of al2(so4)3 and is as asked here to determine the number of moles of Al, S and O. In a mole of aluminum sulfate, there are 2 moles of Al, 3 moles of S and 12 moles of O. Hence in total, there are 14 mol Al, 21 mol S, and 84 mol O.