9.675 g
Since oxygen has an average atomic weight of 15.999 g/mol that would make it 0.21 moles of oxygen.
Ethanol has one atom of oxygen per molecule so that means 0.21 moles of ethanol.
Since ethanol has a molecular weight of 46.07 g/mole, 0.21 moles of ethanol would have a mass of 9.675 g.
Ethanol has the formula C2H6O and has a molar mass of 24+6+16 = 46 g/mol. If there are 3.36 g of O, then first convert that to moles: moles O = 3.36g x 1mol/16g = 0.21 moles O. Since each mole CH3CH2OH contains 1 mole O, there will be 0.21 moles CH3CH2OH. Mass of ethanol will thus be
0.21 moles ethanol x 46 g/mole ethanol = 9.66 grams
The answer is 9,672 g ethanol.
6.09 grams, approx.
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1
it has 6 gram of copper
Because the formula shows that each molecule of ethanol contains 2 atoms of carbon, 6* atoms of hydrogen, and 1 atom of oxygen, the gram molecular mass of one mole of ethanol is the sum of twice the gram atomic mass of carbon, 6 times the gram atomic mass of hydrogen, and the atomic mass of oxygen: 46.07. Therefore, 39.2 grams of ethanol constitutes 39.2/46.07 or 0.851 mole, to the justified number of significant digits. ________________ The subscript 5 immediately after the first appearance of the atomic symbol for hydrogen in the formula, plus the implied subscript 1 of the second appearance of the atomic symbol for hydrogen in the formula.
0.242 t of oxygen; but the value is only an approximate because the pitchblende is not only pure UO2.
23
2
150 (50 x 3)
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If your asking whats the lipids test then it's: Heating a sample with ethanol, if it's cloudy then it contains high levels of lipids.
The pressure is 103,5 at.
1
This mass of barium is 32,33 g.
5
it has 6 gram of copper
Oxygen is just O and water is H2O
Because the formula shows that each molecule of ethanol contains 2 atoms of carbon, 6* atoms of hydrogen, and 1 atom of oxygen, the gram molecular mass of one mole of ethanol is the sum of twice the gram atomic mass of carbon, 6 times the gram atomic mass of hydrogen, and the atomic mass of oxygen: 46.07. Therefore, 39.2 grams of ethanol constitutes 39.2/46.07 or 0.851 mole, to the justified number of significant digits. ________________ The subscript 5 immediately after the first appearance of the atomic symbol for hydrogen in the formula, plus the implied subscript 1 of the second appearance of the atomic symbol for hydrogen in the formula.