A) The limiting reactant is Al
b) Br2 is the excess reactant
c) The amount moles of AlBr3 that get produced will be equal to the number of moles of Al to begin with.
d) 0
Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
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A swimming pool has an inlet pipe that can fill the empty pool in 12 hours with the drain is closed and the inlet pipe is closed the drain can empty the full pool in 20 hours. When the pool is empty how long will it take the inlet pipe to fill the pool drain if it is left open ?
Answer:
the answer is a c and e which is kettle moraine and til ur welcome
Explanation: