The percent composition of each element can be calculated as follows:
% composition = (mass of element / total mass) * 100
The total mass of the quarter is given to be 5.670 grams
Mass of Cu = 5.198 grams
Mass of Ni = 0.472 grams
Substitute in the above equation to get the mass percentage of each element as follows:
% of Cu = (5.198/5.670) * 100 = 91.675%
% of Ni = (0.472/5.670) * 100 = 8.325%
Answer:
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
Explanation:
Given data:
Mass of Fe₂O₃ = 122 Kg ( 122×1000 = 122000 g)
Moles of CO = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of Fe₂O₃:
Number of moles = mass/ molar mass
Number of moles = 122000 g /159.69 g/mol
Number of moles = 764 mol
Now we will compare the moles of Fe₂O₃ with CO.
Fe₂O₃ : CO
1 : 3
764 : 3×764 =2292 mol
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
<h2>Answer:</h2>
True
<h2>Explanation:</h2>
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Answer:
hope it helped you.
Explanation:
The properties of matter that do not depend on the size or quantity of matter in any way are referred to as an intensive property of matter. Temperatures, density, color, melting and boiling point, etc., all are intensive property as they will not change with a change in size or quantity of matter.
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71