Answer:
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Explanation:
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
Answer:
For the first oxide, 1 g gives 0.888 g of copper.
Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.
For the second oxide, 1 g gives 0.798 g of copper.
Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.
So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.
Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.
The law of conservation of mass states that mass can neither be created or destroyed in a chemical reaction.
Measured values are verified and go under a series of experiments before being accepted as the norm. Accepted values are unanimously accepted values, seen as the norm by a qualified group of individuals (scientists or the people of the intellectual sphere).