Well not calculus because that has nothing, well mostly nothing to do with balancing chemical equation, so B or C. Now for me personally B is way faster, though C is sometimes faster if you get lucky the way to solve it is B
1 mol of CO2 is 44.01g/mol
So multiply that by 2 to get 2 mol of CO2, which is 88.02g
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 %
Argon, it's a noble gas in the last group on the periodic table.
