I don’t, but what are Gizmos?
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: oxygen=2.547g
Explanation:
Based on the question,it was observed that the reaction is reversible
2 moles of KClO3 gives 2 moles of KCl and three moles of O2
Molar mass for KClO3 is 245 g/mol
Molar mass for O2 is 96 g/mol
We are to find the mass of O2 and we Are given the mass KCLO3 is 6.50g
245g of KClO3 gives 96g of O2
6.50g of KClO3 gives xg of O2
Cross multiply
245x=624
X=624/245
X=2.547g
Therefore the gram of oxygen is 2.547g
