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:
option B
cementation and compaction
Answer:
We say that the solution is unsaturated.
Explanation:
If the salt solubility is 36 g in 0.1 L of water then we can dissolve 360 g of salt in 1 L of water.
Because the solution contains 200 g of salt in 1 L of water, the solution is unsaturated because more salt can be added until we reach the saturation point.
We call the solution dilute when we compare the concentration of a solution with the concentration of another solution, but here we do not compare different solutions.
Nitrogen is the N and that is what is the element is the reactant in excess.
