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 and Explanation
The isomer picked is the N-Propylamine.
It has a lone pair of electron available on the electron rich Nitrogen and no formal charge.
Since it will be hard to draw the Lewis structure in this answer format, I'll attach a picture of the Lewis structure to this answer.
The lone pair of electron is shown by the two dots on the Nitrogen atom.
Answer:
It would react better than what with dilute and will furnish more hydrogen gas.
Answer:
Ahahaha no but thank you. Have a nice day luv! :-))
Reaction equation:
K₂CrO₄(aq) + PbCl₂(aq) → 2KCl(aq) + PbCrO₄(s)
Concentration = moles / Liter
moles(K₂CrO₄) = 3 x 0.025
= 0.075
By the equation, we see that the molar ratio of
K₂CrO₄ : PbCrO₄ is
1 : 1
moles(PbCrO₄) = 0.075
mass = moles x Mr
mass = 0.075 x (207 + 52 + 16 x 4)
mass = 24.2 grams
