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 %
Explanation:
B more negative charges than positive charges
The mass of the Pbl2 : 1308.87
<h3>Further explanation</h3>
Given
18.2 mL of a 0.156 M Pb(NO3)2
26.2 mL of a 0.274 M KI
Reaction
Pb(NO3)2 (aq) + 2 KI (aq) - Pbl2 (s) + 2 KNO3
Required
the mass of the Pbl2
Solution
mol Pb(NO3)2 = 18.2 x 0.156 = 2.8392 mlmol
mol KI = 26.2 x 0.274 =7.1788 mol
Limiting reactant Pb(NO3)2(smaller ratio of mol : reaction coeffiecient)
mol Pbl2 based on limiting reactant (Pb(NO3)2)
From equation, mol ratio of Pb(NO3)2 : Pbl2 = 1 : 1, so mol Pbl2=mol Pb(NO3)2=2.8392
Mass Pb(NO3)2 :
Answer:
Elements can combine in any proportion to form a compound. -third choice
Answer:
7
8
9
10.
better check book pages for this kinda answer.
