it has 5 electrons in it's outermost shell.
s 2e- , p 5e-
2.4(10^3)
=2.4*10^3
=2.4*(10*10*10)
=2400 <span>milliliters
To </span>centiliters is<span> 2400mL= <u>240.0000cl. </u> </span>
Correct, Was that a question.
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 %
450.0 ML because it has more solution and it would be better if the sciences users it
