Answer:
The mass percentage of bromine in the original compound is 81,12%
Explanation:
<u>Step 1: Calculate moles AgBr</u>
moles AgBr = mass AgBr / molar mass AgBr
= 0.8878 g / 187.77 g/mol
= 0.00472812 moles AgBr
⇒
Since 1 mol AgBr contains 1 mol Br-
Then the amount of moles Br- in the original sample must also have been 0.00472812 moles
<u>Step 2:</u> Calculating mass Br-
mass Br- = molar mass Br x moles Br-
= 79.904 g/mol x 0.00472812 mol
= 0.377796 g Br-
⇒
There were 0.377796 g Br- in the original sample
<u>Step 3:</u> Calculating mass percentage Br-
⇒mass percentage = actual mass Br- / total mass x 100%
% mass Br = 0.377796 g / 0.4657 g x 100 %
= 81.12%
False, it is an example of an Ionic solid
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
Answer:
Mass = 5.33 g
Explanation:
Given data:
Mass of Al = 2.80 g
Mass of Cl₂ = 4.15 g
Theoretical yield of AlCl₃ = ?
Solution:
Chemical equation:
2Al + 3Cl₂ → 2AlCl₃
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 2.80 g/ 27 g/mol
Number of moles = 0.10 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 4.15 g/71 g/mol
Number of moles = 0.06 mol
Now we will compare the moles of AlCl₃ with Al and Cl₂.
Cl₂ : AlCl₃
3 : 2
0.06 : 2/3×0.06 = 0.04
Al : AlCl₃
2 : 2
0.10 : 0.10
Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.
Mass of AlCl₃:Theoretical yield
Mass = number of moles ×molar mass
Mass = 0.04 mol × 133.34 g/mol
Mass = 5.33 g