Answer:
Explanation:
I think you meant a covalent bond, a bond between two non-metal atoms. This image can explain better than I can.
Answer:
2C₂H₆ + [7]O₂ → [4]CO₂ + [6]H₂O
Explanation:
Chemical equation:
C₂H₆ + O₂ → CO₂ + H₂O
Balanced chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Step 1:
2C₂H₆ + O₂ → CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 1
H = 12 H = 2
O = 2 O = 3
Step 2:
2C₂H₆ + O₂ → 4CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 2
O = 2 O = 9
Step 3:
2C₂H₆ + O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 2 O = 14
Step 4:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 14 O = 14
Answer:
It can be removed by acidic chemicals
Explanation:
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 %
