☃️ Chemical formulae ➝ 
<h3>
<u>How to find?</u></h3>
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
<h3>
<u>Solution:</u></h3>
Atomic weight of elements:
Ca = 40
C = 12
O = 16
❍ Molecular weight of
= 40 + 12 + 3 × 16
= 52 + 48
= 100 g/mol
❍ Given weight: 10 g
Then, no. of moles,
⇛ No. of moles = 10 g / 100 g mol‐¹
⇛ No. of moles = 0.1 moles
☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
<h3>1. 10 e⁻</h3>
Oxidation numbers
I₂O₅(s): I (5+); O(2-)
CO(g): C(2+); O(2-)
I₂(s): I(0)
CO₂(g): C(4+); O(2-)
<h3>2. 4 e⁻</h3>
Oxidation numbers
Hg²⁺(aq): Hg(2+)
N₂H₄(aq): N(2-); H(1+)
Hg(l): Hg(0)
N₂(g): N(0)
H⁺(aq): H(1+)
<h3>3. 6 e⁻</h3>
Oxidation numbers
H₂S(aq): H(1+); S(2-)
H⁺(aq): H(1+)
NO₃⁻(aq): N(5+); O(2-)
S(s): S(0)
NO(g): N(2+); O(2-)
H₂O(l): H(1+); O(2-)
Explanation:
In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.
1. I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)
Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)
Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻
2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)
Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻
Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)
3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)
Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻
Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O
Answer:
Airline IATA ICAO
Aerosul Linhas Aéreas 2S ASO
ASTA Linhas Aéreas 0A SUL
Azul Linhas Aéreas AD AZU
Azul Conecta 2F (AD) OWT
If it is already balanced because the compounds aluminum(Al) and oxygen(O) have the same number of atoms
