Answer:Sea Otter and hark
Explanation:
Two Half Cell reactions are as follow,
3e⁻ + Al⁺³ → Al (Reduction Reaction)
Mg → Mg⁺² + 2e⁻ (Oxidation Reaction)
In order to balance the number of electrons, multiply reduction reaction by 2 and oxidation reaction by 3.
So,
6e⁻ + 2 Al⁺³ → 2 Al
3 Mg → 3 Mg⁺² + 6e⁻ (e⁻ cancelled)
_______________________
2 Al⁺³ + 3 Mg → 2 Al + 3 Mg⁺²
Result:
At Anode Magnesium metal is Oxidized and At Cathode Aluminium Ions are reduced.
Answer:
Explanation:
The reaction is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
We see, that the equation is ballanced.
As we have data from only one reactant, we assume the other is in excess.
Ratio is 1:4 (stoichiometry). We solve this, by a rule of three:
1 mol of propane can produce 4 moles of water
Then, 4 moles of propane, may produce
( 4 . 4 ) /1 = 16 moles of water can be made.
Remember, we always have to work with the limiting reactant.
<span>It throws it off, you have to re-do the experiment. The NaOH M will be higher because you poured to much inside and made it pink!</span>
Answer is: <span>the coefficient of phosphoric acid is 12.
</span>Chemical reaction: P₄S₃ + NO₃⁻ + H⁺ → H₃PO₄ + SO₄⁻ + NO.
Reduction half reaction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O /·38
Oxidation half reaction: P₄S₃ + 28H₂O → 4H₃PO₄ + 3SO₄²⁻ + 44H⁺ + 38e⁻ /·3.
38NO₃⁻ + 152H⁺ + 3P₄S₃ + 84H₂O → 38NO + 76H₂O + 12H₃PO₄ + 9SO₄²⁻ + 132H⁺.
Balnced chemical reaction:
3P₄S₃ + 38NO₃⁻ + 20H⁺ + 8H₂O → 12H₃PO₄ + 9SO₄²⁻ + 38NO.