The moles of oxygen gas (O2) that is needed is 4 moles
Explanation
2H2 +O2 → 2H2O
The moles of O2 is determined using the mole ratio of H2:O2
that is from equation above H2:O2 is 2:1
If the moles of H2 is 8 moles therefore the moles of O2
= 8 moles x 1/2 = 4 moles
Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.
P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻
Final reaction :
P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻
A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).
Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.
For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).
Learn more about Half reactions here : brainly.com/question/2491738
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Answer:
i think the answer is yess
Answer:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)
∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
∴ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
∴ Kc = Kc = 1 / PO2∧6
Explanation:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)
∴ O / Al: 0 → +2 ≡ 2e-
Na: +1 → +2
∴ R / H: +1 → 0
2 - Al - 2
2 - Na - 1
8 - O - 8
14 - H - 14
⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
1 - S - 1
4 - O - 4
2 - H - 2
⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
8 - P - 8
12 - O - 12
⇒ Kc = 1 / PO2∧6
Answer: a low 
and low pH.
Explanation:
pH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
Thus as pOH and are inversely related, a solution having higher pOH will have less amount of concentration. And a solution having more pOH will have less pH.
Thus a substance with a high pOH would likely have low concentration and low pH.
