Answer:
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Explanation:
In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.
Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
The half life of carbon-14 is 5700 years. (Half life is the time taken by a radioactive isotope to decay by half of its original mass).
Let A₀ be the initial amount of carbon-14 that is found in living matter (t=0 years), to determine when there was 44.5% of A₀ left.
44.5 = 100 × (1/2)^n, where n is the number of half lives
0.5^n = 0.445
n = log 0.445/log 0.5
n = 1.168
But 1 half life is 5700 years
Therefore, the number of years will be 5700 × 1.168 = 6658.299725 years
≈ 6658.30 years
MgCl2 would be magnesium chloride.
Answer:
NaNO₃ and AgCl are the two products that can be formed.
Sodium nitrate, an aqueous solution and a solid silver chloride (precipitate)
Explanation:
We determine the dissociation of both salts
AgNO₃ (aq) → Ag⁺ (aq) + NO₃⁻ (aq)
NaCl (aq) → Na⁺ (aq) + Cl⁻ (aq)
We make the ionic equation:
Ag⁺ (aq) + NO₃⁻ (aq) + Na⁺ (aq) + Cl⁻ (aq) → NaNO₃(aq) + AgCl (s) ↓
{ 12 mol * 27 g / mol * 0.902 j/gdeg C* ( 658 -72 ) deg C + 12 * 27 * 3.95 }
quantity of energy required to heat 12 mol of aluminium for 72 deg C to its melting point 658 C
so energy applied is 1451 057 . 33 J
