B, the ice blocks demonstrated convection :)
Answer:
reee, here is your answer.
Explanation: CH3OH(l) + 3O2(g) rightarrow CO2(g) + 3H2O(g) CH3OH(l) + O2(g) rightarrow CO2(g) + 2H20(g) CH3OH(l) + 2O2(g) rightarrow 2CO2(g) + 4H20(g) 2CH3OH(l) + 3O2(g) rightarrow 2CO2(g) + 4H20(g) Correct Calculate Delta H degree rxn at 25 degree C.
Answer is: 3,94 of hydrogen gas.
Chemical reaction: 2K + 2HBr → 2KBr + H₂.
n(K) = 9,87 mol.
n(H₂) = ?.
from reaction: n(K) : n(H₂) = 2 : 1.
9,87 mol : n(H₂) = 2 : 1
n(H₂) = 4,935 mol for 100% yield of reaction
n(H₂) = 4,935 · 0,798 = 3,94 mol for 78,9 % yield of reaction.
n - amount of substance
This is an oxidation reaction. The balanced equation is as follows:
6H₂O + N₂ → 2NO₃⁻+ 12H⁺ + 6e⁻
Rules to balance redox reaction in acidic medium
- Write the given equation in ionic form
- Identify elements undergoing oxidation ( charge increase, O.N inc) and reduction (charge dec, O.N dec)
- Break the equation into two halfs
- Balance the half equations
A. Balance all other atoms except Oxygen and hydrogen
B. Balance oxygen by adding H2O to the side deficient in oxygen
C. Balance hydrogen by adding H+ ions
D. Balance charge by adding electrons
5. Add the two half such that electrons gets cancelled
Oxidation number of N in N2 is 0 while in NO₃⁻, it is +5. Thus there is an increase in oxidation number, thus oxidation is taking place.
N₂(g) → NO₃⁻(aq)
N₂ → 2NO₃⁻
6H₂O + N₂ → 2NO₃⁻
6H₂O + N₂ → 2NO₃⁻+ 12H⁺
- balance charge. 0 charge on left, -6 and + 12 on right. add 6e⁻ on right to balance.
6H₂O + N₂ → 2NO₃⁻+ 12H⁺ + 6e⁻
Thus we can conclude that since there is increase in oxidation number, oxidation is taking place.
learn more about balancing redox reactions at brainly.com/question/10203480
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