Od. 0-0 does not pair of the Molecules bid it’s polar bond due to certain pairs of molecules not being present within its presentations
The correct equation
for the overall reaction can simply be obtained by adding the two separate
equations together. Now when you add the two equations together, the overall K can
be calculated by multiplying the individual K values. Therefore:<span>
K(overall) = K1 * K2 </span>
K(overall) = (1.6 x
10^-10) * (1.5 x 10^7)
<span>K(overall) = 2.4 x
10^-3</span>
Answer:
X = Water (H2O) ; Y = Hydrogen ; Z = Oxygen
Explanation:
2(H2O) -------> 2H2 + O2
We are given with
M1 = 1.00 M
M2 = 0.300 M
V2 = 2.00 L
We are asked to get V1
Using material balance
M1 V1 = M2 V2
Substituting the given values
1.00 V1 = 0.300 M (2.00 L)
V1 = 0.600 L or 600 mL
THe volume needed is 600 mL<span />
Explanation:
Reaction:
Cu + 2AgC₂H₃O₂ → Cu(C₂H₃O₂)₂ + 2Ag
The problem is to split the reaction into oxidation and reduction halves:
The oxidation half is the sub-reaction that undergoes oxidation
The reduction half is the one that undergoes reduction:
The ionic equation:
Cu + 2Ag⁺ + 2C₂H₃O₂⁻ → Cu²⁺ + 2C₂H₃O₂⁻ + 2Ag
Oxidation half:
Cu → Cu²⁺ + 2e⁻
Reduction half:
2Ag⁺ + 2e⁻ → 2Ag
C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.
learn more:
Oxidation state brainly.com/question/10017129
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