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nasty-shy [4]
3 years ago
12

In the redox conversion of Ni2+ to NiO4−, the oxidation number of Ni goes from (−2, 0, +2) to (−1, +1, +7, +9). Recall that the

oxidation number of oxygen is typically −2.
1. 0; +9
2. +2; +7
3. +2; −1
4. −2; +9
5. −2; +7
6. 0; +1
7. 0; +7
8. +2; +9
9. −2; +1
Chemistry
1 answer:
inessss [21]3 years ago
7 0
Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero, 
for NiO4-, assuming oxidation number of Ni is x 
 (x ×1) + (-2 × 4) = -1
  x = + 7
Therefore, the oxidation number goes from 0 to +7
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Bond energy of C≡C is 835 kJ/mol

\Delta H= {\text {sum of bond energies of reactants}}-  {\text {sum of bond energies of products}}

\Delta H= {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} -  {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}

=176.6kJmol^{-1}







7 0
3 years ago
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Dmitriy789 [7]
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mol = mass ÷ molar mass

If mass of hydrazine (N₂H₄)  = 5.29 g 
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mole ratio of hydrazine to Nitogen is     1   :  1
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Mass = mol × molar mass

Since mol of nitrogen (N₂)  = 0.165
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3 years ago
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Answer:

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