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mixer [17]
3 years ago
6

Help me with this please! Do not say "I don't know" to get my points!!!!!!! It is not helpful and disrespectful. THANKS!

Chemistry
1 answer:
MariettaO [177]3 years ago
4 0

no- and na- nano5

na- and na+ nana2

hope this helps

Explanation:

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k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

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\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









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