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Phoenix [80]
3 years ago
7

In the Fe/Cu Lab for every 1 mole of Fe reacted, 1 mole of Cu is made, therefor, for every 1 gram of Fe used, 1 gram of Cu is ma

de". Support or refute this statement with SPECIFIC information from your lab/discussion.
Chemistry
1 answer:
Mandarinka [93]3 years ago
3 0

Answer:

Explanation:

Since 1 mole of Fe reacted and 1 mole of Cu is produced.

Mass = molar mass × number of moles

Molar mass of Fe = 56 g/mol

Molar mass of Cu = 63.5 g/mol

Mass of Fe = 56 × 1

= 56 g

Mass of Cu = 63.5 × 1

= 63.5 g

Therefore, the statement is not correct.

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

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

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The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

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Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

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

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

Substitute the value of [NOBr_{2}] in equation (1), we get:

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

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

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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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