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Consider the mechanism. Step 1: A \iff B + C equilibrium Step 2: C + D \to E slow Overall: A + D \to B + E Determine the rate law for the overall reaction, where the overall rate constant is represented as k. rate =

          Consider the mechanism.
Step 1: A \iff B + C equilibrium
Step 2: C + D \to E slow
Overall: A + D \to B + E
Determine the rate law for the overall reaction, where the overall rate constant is represented as k.
rate =
        
Consider the mechanism.
Step 1: A B + C equilibrium
Step 2: C + D →E slow
Overall: A + D →B + E
Determine the rate law for the overall reaction, where the overall rate constant is represented as k.
rate =

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Consider the mechanism. Determine the rate law for the overall reaction, where the overall rate constant is represented as k. rate = Consider the mechanism Learning Step 1: A = B + C Step 2: C + DE Overall: A + DB + E equilibrium slow Determine the rate law for the overall reaction, where the overall rate constant is represented as k. rate =
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Transcript

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00:01 To find out the empirical formula, so we need to first calculate percentage of oxygen present, which will be equals to 100 minus of percentage of nitrogen present plus percentage of hydrogen present.
00:18 Putting the values, it will be 100 minus 42 .41 plus 9 .15, which is equal to 48 .440, which is equal to 48 .4 this is the percentage of the oxygen present.
00:34 Hence we conclude the compositions are nitrogen present is 42 .41 where hydrogen is 9 .15 and oxygen is 48 .44 percent.
00:50 Now we will divide their atomic number.
00:55 They are atomic number.
01:00 Hence, nitrogen will be 42 .41 divided by 14, which is equal to 3 .029.
01:12 Hydrogen is 9 .15 divided by 1, which is 9 .15 and oxygen will be 48 .44 divided 16.
01:26 So it will be 3 .0275.
01:31 Next step we will divide the smallest number, smallest number hence nitrogen value will be divided by 3 .029 divided by 3 .0275 which will be equal to nearly 1.
01:52 This will be nitrogen, hydrogen will be 9.
01:56 5 .15 divided by 3 .0275 that is equal to 3 and oxygen will be 3 .0275 divided by 3 .0275 which is equal to 1.
02:15 Here we got presence of nitrogen is 1, hydrogen is 3, oxygen is 1.
02:22 Hence the empirical formula, empirical formula will be nitrogen 1, hydrogen 3, oxygen 1...
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