A rate is equal to 0.0200M/s. If [A]=0.100 M and rate= k[A]^1, what is the new rate if the concentration of [A] is increased to
0.200M?
1 answer:
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Answer:
There are moles of bactoX present in a 3.0 μL sample volume that contains 7.512×106 bacterial cells
Explanation:
Given that:
The number of molecules present in one bacterial cell is molecules
and the sample contains molecules.
Number of moles = number of molecules /Avogadro's number
where;
Avogadro's number = 6.023 × 10²³
Number of moles =
Number of moles = moles
So; moles is present in one bacteria cell
Similarly; the sample contains molecules.
Therefore; the number of moles present in the bactoX is =
= moles