Question #94cff
1 Answer
Explanation:
Since the problem is providing you with the number of molecules of bromine,
By definition, you can't have
So right from the start, the fact that you have
#2.12 * 10^(23)color(white)(.)"molecules Br"_2 " " < " " overbrace(6.022 * 10^(23)color(white)(.)"molecules Br"_2)^(color(blue)("equivalent to 1 mole of Br"_2))#
means that your sample will contain less than
More specifically, the sample will contain
#2.12 * color(red)(cancel(color(black)(10^(23)))) color(red)(cancel(color(black)("molec. Br"_2))) * "1 mole Br"_2/(6.022 * color(red)(cancel(color(black)(10^(23)))) color(red)(cancel(color(black)("molec. Br"_2)))) = color(darkgreen)(ul(color(black)("0.352 moles Br"_2#
The answer is rounded to three sig figs, the number of sig figs you have for the number of molecules persent in the sample.