I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
Here’s what I got.
For starters, it’s important to note that you’re dealing with sulfur dioxide, ##”SO”_2##, not with “sulfate”.
Now, the problem wants you to find the number of molecules of sulfur dioxide present in ##”6.4 g”## and in ##”640 u”## of sulfur dioxide.
In order to find the number of molecules present in ##”6.4 g”## of sulfur dioxide, you will need to use
So, the first thing to do here is figure out how many moles of sulfur dioxide you have in that ##”6.4-g”## sample. The compound’s molar mass tells you that every mole of sulfur dioxide has a mass of ##”64.064 g”##.
This means that you’ll have
##6.4 color(red)(cancel(color(black)(“g”))) * “1 mole SO”_2/(64.064color(red)(cancel(color(black)(“g”)))) = “0.0999 moles SO”_2##
Now use the fact that
##color(blue)(|bar(ul(color(white)(a/a)”1 mole” = 6.022 * 10^(23)”molecules”color(white)(a/a)|))) ->## Avogadro’s number
to calculate how many molecules you get in that many moles of sulfur dioxide
##0.0999color(red)(cancel(color(black)(“moles SO”_2))) * (6.022 * 10^(23)”molec.”)/(1color(red)(cancel(color(black)(“mole SO”_2)))) = color(green)(|bar(ul(color(white)(a/a)6.0 * 10^(22)”molec.”color(white)(a/a)|)))##
In order to find the number of molecules of sulfur dioxide in ##”640 u”##, you need to use the definition of the unified unit, ##”u”##.
The unified atomic mass unit was defined as the mass of ##1/12″th”## of the mass of an unbound carbon-12 atom in its ground state, and is equivalent to ##”1 g mol”^(-1)##. Keep this in mind for later.
The approximate value of a unified atomic mass unit is
##color(purple)(|bar(ul(color(white)(a/a)color(black)(“1 u” = 1.66054 * 10^(-24)”g”)color(white)(a/a)|)))##
This means that the ##”640 u”## sample will be equivalent to
##640 color(red)(cancel(color(black)(“u”))) * (1.66054 * 10^(-24)”g”)/(1color(red)(cancel(color(black)(“u”)))) = 1.063 * 10^(-21)”g”##
Use sulfur dioxide’s molar mass to determine how many moles would be present in this sample
##1.063 * 10^(-21)color(red)(cancel(color(black)(“g”))) * “1 mole SO”_2/(64.064color(red)(cancel(color(black)(“g”)))) = 1.66 * 10^(-23)”moles SO”_2##
Once again, use Avogadro’s number to find
##1.66 * 10^(-23)color(red)(cancel(color(black)(“moles SO”_2))) * (6.022 * 10^(23)”molec.”)/(1color(red)(cancel(color(black)(“mole SO”_2)))) ~~ color(green)(|bar(ul(color(white)(a/a)”10. molec. SO”_2color(white)(a/a)|)))##
ALTERNATIVE APPROACH
You can get the same result by using the fact that
##color(purple)(|bar(ul(color(white)(a/a)color(black)(“1 u” = “1 g mol”^(-1))color(white)(a/a)|)))##
The unified atomic mass unit tells you the mass of one nucleon, i.e. one proton or one neutron. Take a look at the molar mass of sulfur dioxide, which as you know tells you the mass of one mole of sulfur dioxide.
In essence, you can use the unified atomic mass unit as conversion factor between the mass of a single molecule and the mass of a mole of sulfur dioxide.
##64.064 color(red)(cancel(color(black)(“g mol”^(-1)))) * “1 u”/(1color(red)(cancel(color(black)(“g mol”^(-1))))) = “64.064 u”##
So, if one molecule has a mass of ##”64.064 u”##, it follows that your ##”640 u”## sample will contain
##640 color(red)(cancel(color(black)(“u”))) * (“1 molec. SO”_2)/(64.064color(red)(cancel(color(black)(“u”)))) = 9.99 ~~ color(green)(|bar(ul(color(white)(a/a)”10. molec. SO”_2color(white)(a/a)|)))##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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