Suppose a thin sheet of zinc containing 0.2 mol of the metal is completely converted to zinc oxide, #"ZnO"#, in one month. How would you express the rate of conversion of the zinc?
1 Answer
Here's what I got.
Explanation:
Unless the question is incomplete, all you really have to do here is pick a unit of quantity and a unit of time and use the information given to express the rate at which zinc metal,
For starters, you know that
If you were to stick with moles and months, you can say that the rate at which zinc is concerted is
#"rate" = "0.2 mol month"^(-1)#
This means that with every passing month, you will convert
Alternatively, you can pick any other unit of time you want. For example, the same rate can be expressed as
#0.2color(white)(a)"moles"/color(red)(cancel(color(black)("month"))) * (1color(red)(cancel(color(black)("month"))))/"4.35 weeks" = 0.046color(white)(a) "mol"/"week"#
Therefore, you will have
#"rate" = "0.046 mol week"^(-1)#
Another possibility would be to use grams and seconds. Convert moles to grams first by using zinc's molar mass
#0.2 color(red)(cancel(color(black)("moles Zn"))) * "136.3 g"/(1color(red)(cancel(color(black)("mole Zn")))) = "27.26 g"#
You could use seconds instead of months to get
#1 color(red)(cancel(color(black)("month"))) * (4.35 color(red)(cancel(color(black)("weeks"))))/(1color(red)(cancel(color(black)("month")))) * (7color(red)(cancel(color(black)("days"))))/(1color(red)(cancel(color(black)("week")))) * (24color(red)(cancel(color(black)("hours"))))/(1color(red)(cancel(color(black)("day")))) * (60color(red)(cancel(color(black)("min"))))/(1color(red)(cancel(color(black)("h")))) * "60 s"/(1color(red)(cancel(color(black)("h")))) = "2,630,880 s"#
This means that the rate of the conversion can be expressed as
#"rate" = "27.26 g"/"2,630,880 s" ~~ 1 * 10^(-5)"g s"^(-1)#
As you can see, you can basically use any unit of time you want to express the rate of conversion, provided of course that a unit is not specified to you.
