Question

Which equivalence factor set should you use to convert 2.68 x 1011 atoms of Ag to grams of Ag?

(2.68 x 1011 atoms Ag/1 mol Ag)(1 mol Ag/107.88 g Ag)
(1 mol Ag/2.68 x 1011 atoms Ag)(107.88 g Ag/1 mol Ag)
(1 mol Ag/6.02 x 1023 atoms Ag)(107.88 g Ag/1 mol Ag)
(2.68 x 1011 atoms Ag/6.02 x 1023 atoms Ag)(107.88 g Ag)

Answer 1

Take the # of atoms divide by Avagadaro's Number Multiply by the molecular weight

`2.68x10 x^(11) / 6.02x10 x^(23) = 4.45x10 x^(-13) mol \\ 4.45x10 x^(-13) * 107.86 = 4.79x10 x^(-11)`
So D in this case

Answer 2

Answer:
`(2.68* 10^(11)atoms Ag/6.02* 10^(23)atoms Ag)(107.88 g Ag)`

Step-by-step explanation:

According to Avogadro's law, 1 mole of every substance contains avogadro's number
`(6.023* 10^(23))` of particles.

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given atoms}}{\text {Avogadro's number}}`

For Ag

given atoms =
`2.68* 10^(11)`

avogadro's number =
`6.023* 10^(23)`

Putting values in above equation, we get:

`\text{Moles of}Ag=(2.68* 10^(11))/(6.023* 10^(23))`

1 mole of Ag weighs = 107.88 grams

Thus
`(2.68* 10^(11))/(6.023* 10^(23))` moles of Ag will weigh=
`(107.88)/(1)* (2.68* 10^(11))/(6.023* 10^(23))=43.2* 10^(-12)` grams.