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)

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

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kobenhavn kobenhavn · Answer 2 · 2019-07-01T01:24:39-04:00

Answer: \((2.68\times 10^{11}atoms Ag/6.02\times 10^{23}atoms Ag)(107.88 g Ag)\)

Explanation:

According to Avogadro's law, 1 mole of every substance contains avogadro's number \((6.023\times 10^{23})\) 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 = \(2.68\times 10^{11}\)

avogadro's number = \(6.023\times 10^{23}\)

Putting values in above equation, we get:

\(\text{Moles of}Ag=\frac{2.68\times 10^{11}}{6.023\times 10^{23}}\)

1 mole of Ag weighs = 107.88 grams

Thus \(\frac{2.68\times 10^{11}}{6.023\times 10^{23}}\) moles of Ag will weigh=\(\frac{107.88}{1}\times \frac{2.68\times 10^{11}}{6.023\times 10^{23}}=43.2\times 10^{-12}\) grams.