Asked by Katie
Given that on average, an atom of gold weighs 3.27×10^−22grams, how much
does an Avagadro’s number of gold atoms weigh (take Avagadro’s number to be 6.0221415×10^23).
does an Avagadro’s number of gold atoms weigh (take Avagadro’s number to be 6.0221415×10^23).
Answers
Answered by
Steve
just multiply the mass of each atom by the number of atoms:
3.27*10^-22 * 6.0221415*10^23 = 19.6924*10^1 = 196.924 g
check: mol wt of Au is 196.97. Pretty close
3.27*10^-22 * 6.0221415*10^23 = 19.6924*10^1 = 196.924 g
check: mol wt of Au is 196.97. Pretty close
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