Answer :
Answer: 30.27 g
Explanation:
Since the silver is in units of atoms, we have to use Avogadro's number and molar mass.
Avogadro's number: 6.022×10²³ atoms/mol
Molar Mass: 107.87 g/mol
Now that we have everything we need, we can convert atoms to grams.
[tex]1.69*10^{23} atoms*\frac{1mol}{6.022*10^2^3 atoms} *\frac{107.87 g}{1mol} =30.27 g[/tex]
Final answer:
The mass of a silver coin containing 1.69x10^23 silver atoms can be determined through the use of Avogadro's number and the concept of molar mass. This equates to approximately 30.31 grams of silver.
Explanation:
The subject of this question involves the concept of molar mass in Chemistry.
To find the mass of a silver coin that contains 1.69x10^23 silver atoms, we first need to know the number of atoms in one mole of silver (Avogadro's number), which is 6.022x10^23 atoms/mole.
This means that there are 1.69x10^23 / 6.022x10^23 = 0.281 moles of silver in the coin.
The molar mass of silver is approximately 107.87 grams/mole. Hence, by multiplying the number of moles by the molar mass, we get: 0.281 moles * 107.87 grams/mole = 30.31 grams.
So, the mass of a silver coin with 1.69x10^23 silver atoms is approximately 30.31 grams.
Learn more about Molar Mass here:
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