Answer :
Final answer:
To find how many atoms of Ag are present in 38.4 grams, we first convert the given mass into moles, then we multiply the resultant moles by Avogadro's number. The result is roughly 2.14 x 1023 atoms of Ag.
Explanation:
The question is asking about how many atoms of silver (Ag) are present in 38.4 grams of it. To find this, we need to use Avogadro's number, which states that one mole of any substance contains 6.022 x 1023 atoms. The atomic mass of Ag is approximately 107.87 g/mol. First, we convert the given mass into moles using the equation: moles = given mass / molar mass. Therefore: moles of Ag = 38.4g / 107.87g/mol = 0.356 moles approximately.
Next, to find the atoms, we multiply the moles by Avogadro's number: atoms = moles x Avogadro's Number = 0.356 moles x 6.022 x 1023 atoms/mole = 2.14 x 1023 atoms of Ag approximately.
Learn more about Atomic Composition here:
https://brainly.com/question/30459295
#SPJ11
