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How many atoms are in 41.3 grams of Ag?

User JimmidyJoo
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2 Answers

6 votes
Answer:

number of atoms = 2.31 × 10²³ atoms

Step-by-step explanation:

To calculate the number of atoms in 41.3 grams of Ag (silver), we need to use the Avogadro's number (6.022 × 10²³ atoms/mol).

First, we need to find the number of moles of Ag in 41.3 grams:

moles of Ag = mass of Ag / molar mass of Ag
moles of Ag = 41.3 g / 107.87 g/mol
moles of Ag = 0.383 mol

Next, we can use the Avogadro's number to calculate the number of atoms:

number of atoms = moles of Ag x Avogadro's number

number of atoms = 0.383 mol x 6.022 × 10²³ atoms/mol

number of atoms = 2.31 × 10²³ atoms

Therefore, there are 2.31 × 10²³ atoms in 41.3 grams of Ag.
User Ssindelar
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8.0k points
3 votes

Step-by-step explanation:

41.3 × (6.022 × 10²³)

248.7 × 10²³

2.487 × 10²⁵

User LiquidDeath
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