Question

Given 88.0 grams of Boron, how many atoms of Boron (B) are present?

Answer 1

If you have 88 grams, you need to first work out how many moles of Boron you have: mass = mr x moles so 88/ 10.8 (the mr of boron) = 8.15 moles. Using avogadros constant you can then do 8.15 x ( 6.02x 10^23 ) to get the answer.
Sorry I don’t have a calculator right now so I can’t do the final calculation.
Let me know if you have any questions

Answer 2

There are 4.9 × 10²⁴ atoms in 88.0 grams of Boron.

HOW TO CALCULATE NUMBER OF ATOMS:

• The number of atoms in a substance can be calculated by multiplying the number of moles in the substance by Avogadro's number (6.02 × 10²³).

• The number of moles in 88.0 grams of Boron can be calculated as follows:

no. of moles = 88/10.81

no. of moles = 8.14 moles

• The number of atoms in Boron is calculated as follows: 8.14 moles × 6.02 × 10²³

= 4.9 × 10²⁴ atoms of Boron.

Therefore, there are 4.9 × 10²⁴ atoms in 88.0 grams of Boron.

Learn more about how to calculate number of atoms at: https://brainly.com/question/4206478