Ask Question
13 October, 12:58

How many atoms of ca are present in 8.6 mg? the atomic weight of ca is 40.08 amu?

+2
Answers (1)
  1. 13 October, 13:04
    0
    One mole of any element is Avogadro's number, which is 6.02 * 10^23 atoms. One mole of Calcium is equal to its atomic weight in grams, 1 mole Ca = 40.08 x 1000 mg = 6.02 * 10^23 atoms. We have 8.6mg of Ca which means 8.6mg / 40,080mg =.00021457 of a mole of calcium. Now multiply that by Avogadro's number which gives number of atoms = 0.00021457 * 6.02 * 10^23 atoms. So the answer is 1.292 * 10^20 atoms of Ca.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “How many atoms of ca are present in 8.6 mg? the atomic weight of ca is 40.08 amu? ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers