1.907g of M contains 2.87 x 10^22 atoms of M.
M is a Group 0 element. Caculate the relative atomic mass of M, given that the Avogadro constant = 6.02 x 10^23 mol^ -1
I'm not sure how to tackle this as I haven't done one before - could you please provide the steps in the calculation?
Thanks
2007-11-21 05:26:50 · 4 answers · asked by Ella B 2 in Science & Mathematics ➔ Chemistry
It's so simple here's how:
1.907 g/2.87*10^22molecules * 6.023*10^23molecules/mole =
= 40.02 g/mole
Looks like it a Group II element Calcium
2007-11-21 05:33:17 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
a million) bal. the rxn: 2 P + 3 Cl2 -----> 2 PCl3 2) calc. moles of P: 0.ninety 3 g P/30.ninety seven g/mole P=0.03 3) from bal. rxn., 2 moles of PCl3 are shaped in step with 2 moles of P (a million:a million) 0.03 x 137.33 g/mole PCl3= 4.12 g PCl3
2016-10-17 15:21:13 · answer #2 · answered by goldthorpe 4 · 0⤊ 0⤋
divide your atoms by avagadro to get your mols then divide your grams (1.907) by your mols then you will have your molecular wt of m
here are the numbers
1.906gM/2.87x10^22atomM x 6.02x1023atomM/1moleM = 40.0 g/mol
2007-11-21 05:37:29 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1.907gM/2.87x10^22atomM x 6.02x1023atomM/1molM = 40.0 g/mole
Calcium
2007-11-21 05:36:35 · answer #4 · answered by steve_geo1 7 · 0⤊ 0⤋