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1. How many cobalt atoms in mole ofCo? 2. How many germanium atoms in one mole o

ID: 808488 • Letter: 1

Question

1. How many cobalt atoms in mole ofCo? 2. How many germanium atoms in one mole of Ge? Given that the atomic mass of Ni is 58.69g /mol and that of Sn is 118.71g/mol, answer the following questions: 3. If you had 5.0g of Ni and 5.0g of Sn, which would contain more atoms? 4. If you had 5.0 mol of Ni and 5.0 mol of Sn, which would have more mass? 5. If you had 5.0 mol ofNi and 5.0 mol of Sn, which would have the most atoms? 6. What is the mass of 0.417 mol ofNi? 7. How many moles of Sn in 4.435g of Sn? 8. How many atoms of Sn in 256.88g of Sn? 9. Calculate the molar mass for each of the following compounds: a. H2S b. Na2O c, Pbl2 d. Ba(OH)2 e. Sn (PO 10. How many molecules of AlzO in 1 mole of AlyO? 11. How many atoms of Al in one molecule of AlzO? 12. How many moles of O in 1 mole of Alzos? 13. How many atoms of Cin 1 mole of C3Hs? 14. Convert the following to moles: 15. How many molecules in 7.862g SO? 16. How many atoms of sulfur in 3.215g of Fe (SO) ?17. How many atoms of iodine in 1.852g of Cla 18. Find the percent, by mass, of each atom in the following compounds: a. Cu b. Na3P c, FeCl d, Ba (NO)

Explanation / Answer

1) 6.023 x 1023 atoms

2) 6.023 x 1023 atoms

3) Ni contain more atoms

4) Sn have more mass

5) same no.of atoms in Ni and Sn

6) 24.47gm

7) 0.0375moles

8) 2.17 x 6.023 x 1023 atoms

9) a. H2S---2+32 = 34

     b.Na2O--46+16=62

     c.PbI2 ----253+207 = 461

      d.Ba(OH)2 --- 171

     e.Sn3(PO4)2 ---- 9133

10) 6.023 x 1023 molecules

11) 2 x 6.023 x 1023 atoms

12) 3 moles

13) 3 x 6.023 x 1023 atoms

14) a. 0.032moles

      b. 0.416 moles

      c. 0.104 moles

      d. 0.681 moles

15) 0.12 x 6.023 x 1023 molecules

16) 3 x 0.08 x 6.023 x 1023 atoms

17) 4 x 0.0035 x 6.023 x 1023 atoms

18) a. CuI ---- Cu% = (Cu mass/ CuI mass) x 100 = (63.5/190) x 100 = 33.42%

                         I % = 100 - 33.42 = 66.57%

     Similarly

      b. Na3P --- Na% = (3 x 23/99.9) x 100 = 69.06%

                          P% = 100 - 69.09 = 30.9%

     c.FeCl3 ---- Fe% = (55/162) x 100 = 33.95%

                         Cl% = 100 - 33.95 = 66.05%

     d. Ba(NO3)2 --- Ba% = (137/261) x 100 = 52.49%

                            NO3% = 100 - 52.49 = 47.5%