Determine the number of atoms in each sample.

a.) 98.3 g mercury, hk

c.) 10.7 g lithium, li

determine the number of moles in each sample.

a.) 6.84g sucrose c12h22)11

c.) 68.0g ammonia, nh3

determine the mass of the following molar quantity.

a.) 3.52 mol si

c.) 0.550 mol f2

chemistry, year1, honors. can *someone* me?

So there are three formulas that you must be aware of hen attempting these questions (by technicality, only two because the third is a transposition of the second).

These formulas are :

   mole x L =  # of atoms   (where L = Avogadro's Constant)

   mass / Mr  = mole  (where Mr = Molar Mass)

   Mole  x  Mr  = mass 

So no we can easily ace these questions

1)  a.  98.3 g  of  Hg

             mass of Hg  =  98.3g

             Mr      "   "    =  201 g / mol

                   mole = 

                           =  0.489 mol

       ∴ # of atoms  = 0.489 mol  *  (6.03 * 10²³)

                             = 2.95  *  10²³  atoms

      c.  10.7g of Li

            mass of Li  =  10.7 g

            Mr      "   "  =   7 g / mol

               thus  mole = 

                                 = 1.529 mol

           # of atoms  =  1.529 mol  *  (6.03 * 10²³)

                              =   9.22  *  10²³

2)  a.   6.84 g   of    C₁₂H₂₂O₁₁

          mass of sucrose  =  6.84 g

          Mr      "     "  =  ∑ (Mr of each element  *  amount of each element's atom)

                              =  ∑ [(12 * 12)+(1 * 12)+(16 * 11)]  g / mol

                              =  342 g / mol

          ∴   mole of sucrose  = 

                                           =  0.02 mol

    c)  68.0g    of  NH₃

               mass of NH₃   =   68.0 g

               Mr       "   "       =   [(14 * 1) + ( 1 * 3)] g / mol

         ∴  mole  of  NH₃    =  

                                       =    4  mol

3)     a.   3.52 mol of Si

                   mole of Si  =  3.52 mol

                   Mr     "   "   =  28 g / mol

               Mass =  3.52 mol  *  28 g / mol

                        =   98.56 g

        c.   0.550  mol  of F₂

              mol  of F₂   =  0.55 mol

               Mr   "  "     =  (19  *  2 )  g / mol

          Mass =  0.55 mol  *  38 g / mol

                   =  20.9 g

Hope I as clear in my answers and that they helped