Chem 111 and Chem 111N Unit 2 Review Answers

Part A: Multiple choice questions:

1) B                                                         6) A                                        11) E

2) A                                                        7) E

3) A                                                        8) B

4) B                                                         9) C

5) A                                                        10) D

 

Part B: Short answers and calculations:

2) a) CBr4                                               b) SO2                                    

3)             molar mass is 132 g/mole.

a) 30.0 g   x     1 mole                            =  0.227 mole

                                132 g  

b) 0.227 mole   x    6.02 x 10 23 molecules  =  1.37  x   1023  molecules

                                                1 mole

c) 1.37  x   1023  molecules   x    12  H atoms  =  1.64  x  1024   H atoms

                                                                                                      molecule

 

4)  a)  phosphorus tribromide                                             b) diphosphorus pentoxide

5)

6) 9.13  x   10 19 molecules HCl      x      1 mole                                     =   1.52  x  10 –4 moles  HCl

                                                                                6.02 x 10 23 molecules

1.52  x  10 –4 moles  HCl   x    36.46 g   =  5.53 x 10 –3 g

                                                                                1 mole

7) molar mass of sodium sulfite (Na2SO3) is 126.048 g/mole.

     0.0141 mmol  x                  1 mol                        x               126.048g   =   1.78 x 10-3 g

                                        1000 mmol                      1 mole

8) Fat soluble.  There is only one polar group, OH in this molecule with 20 carbons and many C-H bonds.  Overall the molecule will be non-polar therefore will be fat soluble.

 

9) The molar mass of calcium hydroxide is 74.08 g/mole.

                 500 mL   x      1L       x     0.5 moles   =   0.25 moles of Ca(OH)2 needed

                                        1000 mL          L

0.25 moles of Ca(OH)2              x              74.08 g    = 18.52 g of Ca(OH)2 and add

                                                                                1 mole                          enough water to make 500 mL

VM = VM                              (0.5M)(x)   =   (0.1M) (100 mL)  solve for x and get 20 mL.

Take 20 mL of 0.5M solution and dilute to 100 mL.

 

10) H2: covalent bonding – non polar.  The two electrons between hydrogen atoms are shared equally.

HF: polar covalent bonding.  F is very electronegative and pulls electron density towards it, causing it to have partial negative charge.  Hydrogen will have partial positive charge because it is losing electron density to fluorine.  The compound will have hydrogen bonding.

NaF: no sharing of electrons- ionic bonding.  Na has given up an electron to F and “bonding” is due to the charge attraction between the sodium cation and fluoride anion.