Tuesday, March 29, 2011

Lesson 28 Bonding

Bonds & Electronegativity


TYPES OF BONDS

  • 3 main types of bonds 
1. Ionic ( metal - nonmetal )
        ~ e- are transferred from metal to non-metal
2. Covalent (nonmetal to nonmetal)
        ~ e- are shared between non-metals
3. Metallic (metal)              
        ~ holds pure metals together by electrostatic attraction

Electronegativity

  • Electronegativity (en) is a measure of an atom's attraction for electrons in a bond 
               Fluorine = 4.0
               Chlorine = 3.0
               Cesium = 0.8

  • Atoms with greater en attract e- more 
  • Polar covalent bonds form from an unequal sharing of e- 
  • Non-polar covalent bonds form from equal sharing 

Bonds 

  • The type of bond formed can be produced by looking at the difference in electronegativity (en) of the elements 
                 en  > 1.7 = ionic bond
                 en  < 1.7 = polar covalent bond
                 en  =  0  is non-polar covalent bond

Example 1
Predict the type of bond ' formed' 
C - C 
2.55 - 2.55 = 0          non-polar covalent bond! 


Pb - B 
2.33 - 2.04 = 0.29     polar covalent bond! 


Ba - F
0.89 - 3.99  =  3.1     ionic bond! 


Identify the positive and negative side of the polar bonds 
B - Cl 
2.04 - 3.18 = 1.14     polar covalent bond!    
B+       Cl-


C - N 
2.55 - 3.04 =  0.49   polar covalent bond!
C+        N-


- kim 

posted by chemistry11 @ 6:38 PM 0 Comments

Monday, March 28, 2011

LESSON 27 DILUTION

Diluting Solutions
  • when 2 solutions are mixed the concentrations changes
  • dilution is process of decreasing the concentration by adding a solvent(usually water)
  • the amount of solute doesn't change
    •  nB   =   nA       (n = the # of moles)         (B - before ,  A - after)
  • since concentration is mol/L you can write:
    • C = n/v
    • n = CV
    • C1V1  =  C2V2
eg. A 45.0 mL of 12.1 M of HCl is diluted to a final volume of 250mL. What is the concentration 
      of the new solution?
C1V1 = C2V2
1. (12.1M)(0.045L)  =  C2 (.250L)

2. (12.1)(0.045L)  =  C2 (.250)     -->   C2 = 2.18 mol/L
           .250                   .250 

eg. 150 mL of 0.266 M  2NaSO4 is added to 525 mL of 0.135 M 2NaSO4. What is the [2NaSO4].

(.150 L  x  0.266 mol)  =  0.0399  mol
                         L           +
(.525 L  x  0.135 mol)  =  0.07087 mol
                         L              0. 1108  mol

 [2NaSO4]  -->   0.1108 mol   =  0.16 M
                             .675 L

(150ml+525ml = 675ml)

eg. H2CO3 has a concentration of 16.5 M. How much concentrated do you need to make 2.00 L of 2.00M H2CO3? 
C1V1 = C2V2

(16.5)V1 =  (2.00)(2.00)    --->   V1 = 0.242 L
   16.5              16.5
 -sherilyn





posted by chemistry11 @ 11:20 PM 0 Comments

LESSSON 24 SOLUTION STOICHIOMETRY

eg. 250 mL of .300M chromium (II) fluoride reacts with excess zinc. How many grams of chromium are produced?
CrF2  +  Zn  -->  ZnF2  +  Cr

.250L  x  .300 mol  x 1  x  52.0g  =    3.9 g
                    L            1       mol

- how many moles of zinc flouride are produced?
.250L  x .300 mol  x  1  =    0.075 mol      
                    L            1

determine [ZnF2]  0.075 mol   x  1      =     0.3 M
                                                  0.250L

eg. A beaker contains 150 mL of 2.5 M HBr. Extra cadmium is added to the beaker.
Determine how many litres of hydrogen gas should be produced @ STP.

HBr  +  Cd   -->   H2   +   CdBr2

0.150 L  x  2.5 mol  x  1   x  22.4 L  =   4.2 L
                       L           2        mol
- if 2.50 L of hydrogen gas is actually produced what is the % yield?
                3.5 L  x  100  =  83%
                4.2 L

eg. A 4.50g piece of chromium is added to a beaker containing 300mL of 0.800 M RbNO3.
Determine L.R
Cr  +  2RbNO3   -->   2Rb  +   Cr(NO3)2

.300L  x  .800 mol  x  1   x   52.0g   =   6.34 g of Cr   (excess) 
                   L              2
                                                                 L.R =  RbNO3

-sherilyn

posted by chemistry11 @ 10:34 PM 0 Comments

Wednesday, March 16, 2011

LESSON: 26 Titration


  •        A titration is an experiment technique used to determine of the concentration of an unknown solution

    TERMS AND EQUIPMENT
  •   Buret: contains the known solution. Used to measure how much is added.
  •   Stopcock: Valve used to control the flow of solution from the buret
  •   Pipet: used to accurately measure the volume of unknown solution
  •   Erlenmeyer flask: Container for unknown solution.
  •  Indicator: used to identify the end point of the titration
  •   Stock Solution: Known solution
Pipet


Example: Mike the mighty chemist wants to determine the concentration of a Sodium Hydroxide sample so he does a titration with HCl. He gathers the follow data. Use the information to determine [NaOH].  
NaOH samples= 25.00mL          [HCl]= 0.85M

Average Value = 11.75 mL

NaOH + HCl® HOH + NaCl
0.85 mol x 0.01175L x 1 = 0.00998 = 0.0010 mol
       L                               1

0.00998 mol x   1       = 0.3995 = 0.40 M
                        0.025        


~Kelly

posted by chemistry11 @ 5:08 PM 0 Comments

LESSON: 25 Solution Stoichiometry

Example: 200mL of .362M Iron(II) Chloride reacts with excess copper. How many grams of Iron are produced?
·         Balanced Equation: FeCl2 + Cu ®CuCl2 + Fe
                           0.362 mol x .200L x 1 x 55.8g   = 4.04 g
                                   L                          1    1 mol
·         How many mole of Copper(II) chloride are produced?
             ­0.362 mol  x  0.200 x  1   = 0.0724 mol
                     L                             1
·         Determine [CuCl2]
             0.0724 mol x    1 = 0.362 M
                                    0.200L

Example: A 0.666 M solution of Ba(OH)2 reacts with 32mL of 1.25M HCl. What volume of Ba(OH)2 is required for a complete reaction?
·         Balanced Equation:  2HCl + Ba(OH)2 ® BaCl2 + 2HOH
·         1.25 mol x 0.032L x 1  = 0.020 mol
     L                            2
              1 L x 0.020 mol = 0.030 L
0.666 mol


~Kelly  

posted by chemistry11 @ 4:26 PM 0 Comments