Natalie- Tuesday, 2/22

1) We picked up one sheet: ChemThink Precipitate Lab

2) We went over lab quiz we took on friday

• Molarity= unit of concentration
• Use formula to determine how many mols of solute needed to make 0.15 M of solution
• If you need help with the formula use this video as a reference: http://www.youtube.com/watch?v=8oTqwBAvbnY&feature=fvwrel
• 
  

  .015 mol NaCl	58.44g NaCl
  	1 mol NaCl

  
  
• Fill beaker with 50 mL of water
• Then mix in .88g of Solute (NaCl)
• Then fill rest of the beaker to 100 mL

3) Went through IV. Solubility Notes

• 
• Soluble in water= dissolves in water (aqueous)
• In soluble= does not dissolve in water (solid)
• There are some exceptions though, mentioned on the sheet

4) Went through V. Solubility Practice #1-3:

• 
• If the balanced equation has aqueous on both sides, then no reaction has taken place

5) Homework are questions #1 and #2 on the ChemThink precipitate lab sheet we picked up in

class. Use www.chemthink.com to answer the questions under Solubility

Shayne - Friday, 2/18

I. We picked up two sheets at the beginning of class - (Molarity 2 and Reaction Stability 1)

II. We went through the Molarity 2 notes:

EXAMPLES:

1.) What will be actually present in the solution when each of the following is added to water?

a. KCL ----------> K+ (aq) plus Cl- (aq)
b. NaNO3 ----------> Na+ (aq) plus NO3- (aq)
c. C2H6O ----------> C2H6O (aq)

3.) Calculate the mlarity (M) of the following solutions. (M = Mol/L)

a. M = 0.0346 Mols/0.250 Liters = 0.138M
b. (405 g NiCl2 X 1 mol NiCl2)/129.59 g NiCl2 = 3.12524 mols NiCl2
M = 3.12524 Mols/7.54 Liters = 0.414M
c. (22.57 g H2SO4 X 1 mol H2SO4)/98.09 g H2SO4 = 0.23009 mols H2SO4
100 Ml = 0.1 L so.....
M = 0.23009 Mols/0.1 L = 2.301M

5.) Calculate the mass of the solute present in the following solutions.

a. Mol = M x L so...
0.125M x 0.200L = (0.25 Mols Kbr x 119 g KBr)/1 Mol KBr = 2.98 g KBr
b. Mol = M x L so...
Mol = 0.150M x 0.1L
(0.015 Mols Na2SO4 x 126.05 g Na2SO4)/1 mol Na2SO4 = 1.89 g Na2SO4

III. We took a lab test in groups of two, having to make a solution of 0.15M of NaCl and show calculations