Tuesday, December 15, 2009 6:54 PM
12/15/09 - DAY 1CDBAToday we learned about Chemical Equations
Chemical Equations
You MUST have correct equations when balancing Chemical Equations !
For Example:
Lead (IV) Nitrate = Pb(NO3)2
Copper (II) Chloride = CuCl2
2Mg (s) + 1O2 (g) → 2MgO (s)
^ REACTANTS ^ ^ FORMS ^ PRODUCTS
Examples:
1) 4 K + O2 → 2 K2O
2) N2 + 2 H3 → 2 NH3
3) 2 Al + 3 Cl2 → 2 AlCl3
4) 2 AlBr3 + 3 SrCO3 → Al(CO3)3 + 3 SrBr2
5) Fe2O3 + 3 H2SO4 → Fe2(SO4)3 + 3 H2O
6) Ca(OH)2 + 2 NH4Cl → 2NH3 + CaCl2 +2 H2O
WATCH THIS VIDEO !
Thursday, December 10, 2009 4:51 PM
Well today(yesterday) we had a big test on moles like Molarity and Molar Concentration. To find out what happened click
HERE . there weren't any good videos out there to explain what we did but this one helps out fine
Monday, December 7, 2009 8:18 PM
Giving Directions
outline for experimental procedures
-->Find the mass you need
Example 1: Jeremy is asked to make a 0.55M solution of K2SO4. If he needs 250mL what procedure should he use?
Answer 1:con'c(concentration)->mol>mass
2K=2(39.1)=78.2
1S=1(32.1)=32.1
40=4(16.0)=64.0
[174.3]
0.55mol/L x 0.25L= 0.1375mol
0.1375mol x 174.3g/mol= 24g
(1) Measure 250mL of water
(2) Weigh 24.0g of K2SO4
(3) Add K2SO4 to water
(4) Stir until it all dissolves
Example 2: Give Directions to make 2.00L of 6.0M NaOH.
Answer 2:
1 Na=23.0
1 O=16.0
1 H=1.0
[40.0]
6.0mol/L x 2.00L= 12 mol
12 mol x 40.0g/mol= 480g
(1) Measure 2.00L of water
(2) Weigh 480g of K2SO4
(3) Add NaOH to water
(4) Stir until it all dissolves
Dilution of Solutions
When you add water con'c(concentration) decrease.
If the volume is doubled con'c is halved.
Volume(L)
6.0
12.0
48.0
Con'c(mol/L)
2.0
1.0
0.25
# of moles
12.0
12.0
12.0
# moles1=# moles2
C1V1=C2V2
Example 1: Karolyn adds 150.0mL of water to 50.0mL of 0.60M HCl. Find [HCl]
Answer 1:
V2= 200.0mL
V1= 50mL
C2= ?
C1= 0.60M
C1V1=C2V2
(0.60M)(50mL)/200.0mL=C2(200.0mL)/200.0mL
C2= 0.15M
Example 2: Jesse adds water to 100.0mL of 0.25M HF to a final volume of 400.0mL. Find the [HF].
Answer 2:
V2= 400.0 mL
V1= 100.0 mL
C2= ?
C1= 0.25M
C1V1=C2V2
(0.25M)(100.0mL)/400.0mL=C2(400.0mL)/400.0mL
C2= 0.0625M
Example 3: Cheyenne dilutes 60.0mL of 0.40M HNO3 to 0.15M. What is the final volume? How much water did she add?
Answer 3:
V2= ?
V1= 60.0mL
C2= 0.15M
C1= 0.40M
C1V1=C2V2
(0.40M)(60.0mL)/0.15M=(0.15M)V2/0.15M
V2= 160mL
160mL-60mL= 100mL
She added 100mL of water.
