Today, we had a sub. Her name was Ms. Wallace. We went to the computer lab to do our virtual lab. The website we were given is
http://chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redox/home.html
The point of this lab was to see the virtual simulation of putting strips of metals into different compounds to observe the different chemical reactions.
Monday, January 31, 2011
Friday, January 28, 2011
Jan 27 2011 - Types of Reactions
Today we went over types of reactions.
We learned about three types of reactions that are synthesis, decomposition and single replacement.
1. Synthesis
This reaction is combination of two or more reactants to form one product.
A + B -> C
Ex. 2H2 + O2 => 2H2O
2. Decomposition
In this reaction, a compound is broken down into two or more products.
A -> B + C
Ex. 2H2O => 2H2 + O2
3. Single Replacement
Single replacement is when an element replaces another in a compound. (metal elements replace positive ions and non-metal elements replace negative ion)
A + BC -> AC + B (A=metal)
A + BC -> BA + C (A=non-metal)
Ex.
Metal - Zn + CuCl2 => ZnCl2 + Cu
Non-Metal - Cl2 + NaBr => NaCl + Br2
Predicting Single Replacement Reactions
An element highes up on the series replaces the ion below it on the table.
Ex. 2Li + MgCl2 => 2LiCl + Mg
Zn + CuCl2 => ZnCl2 + Cu
Tuesday, January 25, 2011
January 25, 2011 -- Balancing Equations
Today we reviewed something we learned many times before: how to balance equations.
To balance an equation, all you have to do it multiply each side of the equation until both sides of the equation, the reactant and the product, have the same number of atoms.
For example, Al + O2 -> Al2O3 would become 4Al + 3O2 -> 2Al2O3.
To balance an equation, all you have to do it multiply each side of the equation until both sides of the equation, the reactant and the product, have the same number of atoms.
For example, Al + O2 -> Al2O3 would become 4Al + 3O2 -> 2Al2O3.
Sunday, January 16, 2011
January 11 - Molar Volume of a Gas at STP (Standard Pressure & Temperature)
We learned about STP, which is Standard Pressure and Temperature. We have STP because with it, we can compare the volume of gases.
STP is 1 atmosphere of pressure and a temperature of 0 degrees Celsius or 273.15 Kelvin.
At STP, 1 mole of gas occupies 22.4L.
STP is 1 atmosphere of pressure and a temperature of 0 degrees Celsius or 273.15 Kelvin.
At STP, 1 mole of gas occupies 22.4L.
Thursday, January 13, 2011
January 13, 2010
Today, we did some reviewing for the up coming test! on Monday!
SO START STUDYING!
-the test would be on ch.4 materials
-Mole conversion
-Molar Volume
-Molar Mass
-Avogardro's Number
-Molarity
-Finding out Mass
-Finding out Molarity
-Finding the # of moles
-Finding the Total Number of Atoms (or specific element)
-Empirical Formula
-% Composition of Elements in Compound
Be sure to do Ms.Chen's review sheets because PRACTICE MAKES PERFECT!
a good site to check out:
a very popular Youtube Video Channel with Many helpful Vidoes:
Monday, January 10, 2011
Jan 07 2011 - Diluting Solutions to Prepare Workable Solutions
We learned how to calculate diluting solutions.
Key idea - the moles of solute is constant (i.e. the only difference is that there is more water in the less concentrated solution)
moles before = moles after
Equation : M1L1=M2L2 (1 denotes "before")
For example, 20.0 mL of 0.200 M NaOH solution is diluted to a final volume of 100.0 mL, calculate the new concentration.
M1 = 0.200 M NaOH
Key idea - the moles of solute is constant (i.e. the only difference is that there is more water in the less concentrated solution)
moles before = moles after
Equation : M1L1=M2L2 (1 denotes "before")
For example, 20.0 mL of 0.200 M NaOH solution is diluted to a final volume of 100.0 mL, calculate the new concentration.
M1 = 0.200 M NaOH
L1 = 20.0mL
M2 = ?
L2 = 100.0 mL
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaM1L1=M2L2
aaaaaaaaaaaaa 0.200 M NaOH x 20.0 mL = M2 x 100.0 mL
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa0.200 x 20.0
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaM2 =__________________
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa100.0
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaM2 = 0.200 M
The video for practice problems on dilution calculations
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaM1L1=M2L2
aaaaaaaaaaaaa 0.200 M NaOH x 20.0 mL = M2 x 100.0 mL
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa0.200 x 20.0
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaM2 =__________________
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa100.0
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaM2 = 0.200 M
The video for practice problems on dilution calculations
Wednesday, January 5, 2011
January 5th 2011 - Molarity
Today we learned how to find the molarity of a substance when given the component of it.
The formula for finding molarity is moles of solute / liters of solution.
For example, let's do this question.
What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 L?
When given different pieces of information, just find the number of moles and the number of liters to solve for molarity again.
What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water to make 125 ml of solution?
The formula for finding molarity is moles of solute / liters of solution.
For example, let's do this question.
What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 L?
| molarity = | 0.32 moles NaCl 3.4 L | ||
| = | 0.094 M NaCl |
When given different pieces of information, just find the number of moles and the number of liters to solve for molarity again.
What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water to make 125 ml of solution?
2.5 g NaCl x | 1 mole NaCl | = 0.0427 mole |
molarity = | 0.0427 mole NaCl 0.125 L |
= | 0.34 M NaCl |
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