KEY CONCEPTS FROM ST LABS
1.
How to Distinguish Between Metals, Metalloids and Nonmetals.
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Metals usually react with acid to produce a gas; metalloids and non metals do not.
•
If a metal reacts with water or oxygen, it will leave behind a base
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Non metals are poor conductors; metalloids conduct but usually not as well as most metals.
•
Metals are malleable; metalloids and non metals are not.
•
Some nonmetals and metalloids can sparkle but they won’t have that almost mirror-like luster
that most metals have.
Example:
Match column 1 with column 2
COLUMN 1
A solid that conducts electricity__A_
A yellow-gold, shiny crystal____A___
Malleable and bubbles when placed in acid B_
Not malleable___A_
Does not leave a base after being placed in
water___A_
2.
Electrolysis of Water
2 H2O 
a)
b)
c)
d)
COLUMN 2
Inconclusive; need more tests
Definitely a metal
Definitely a metalloid
Definitely a non metal.
2 H2 + O2
•
Water can be split with electricity and the help of acid or salt
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Electrodes connected to a battery are inserted into separate test tubes. Each one will collect a
different gas.
•
When it dissociates, in theory, it should give a 2: 1 ratio of H2 and O2 gases to reflect the
balanced equation
•
But because of competing impurities inthe water, usually less oxygen is produced.
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Examples:
1. Identify the test tube containing oxygen and find the
approximate percent of oxygen in that test tube. Use a
ruler.
0.53cm/3.30 cm =
17%
Notice that the more meaningful ratio of hydrogen to oxygen can also be measured.
2.
A)
Water can be split with electricity and the help of ____salt____ or acid or
base____
B)
Electrodes connected to _____a power source___ are inserted into separate _test
tubes_. Each one will collect a different gas.
C)
When it dissociates, in theory, it should give a ____2:1__ ratio of H2 and O2 gases to
reflect the balanced equation
D)
But because of competing impurities inthe water, usually_less___ oxygen is produced.
3.
Preparing a Solution
WDTA for a solid after using m = CV
and PTA for dilution
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Examples:
1.
a)
How would you prepare 5.0 ml of a 3.0 g/L solution?
•
•
•
b)
What if there was no 5.0 ml volumetric flask. What could you use instead?
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c)
Weigh 0.005L* 3.0 g/L = 0.015 g of solute
Dissolve in a beaker with less than 5.0 ml
Transfer to a volumetric flask and top to 5.0 ml mark.
(In this case since the volume is so small you could use a 5.0 ml graduated
cylinder.)
What would you have to do to make the solution 1000 times more dilute? Assume that there is
no micropipette in the lab but that there is an IV medicinal drip, which is adjusted to deliver 10
drops from each 1.0 ml.
1000 times more dilute means we want C2 = 3/1000 = 0.0030 g/L
V1 =?
C1= 3.0 g/L
V2 = not mentioned, so we can choose, say, 100.0 ml = 0.100 L
C1V1=C2V2
3 V1= 0.0030 g/L(0.100)
V1 = 0.00010 L = 0.10 ml
Since 1.0ml /10 drops = 0.10 ml/drop, you would need only 0.10 ml/ (0.10ml/drop) = 1 drop of
the original solution, and that would have to be diluted to 100.0 ml with water
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4.
Carbon Cycle Lab
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By blowing into water or any aqueous solution we
represented what happens when carbon dioxide from the
atmosphere encounters water:
H2O + CO2 H2CO3
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In the presence of base (high pH) and calcium ions, the
ions from the H2CO3 react and form calcium carbonate, which is
found in shells.
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Both vinegar and H2CO3 are acids, so both change the
colour of bromothymol blue towards yellow. But since H2CO3 is
a weaker acid, it only goes to the in-between colour of green.
Example:
a) How do you test for the presence of carbon dioxide in breath?
Use limewater. It will go cloudy.
b) Why does the cloudiness go away if you keep blowing into it?
Extra breath introduced more carbon dioxide, which makes the solution eventually acidic and it
dissolves the calcium carbonate precipitate that made id cloudy in the first place.
5.
How to Build Circuits
Case 1: Series Circuit (Note how the voltmeter is connected to each end of the resistor. The ammeter is
only connected to one end.
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Case 2: Parallel Circuit
a. Ammeter Positioned to Measure Total Current
b. Ammeter Positioned to Measure I2
c. Ammeter Positioned to Measure I1
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6.
A)
B)
C)
Gears: Practical Concepts
In order to have a velocity ratio greater than 1, you need to have the larger gear as the input.
To further increase the ratio, a gear box can be used.
In order to maintain same direction for the output as input, you need:
(1) An odd number of gears in a gear train.
(2) Or use a chain between two gears
(3) Or use a belt between two grooved wheels.
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Example:
Draw a set up in which a pair of 6-toothed gears and a pair of 2-toothed gears are
rearranged to create a gear ratio of 9.
You need a gear box.
7.
Transformation Systems
You can transform circular motion to linear motion with
(1)
(2)
(3)
Rack and pinion
Cam and follower
Crank-slider
Example:
a) Which transformation system is
this?
Crank and slider.
b) What can be attached to make it
more practical?
A handle
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