Additional Pretest Questions
The following table can be used to determine the relative concentrations of the acid and base
1.
Indicator
Drops of Acid
Color of indicator
After adding acid
red
Drops of base added to
change color of
indicator
12
Color of indicator
after adding
sufficient base
green
5 drops of blue
indicator
10
Indicator
Drops of base
Color of indicator
After adding
base
Drops of acid
Added to change color
of indicator
Color of indicator
after adding
sufficient acid
5 drops of blue
indicator
10
green
9
red
Calculations – Conclusions – Determine the relative concentrations of the acid and base
Since 12 drops of the base is needed to neutralize the 10 drops of acid the acid is more
concentrated. It is between 11/10 ( 1.1) and 12/10 ( 1.2 ) times more concentrated. 11
drops is not quite enough base to neutralize the 10 acid and the 12 drops is more than
enough base to neutralize the 10 drops acid
Since 9 drops of acid are needed to neutralize 10 drops of base the acid is between
10/9 ( 1.11) and 10/8 ( 1.25 ) times more concentrated. 8 drops is not quite enough acid to
neutralize the 10 base and the 9 drops is more than enough acid to neutralize the 10
drops base
2.





A column cartridge is composed of a polar stationary phase.
Isopropyl alcohol is a relatively non-polar solvent.
Water is a polar liquid.
A maroon pigment is extracted using isopropyl alcohol.
The pigment is then injected into the column cartridge and the maroon pigment stays in
the cartridge and the eluent ( product) is clear.
 A 5 % water 95 % isopropyl solution is then injected into the cartridge containing the
maroon extract. The column cartridge turns red and the eluent ( product ) is orange.
 A 25 % water 75 % isopropyl solution is then injected into the cartridge containing the
red pigment. The column cartridge turns clear and the eluent ( product ) is red.
What conclusions can you make given the above information about the maroon pigment.
 The maroon pigment is a mixture of orange and red pigments. The eluent
after adding the 5 % water and 95 % isopropyl alcohol to the maroon mixture in the
column’s polar stationary phase is orange and the eluent is red after adding the
25 % water 75 % isopropyl alcohol solution to the polar stationary phase after extracting
the orange pigment. The column is clear after the second extraction indicator that the
pigment contains two pigments.
 The orange pigment is more non-polar than the red pigment. It was extracted
from the polar cartridge stationary phase with the more non polar mobile phase
mixture of 5 % water and 95 % isopropyl alcohol.
 The red pigment is less non-polar or more polar than the orange pigment. It was
extracted from the polar cartridge stationary phase with the more polar mobile
phase mixture of 25 % water and 75 % isopropyl alcohol.
3.
A solution of gold nitrate is obtained. Explain in detail using the following terms how the
Gold can be extracted from the solution.
Electrons, reduction, cathode, anode, oxidation, red (+) wire, black (-) wire, water, oxygen
Hydrogen ions.
Metal ions such as gold ions found in gold nitrate are attracted to the negative black lead
where the gain electrons to form gold metal
Significant figures
How many significant figures are in each of the following. If the number should be written in
scientific notation in order to determine the number of significant figures please state that. 1pt
each
4a. 6021 km
4
b. 17,800 km
sci notation
c.
.409m
3
d.
.00087 m
2
e. 8.070 g
4 4
f. Write the above number that should be written in scientific notation to 4 significant figures in
scientific notation?
1.7080 x 105 km 2pts
1.780 x104 km
g) Round off the following two calculations two the correct number of significant figures? 1pt each
i. m / D = V = 834.5 g / 1.26
ii.
g
= 662. 3 ml = 662 ml ( Least # of digits)
---ml
8 g + 5.036 g + 9.811 g = 22.847 g = 23 g ( Least # of decimal places )
Temperature, Specific Heat Capacity, Melting Pts, Boiling Pts, Heat of Fusion, Heat of
Vaporization, Phase Changes, Potential Energy Changes, Kinetic energy changes
5.
The following data can be used to predict the shape of a heating curve
Mass is 10g Rate = 5 cal /s
Specific heat capacity of the solid = .25 cal / g oC
Specific heat capacity of the liquid = .50 cal / g o C
Specific heat capacity of the vapor = .125 cal / g oC
Heat of Fusion = 50 cal / g
Heat of Vaporization = 150 cal / g
The melting point of the substance is – 40 C
The boiling point of the substance is 20 C
The substance starts out at – 60 C
The substance ends at + 80 C
Sketch the curve
Indicate on the graph changes in kinetic energy and changes in potential energy
Temperature as a function of
time heating 10g of a substance
at a rate of 5cal/s
100
80
60
temperature oC
40
20
0
-20 0
100
200
300
400
-40
-60
-80
time heating in seconds
500
Show yoru work
-60oC to -40oC Q=mct
time = Q
rate
Q=10g(-40oC-(-60OC))(.25cal/goC)=50cal
50 cal = 10s
5 cal /s
-40 oC to -40 oC
Q=mLf time = Q
rate
Q = 10g ( 50 cal /g ) =500 cal
500 cal = 100s
5 cal/s
-40oC to 20oC Q=mct
time = Q
rate
Q=10g(20oC-(-40OC))(.50cal/goC)=75cal
300 cal = 60s
5 cal /s
20 oC to 20oC
Q=mLf time = Q
rate
Q = 10g ( 150 cal /g )= 1500 cal
1500 cal = 300s
5 cal/s
20 oC to 80 oC Q=mct
time = Q
rate
Q=10g(80oC-20OC))(.125cal/goC)=75cal
75 cal = 15s
5 cal/s
6. Heat is added to a solid at its melting point of 200 C at a rate of 4000 J per
minute. It takes 3.00 minutes to melt 11 grams of the solid. What is the heat of
fusion of the solid at 200 C in J per Kilogram. (1000g = 1kg)
Q = 4000 J / min ( 3.00 min ) = 12,000 J to melt the solid
Q = 12,000 J = 1090 J
11 g
g
( 1000 g ) = 1.09 x10 6 J
( Kg )
kg
7. The specific heat capacity of aluminum is .88
The specific heat capacity of lithium is
Joules
----------g C
2.78 Joules
----------g C
If 500 calories of heat were absorbed by the 20.00 g of lithium and 20.00 g aluminium which
would at 20 oC which sample would have the highest temperature and why?
The aluminium will experience the greatest change in temperature because it requires the
least amount of energy ( .88 J / g C) to change its temperature compared to the lithium with
its higher specific heat capacity (2.78 J / g C)
8. What is the difference between the measurement of temperature and the
calculation of heat? How are they related?
Temperature is measured relative to known standards. It indicates how hot or cold a sample is
relative to the known standards.
Heat is related to temperature in that in order for heat to be transferred between samples
there must be a temperature difference between the samples.
One way heat can be calculated is to multiply the mass of the sample by the temperature
change it undergoes multiplied by the specific heat capacity of the substance.
9. What is the difference between the specific heat capacity of a liquid and the
latent heat of of vaporization liquid? The specific heat capacity of a substance is calculated
by determining the amount of energy
required to raise the temperature of 1 kilogram 1 degrees Celsius.
The latent heat of vaporization involves no temperature change.
The latent heat of vaporization is calculated by determining the amount of energy per
kilogram to vaporize 1 kg of the liquid at its boiling point.
10. Why is it important that it takes 2.24x106 J / kg or 540 cal / g to vaporize water
to life on this planet?
It is important that it takes a very large quantity of energy to vaporize water because it
allows most of the water in this planet to stay in the liquid phase at the range of temperatures
found on this planet. If it were less than there most likely would be more precipitation and
drastic changes in climates.
Negative electrons attract the positive gold ions of the gold nitrate solution
Gold metal forms as the electrons reduce the positive gold ions to metal atoms
The nitrate ions migrate to the positive electrode
At the positive electrode – red lead – anode water decomposes