Unit Test Review
Q#
1
2
3
answer #
D
B
B
Q#
4
5
6
answer #
A
C
B
Q#
7
8
9
answer #
D
B
D
Q#
10
11
answer #
C
C
1. Identify as Endothermic or Exothermic
exothermic a) a substance was dissolved in water and the temperature of the water increased
endothermic b) liquid  gas
exothermic
c) liquid  solid
exothermic
d) in an _________-thermic reaction, the products always possess more potential
energy than the reactants
exothermic
e) gas  solid
endothermic f) 2H2O(l)  2H2(g) + O2(g)
endothermic g) melting of solid salts
endothermic h) evaporation of water
exothermic
i) condensation of water
_A___2. As ice cools from 273K to 263K, the average kinetic energy of its molecules will
a) decrease
b) increase
c) remain the same
__D___3. The molecules of which substance has the highest average kinetic energy:
a) He(g) at O0C
b) CO2(g) at 200C
c) HCl(g) at 400C
d) N2(g) at 600C
__C___4. As ice at O0C changes to water at O0C, the average kinetic energy of the molecules
a) decrease
b) increase
c) remain the same
__A___5. As water vapor condenses at 1000C the potential energy of the molecules
a) decrease
b) increase
c) remain the same
__B___6. Calculate the total number of joules of heat that must be absorbed to change the temperature
of 100 g of water from 250C to 300C
a) 420 J
b) 2100 J
c) 10,500 J
d) 12,600 J
__A___7. The term below that represents a form of energy is:
a) heat
b) degree
c) joule
d) temperature
__D___8. The formula that represents a compound is:
a) C
b) Cl2
c) N2
d) HF
__D___9. The substance that cannot be decomposed by a chemical change is:
a) H2SO4
b) NH3
c) H2O
d)Ar
__C___10. 840 J are added to 20g of water at 300C, calculate the final temperature of the water?
a) 200C
b) 350C
c) 400C
d) 500C
__C___11. How many grams of water will absorb a total of 2520 J of energy when the water’s
temperature water changes from 100C to 300C
a) 10 g
b) 20 g
c) 30 g
d) 60 g
__D___12. The material that is a pure substance is:
a) air
b) earth
c) fire
d) water
__D___13. The pair of substances that can be decomposed by chemical means is:
a) CaO & Ca
b) MgO & Mg
c) Co & Mg
d) CaO & MgO
__A___14. An example of a heterogeneous mixture is
a) soil
b) sugar
c) carbon dioxide
d) carbon monoxide
__D___15. A homogeneous mixture is represented as:
a)NaCl(s)
b)NaCl(l)
c)NaCl(g)
d) NaCl(aq)
_C____16. At which temperature would the molecules in a one gram sample of water have the lowest
average kinetic energy:
a) 50C
b) -1000C
c) 5K
d) 100K
__C___17. As electrical energy is converted into heat energy, the total amount of energy in the system:
a) decreases
b) increases
c) remains the same
__B___18. The following statement best describes exothermic chemical reactions:
a) they never release heat
c) they never occur spontaneously
b) they always release heat
d) the always occur spontaneously
__B___19. If Joules of energy are added to a sample of water, the temperature will:
a) decreases
b) increases
c) remains the same
_C____20. The type of energy stored in a chemical bond is:
a) electrical
b) nuclear
c) chemical
d) heat
__A___21. The temperature of a substance changes from 100C to 200C. Calculate the number of Kelvin
degrees that the temperature changes:
[remember: 1 degree change in Celsius = 1 degree change in Kelvin]
a) 10
b) 20
c) 273
d) 283
_______22. A 60g sample of water at 100C absorbs 1.3KJ of heat. Calculate what the temperature
changes to.
Q=mcΔT
1300J = (60g)(4.18J/gºC)( ΔT)
5.18ºC = ΔT
ΔT = Tf - Ti
5.18ºC = Tf - 10ºC
15.18ºC = Tf
_______23. 5 g of water is raised to 150C by adding 200J of heat. Calculate the initial temperature.
Q=mcΔT
200J = (5g)(4.18J/gºC)( ΔT)
9.57ºC = ΔT
ΔT = Tf - Ti
9.57ºC = 15ºC – Ti
-5.43ºC = -Ti
5.43ºC = Ti
_______24. 50g of water is raised to 600C by adding 3150J of heat. Calculate the initial temperature.
Q = mcΔT
ΔT = Tf - Ti
3150J = (50g)(4.18J/gºC)( ΔT)
15.07ºC = 60ºC – Ti
15.07ºC = ΔT
-44.93ºC = -Ti
44.93ºC = Ti
MATTER
Matter is anything that has mass and occupies space
The three states of matter are: solid, liquid and gas
Describe physical properties of matter. Explain & give examples.
 use senses to observe physical properties: color, texture, odor, state of matter
 measure:
o physical constants: bp, mp, density
o mass, volume
Describe chemical properties of matter. Explain & give examples.
 how matter behaves in presence of other matter (chemical reactions)
How do you distinguish between a physical change and a chemical change?
 physical change: change in form or appearance
o maintains same identity
o no new matter is created
o properties do not change
o can go back
o can be separated according to physical properties
 chemical change: change in identity
o create something new
o new matter has different properties than matter formed from
According to the Law of Conservation of Matter, the mass of substances involved in a
chemical reaction cannot be created or destroyed.. The total mass of reactants equals the
total mass of products. Atoms are just rearranged, not created or destroyed.
Ex:
2KClO3 → 2KCl + 3O2
500g
300g
?
If 500 g of KClO3 decomposes and produces 300g of KCl, how many grams of O2 are
produced?
500g = 300g + x
200g = x (grams of oxygen produced)
Sketch in the appropriate boxes below pictures of molecules, showing their different
arrangement when found in solid, liquid and gas phases.
solid
liquid
gas
What are the two broad categories of matter? pure substances and mixtures
What’s the difference between an element and a compound?
 element is the simplest form of matter that cannot be broken down further and still
maintain the properties of that element
 compound is two or more DIFFERENT elements chemically combined together
What’s the difference between a mixture and a compound?
 mixture is a physical combination of two or more different types of pure substances
 compound is a chemical combination of two or more different types of pure
substances
Explain the difference between homogeneous and heterogeneous mixtures.
homogeneous mixture:
heterogeneous mixture:
 uniform composition
 variable composition
 transparent/translucent
 cloudy/opaque
 individual particles too small to
 individual particles are big enough
scatter light so light passes
to scatter light so light does not
straight through
pass through
 solutions
 suspensions (including colloids)
Mixtures can be separated according to their physical and chemical properties. Explain how you
would separate the following mixtures into their individual components.
1. sand and water
2.
sugar and water
filtration
1. evaporate/boil off water (bp water much lower than bp of sugar)
2.crystallization
3. mixture of heptane (bp = 98˚C) and heptanol (bp = 176˚C) distillation
4. mixture of sand and iron pellets magnet (iron is magnetic and sand is not)
ENERGY
The SI unit of energy is called the JOULE.
You put a rock in the oven and heat it up. Then you put the rock in a large pail filled with cold water
(about 15˚C). You observe the temperature of the water before and after the rock is placed in it.
What happens? Why?
essentially placing a HOT rock in cooler water; heat always travels from hotter object to
colder object so heat travels from rock to water causing the water temp to rise and the
temp of the rock to decrease; eventually the temp of the water stops changing indicating
the water and the rock are the same temperature.
According to the KMT, atoms and molecules are in constant motion. How does temperature affect the
movement of particles in a substance? [temperature  motion of particles]
Increasing temperature causes particles in a substance to move faster, while decreasing the
temperature causes the particles to slow down.
Energy changes accompany all chemical and physical changes. If energy leaves the “system” we say the
energy change is EXOTHERMIC because the system had more energy before the change than after. If
energy enters the “system” we say the energy change is ENDOTHERMIC because the system had more
energy after the change than before.
Identify the following according to their being a physical/chemical change; identify each
change as being an endothermic/exothermic change.
change
burning coal
boiling water
burning
propane in
BBQ
drying clothes
in clothes
dryer
physical/chem
chemical
physical
exo/endo
exo
endo
chemical
exo
physical
endo
change
melting ice
activation of a
light stick
evaporation of
sweat from skin
lg
hammering
copper into flat
sheet
physical/chem
physical
chemical
exo/endo
endo
exo
physical
endo
physical
exo: hammer
endo: copper
A cold pack is placed on your injured leg. Indicate the direction of the flow of energy between your leg
and the cold pack. heat travels from leg to cold pack
What is the difference between potential and kinetic energy?
 potential energy (PE) is stored energy of position
 kinetic energy (KE) is the energy of motion
Name the six phase changes and state whether they are endothermic or exothermic.
ENDOTHERMIC phase changes
sl
melting
EXOTHERMIC phase changes
gl
condensation
lg
vaporization (boiling)
ls
solidification (freezing)
sg
sublimation
gs
deposition
Show in the particle diagrams below what happens to the particles as they change phase:
need to draw the same number of atoms/molecules in each diagram
solid
liquid
gas
Energy changes can be calculated using:
Q = mcΔT
where:
Q = energy
m = mass
c = specific heat
ΔT = change in temperature
[ΔT= Tfinal – Tinitial]
Solve the following:
1. What quantity of heat would have to be added to 5000g of water to change its
temperature from 20˚C to 80˚C?
Q = mcΔT
= (5000g)(4.18J/gºC)(60ºC)
= 1,254,000J
2. How much heat energy is required to raise the temperature of 8.0g of aluminum from 293K to
298K? [specific heat of aluminum is 0.90J/gK]
Q = mcΔT
= (8g)(0.90J/gK)(5K)
= 36J
3. An exothermic reaction releases 800 Joules of heat to a calorimeter containing 40 grams of
water. Calculate the temperature change of the water.
800J = mcΔT
= (40g)(4.18J/gºC)(ΔT)
4.78ºC = ΔT
4. What is the final temperature when 168 joules of heat is added to 4 grams of water at 283K?
168J = mcΔT
= (4g)(4.18J/gK)(ΔT)
10.0K = ΔT
5.
ΔT= Tfinal – Tinitial
10.0K = Tf – 283K
293K = Tf
500 gram sample of water loses 1.05 x 104 joules of heat. The temperature of the water, after
the heat loss, is found to be 25˚C. What was the initial temperature of the water?
Q = mcΔT
1.05 x 104J = (500g)(4.18J/gºC)(ΔT)
5.02ºC = ΔT
ΔT= Tfinal – Tinitial
5.02ºC = 25ºC - Ti
30.02ºC = -x
-30.02ºC = x