exam 2 practice problems packet answer key … · exam 2 practice problems packet – answer key 1)...

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EXAM 2 PRACTICE PROBLEMS PACKET – ANSWER KEY

1) Write the conversion factor between milliliters and liters (use symbols): 1000 mL = 1 L

2) Write the conversion factor between meters and centimeters (use symbols): 1 m = 100 cm

3) Write the conversion factor between grams and kilograms (use symbols): 1000 g = 1 kg

Metric Prefix (name) Metric Prefix (symbol) Exponent

Question 4) micro− μ 10−6

Question 5) giga− G 109

Question 6) nano− n 10−9

Question 7) kilo− k 103

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Questions 8 – 17: Write the name Questions 18 – 27: Write the chemical symbol

Each name is worth 2 points. Each chemical symbol is worth 2 points.

Every three spelling errors = 3% deduction Write capital and lower case letters clearly

8) Ne neon 18) potassium K

9) N nitrogen 19) silver Ag

10) Ca calcium 20) sulfur S

11) U uranium 21) copper Cu

12) Ga gallium 22) mercury Hg

13) Li lithium 23) boron B

14) Sr strontium 24) krypton Kr

15) Co cobalt 25) bromine Br

16) W tungsten 26) plutonium Pu

17) Mg magnesium 27) carbon C

Note: For Astatine (At), I will not take points for including or omitting it. Different sources conflict

on whether astatine is a metalloid or not.

Question 28) In the table below, shade in the metalloids.

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Question 29) In the table below, circle the alkali metals. Question 30) In the table below, circle the halogens.

Some notes

Alkali Metals: Hydrogen is not considered to be an alkali metal. No points will be deducted from me if you included. However, CHEM 120

you will be expected to know this.

Halogens: Astatine and Element 117: Please include these in the halogens.

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PART 2 (NON-CALCULATOR SECTION); SUGGESTED TIME ALLOTMENT: 15 MIN – 25 MIN

Approximately 25 to 35 questions.

31) Rewrite the following types of radiation in order of increasing (shortest to longest) wavelength.

blue light gamma radiation infrared yellow light

gamma blue yellow infrared

32) P: 1s2 2s2 2p6 3s2 3p3 How many valence electrons does phosphorus have? 5

33) Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 How many valence electrons does bromine have? 7

C 34) WOTF (which of the following) is NOT an example of an intensive property?

D 35) WOTF is an NOT an example of a chemical change?

A 36) WOTF is an NOT a metalloid?

C 37) WOTF is an NOT considered a metal?

38) Plum Pudding Model

Question 39 (Answers will vary):

Category 1 (any one of the phase changes): (list a substance) with freezing, melting, vaporization,

condensation, sublimation, or deposition

Category 2 (solubility/miscibility): dissolving salt in water, water and alcohol mixing

Category 3 (physical mutilation): bending, breaking, tearing, flattening (malleability),

stretching (ductility)

So, appropriate answers to Question 39:

1) Water freezing

2) Sugar dissolving in water

3) Cutting a piece of paper in half.

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40) deposition

41) gas

Answer to Question 42

Malleability is the ability to flatten a substance (1 pt). Malleability is a physical property

because a chemical change is not necessary to observe this property (1 pt). Malleability is

an intensive property because this property is not dependent upon the amount of

substance present (1 pt).

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Questions 43 through 48 are True/False questions.

If the statement is true, write a T in the answer box.

If the statement is false:

Write an F in the answer box

In the box, write a word or phrase to replace the bold/italicized word to make the statement true.

F 43) “Atoms of the same element are identical with the same physical properties.”

varying (or different)

F 44) Rutherford conducted the cathode ray tube experiment.

Gold Foil Experiment

F 45) The symbol for atomic mass is Z.

atomic number

T 46) A heterogeneous mixture has inconsistent composition throughout the sample.

T 47) Elements cannot be broken down into simpler substances through chemical reactions.

T 48) An electron has a relative mass of 1/1840 when compared to the mas of a proton.

T 49) The SI base unit for mass is the kilogram.

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D 50) Iodine gas cooling down to become iodine solid. A) chemical change

G 51) Water heating up to become steam. B) chemical property

E 52) Red dye added to water turns the water red. C) condensation

A 53) Two clear colorless liquids turn red when added together. D) deposition

B 54) Iron does not react with nitrogen gas. E) dissolving

H 55) Alcohol is miscible with water. F) kinetic energy

F 56) The energy associated with a ball brick falling to the ground. G) physical change

I 57) Energy stored within the chemical bonds of an explosive. H) physical property

I) potential energy

J) sublimation

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If the Plum Pudding Model were correct (diffuse positive cloud with electrons scattered

throughout), then 100% of the alpha particles should have passed straight through the gold foil (3

pt).


A cathode ray results only when the gas tube is filled with gaseous vapor of any substance (3 pt).

The beam was found to be negatively charged because it was attracted to the positive plate of the

external circuit (2 pt).

alpha particle

source () Au

POWER

SOURCE

− cathode

+ anode

+

−

external positive plate

external negative plate

EXTERNAL

POWER

SOURCE

Partially evacuated vacuum

tube, containing Hg vapor

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61) B

62) A

63) D

64) F

65) C

66) 3 Hz


The Unknown contains Element A and Element E.

Element A: All of the lines in Element A are present in the same position as the lines in the

Unknown.

Element B contains two green lines that are not in the Unknown in the same position. Therefore,

Element B can be eliminated.

Element C contains two orange lines that is completely absent in the Unknown. There are also

two red lines that are not in the same position as the red lines in the Unknown. Therefore,

Element C can be eliminated.

Element D contains a teal (aqua) line that is completely absent in the Unknown. Therefore

Element D cannot be in the unknown. There is also a purple line in Element D that is not in the

same position as any of the purple lines that are in the Unknown.

Element E: All of the lines in Element E are present in the same position as the lines in the

Unknown.

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PART 3 (CALCULATOR SECTION): SUGGESTED TIME ALLOTMENT: 15 MIN – 25 MIN

Question 68) Equation 2: hydrogen(g) + nitrogen(g) ammonia(g)

6.05 g 14.01 g 20.06 g

Use the Law of Conservation of Mass to determine the mass of ammonia. You do not have to show your work.

Question 69) Equation 1: methanol(l) + salicylic acid(g) methyl salicylate(g) + water(l)

32.04 g 138.13 g 152.15 g 18.02

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Composition of Various Isotopes – Table 1

#no Z Atomic Symbol #p+ A #e− Isotope Name

70) 33 27 Co2760

27 60 27 Cobalt-60

71) 115 77 Ir77192 77 192 77 Iridium-192

72) 32 30 Zn3062 30 62 30 Zinc-62

73) Al: 1s2 2s2 2p6 3s2 3p1

74) Ba: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2

75) Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

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76) The table below contains information about the four naturally occurring isotopes of chromium metal. Use

the information to determine the atomic mass of chromium. Remember: for these problems, I want you

to ignore significant figure rules. The final answer should have a total of four significant digits. Show your work.

Isotope Atomic Mass of Isotope Natural Abundance of Isotope

Chromium-50 50 amu 4.34%




Chromium-50: (4.34

100) × 50 amu = 2.17 amu 2.1700 amu

Chromium-52: (83.79

100) × 52 amu = 43.5708 amu 43.5708 amu

Chromium-53: (9.50

100) × 53 amu = 5.035 amu 5.0350 amu

Chromium-54: (2.37

100) × 54 amu = 1.2798 amu + 1.2798 amu

52.0556 amu

Remember the instruction: regardless of significant figure rules, I always want four significant digits only

for atomic mass values.

So: x52.06 amux

You should have confidence that your answer is correct if it is very close to the atomic mass value on the

periodic table.

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77a) Nitroglycerin is a powerful contact explosive. It is 18.53% nitrogen by mass. Write the conversion factor

that can be derived from the percent-by-mass value of nitrogen in nitroglycerin. (The chemical formula

for nitroglycerin is C3H5N3O9. Assume mass values are measured in grams.

18.53 g N = 100 g C3H5N3O9

77b) If a sample of nitroglycerin contains 0.400 grams of nitrogen. Calculate the mass (in grams) of the

entire sample. Show your work.

(0.400 g N

1) (

100 g C3H5N3O9

18.53 g N) = x2.16 g C3H5N3O9x

77c) Nitroglycerin has a density of 1.6 g/mL. Write the conversion factor that can be derived from this

density value.

1.6 g C3H5N3O9 = 1 mL C3H5N3O9

77d) Calculate the mass of nitrogen (in grams) of a 50.00-mL sample of nitroglycerin. Hint: You may need to

use both conversion factors (from Question 77a and Question 77c).

(50.00 mL C3H5N3O9

1) (

1.6 g C3H5N3O9

1 mL C3H5N3O9) (

18.53 g N

100 g C3H5N3O9) = x14.82 g Nx

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77e) A sample of nitroglycerin is poured into a spherical delivery apparatus with a radius of 2.50 cm.

Determine the mass of nitrogen (in grams) of the amount of nitroglycerin that will fill the delivery

apparatus.

Vsphere = (4

3) π(r)3

Helpful Hints

Solve this in pieces

First, determine the volume of the sphere.

The volume of sphere will be equal to the volume of nitroglycerin. Assume 1 cm3 = 1 mL

The volume of nitroglycerin can be converted to mass of nitroglycerin (through density conversion factor)

The mass of nitrogen can then be determined through the %-by-mass conversion factor.

Vsphere = (4

3) π(r)3 = (

4

3) π(2.50 cm)3 = (

4

3) π (15.625 cm3) = 65.4 cm3

(65.4 cm3 sphere

1) (

1 mL sphere

1 cm3 sphere) (

1 mL C3H5N3O9

1 mL sphere) = 65.4 mL C3H5N3O9

(65.4 mL C3H5N3O9

1) (

1.6 g C3H5N3O9

1 mL C3H5N3O9) (

18.53 g N

100 g C3H5N3O9) = x19.4 g Nx

EXTRA CREDIT

I am leaning against extra credit questions for this Exam, but I’m not 100% decided. And the test will be shorter than this practice packet.

exam 2 practice problems packet answer key … · exam 2 practice problems packet – answer key 1)...

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