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Seminar:Test 1 Review
Freddie Arocho-PerezKaplan UniversitySC155: Introduction to Chemistry
General Information
Access Dates:
Wednesday, Oct 13 – Tuesday, Oct 26 During the time that the test is available, you may
take it anytime you wish and as many times as you want. Only the last submission will be saved and graded.
I made it available for more time than I originally noted in the syllabus. The only reason I did this was because I assume most of you may want to work on this test on evenings and weekends.
General Information
The questions have been formulated by me (Instructor) and the Science Instructional Committee. I did apply this change in order to make the test content more related to our seminars and discussions.
The test consists of 30-multiple choice questions.
Study Guide sent by e-mail and posted in DocSharing.
General Information
The test questions will account for reading comprehension, applications, definitions, and mathematical analysis.
There will be 9 math-related questions, coming specially from the material included in:
– “Seminar: Math Review Session”– “Seminar: Elements, Compounds, and Chemical Reactions”
Useful Materials:– Calculator– Periodic Table (see DocSharing for a hand-out)– Conversion Factors (see hand-outs “Metric and English
Conversion Factors” and “Math Review: Metric System / Dimensional Analysis” in DocSharing)
Practice Question
Which of the following element/chemical symbol pairs is incorrectly matched?
– Chromium: Cr– Rubidium: Rb– Scandium: S– Argon: Ar– Gold: Au
Practice Question
Which of the following element/chemical symbol pairs is incorrectly matched?
– Chromium: Cr– Rubidium: Rb
– Scandium: S– Argon: Ar– Gold: Au
Solution: Practice Question
From the Periodic Table:– Scandium: Sc– Atomic Number: 21
– Sulfur: S– Atomic Number: 16
Practice Question
Which of the following statements is incorrect?
– All atoms of an element have the same atomic number.
– All atoms of an element must have the same mass.
– All atoms of an element have the same number of protons.
– Atoms of an element may have different numbers of neutrons.
Practice Question
Which of the following statements is incorrect?
– All atoms of an element have the same atomic number.
– All atoms of an element must have the same mass.
– All atoms of an element have the same number of protons.
– Atoms of an element may have different numbers of neutrons.
Solution: Practice Question
All atoms of an element have the same atomic number = TRUE
All atoms of an element must have the same mass = FALSE
All atoms of an element have the same number of protons = TRUE
– (# Protons = Atomic Number) Atoms of an element may have different numbers of
neutrons = TRUE– Isotopes: different forms of an element each having
different atomic mass. Isotopes of an element have the same number of protons (the same atomic number) but different numbers of neutrons.
Practice Question
Which of the following involves chemistry?
– automobiles– detergents– cooking – all of the above
Practice Question
Which of the following involves chemistry?
– automobiles– detergents– cooking
– all of the above
Solution: Practice Question
Chemistry is EVERYWHERE!!!!
Practice Question
Tap water consists of water, sodium and chloride ions, and possibly bacteria, chlorine, and other ingredients. Which choice best defines what tap water is?
– a molecule– an element– a mixture– a compound
Practice Question
Tap water consists of water, sodium and chloride ions, and possibly bacteria, chlorine, and other ingredients. Which choice best defines what tap water is?
– a molecule– an element
– a mixture– a compound
Solution: Practice Question
Pure Substance: A sample of matter, either an element or a compound, that consists of only one component with definite physical and chemical properties and a definite composition.
Element: Substance consisting of only one type of atom.
Compound: Two or more atoms joined together chemically (with bonds).
Mixture: Composed of two or more substances, but each keeps its original properties.
Practice Question
Calculate how many liters of liquid are in a container that has 579 mL of the liquid.
(1 L = 1,000 mL)
– 5.79 L– 0.0579 L– 0.579 L– 57.9 L
Practice Question
Calculate how many liters of liquid are in a container that has 579 mL of the liquid.
(1 L = 1,000 mL)
– 5.79 L– 0.0579 L
– 0.579 L– 57.9 L
Solution: Practice Question
Conversion Factor:
1 L = 1,000 mL Use Dimensional Analysis:
LmL
LmL 579.0
000,1
1579
Practice Question
How many grams are contained in 3.20 pounds?
(1 pound = 453.6 g)
– 678.2 g– 964.3 g– 1,322.2 g– 1,451.5 g
Practice Question
How many grams are contained in 3.20 pounds?
(1 pound = 453.6 g)
– 678.2 g– 964.3 g– 1,322.2 g
– 1,451.5 g
Solution: Practice Question
Conversion Factor:
1 pound = 453.6 g Use Dimensional Analysis:
gramspound
gramspounds 5.451,1
1
6.45320.3
Practice Question
Which component of the atom has no charge?
– electrons– protons– neutrons– all components have a charge
Practice Question
Which component of the atom has no charge?
– electrons– protons
– neutrons– all components have a charge
Solution: Practice Question
Sub-atomic Particles:– Protons: Positive (+) Charge– Electrons: Negative (-) Charge– Neutrons: Neutral (0) Charge
Center of the Atom:– Protons– Neutrons
Surrounding the Center of the Atom:– Electrons
Practice Question
Which of the following is an example of a chemical property of water? It
– boils at 100 C.– is transparent.– has no odor.– reacts with calcium.
Practice Question
Which of the following is an example of a chemical property of water? It
– boils at 100 C.– is transparent.– has no odor.
– reacts with calcium.
Solution: Practice Question
A chemical property of a substance is how the substance reacts with other substances and what new products may form.
A physical property describes the substance’s characteristics.
An example of a chemical property is how substances react to heat (flammability) or acid.
An example of a physical property is the melting point or density of a substance.
Solution: Practice Question
Water:– Physical properties: it melts at 0 C, boils at 100 C
(at sea level), transparent, odorless, density is 1.0 g/cm3.
– Chemical properties: reacts with some metals like Calcium quickly, reacts with other metals slowly to form rust. Doesn’t burn.
Salt:– Physical: white, crystalline, high melting point.– Chemical: doesn’t burn, doesn’t react with acid.
Practice Question
Which of the following is an example of a physical property of hydrogen? It
– is less dense than air.– reacts with oxygen.– is highly flammable.– forms hydrochloric acid.
Practice Question
Which of the following is an example of a physical property of hydrogen? It
– is less dense than air.– reacts with oxygen. – is highly flammable.– forms hydrochloric acid.
Mass Relations
Atomic Mass vs. Molecular Mass:– Atomic Mass: Mass of the atom of an element.– Molecular Mass: Mass of a molecule; Sum of the atomic
masses of the atoms in a molecule.– amu: atomic mass units.
From the Periodic Table, we can obtain atomic masses:C = 12.01 amuCa = 40.08 amuH = 1.008 amuK = 39.10 amuP = 30.97 amuS = 32.07 amuO = 16.00 amu
Mass Relations
Atomic masses:C = 12.01 amuCa = 40.08 amuH = 1.008 amuK = 39.10 amuP = 30.97 amuS = 32.07 amuO = 16.00 amu
Calculate the molecular mass of H2O:H: 2 atoms X 1.008 = 2.016O: 1 atom X 16.00 = 16.00
Total: 18.016 ~ 18.02 amu
Mass Relations
Atomic masses:C = 12.01 amuCa = 40.08 amuH = 1.008 amuK = 39.10 amuP = 30.97 amuS = 32.07 amuO = 16.00 amu
Calculate the molecular mass of CO2:C: 1 atom X 12.01 = 12.01O: 2 atoms X 16.00 = 32.00
Total: 44.01 amu
Practice Question
Atomic masses:C = 12.01 amuCa = 40.08 amuH = 1.008 amuK = 39.10 amuP = 30.97 amuS = 32.07 amuO = 16.00 amu
Calculate the molecular mass of CH4:
Practice Question
Atomic masses:C = 12.01 amuCa = 40.08 amuH = 1.008 amuK = 39.10 amuP = 30.97 amuS = 32.07 amuO = 16.00 amu
Calculate the molecular mass of CH4:C: 1 atom X 12.01 = 12.01H: 4 atoms X 1.008 = 4.032
Total: 16.042 ~ 16.04 amu
Practice Question
48 C is approximately the same as
– 341 K.– 321 K.– 285 K.– 205 K.– 158 K.
Practice Question
48 C is approximately the same as
– 341 K.
– 321 K.– 285 K.– 205 K.– 158 K.
Solution: Practice Question
This is a temperature conversion problem:
K = C + 273.15 K = 48 + 273.15
= 321.15
= 321 (rounded)
Practice Question
120.0 C is approximately the same as
– 100 F.– 159 F.– 207 F.– 248 F.– 306 F.
Practice Question
120.0 C is approximately the same as
– 100 F.– 159 F.– 207 F.
– 248 F.– 306 F.
Solution: Practice Question
This is a temperature conversion problem
F = (1.8 x C) + 32 F = (1.8 x 120.0) + 32
= 216 + 32
= 248 (rounded)
Practice Question
Atomic masses:C = 12.01 amuCa = 40.08 amuH = 1.008 amuK = 39.10 amuP = 30.97 amuS = 32.07 amuO = 16.00 amu
Calculate the molecular mass of K2SO4:
Practice Question
Atomic masses:C = 12.01 amuCa = 40.08 amuH = 1.008 amuK = 39.10 amuP = 30.97 amuS = 32.07 amuO = 16.00 amu
Calculate the molecular mass of K2SO4:K: 2 atoms X 39.10 = 78.20S: 1 atom X 32.07 = 32.07O: 4 atoms X 16.00 = 64.00
Total: 174.27 amu