Exam 1 Practice
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Scientific Method, Properties, Classification of Matter
Classify each statement as being either qualitative or quantitative.
- This recitation worksheet took a good amount of time to complete
- This recitation worksheet took 50 minutes to complete
- The University of Georgia is a very large school
Solution
Answer:
- qualitative
- quantitative
- qualitative
Concept: Classificying data
Which statement best explains the difference between a law and a theory?
- A law is truth whereas a theory is mere speculation.
- A law summarizes a series of related observations, while a theory gives the underlying reasons for them.
- A theory describes what nature does; a law explains why nature does it.
Solution
Answer: B
Concept: Scientific method
A law is the “what” and a theory is the “why”.
A scientific law…
- is a statement that summarizes past observations and makes predictions.
- is subject to change via legal action.
- contains an explanation of observations.
- must have a mathematical formula.
- cannot be modified.
Solution
Answer: A
Concept: Scientific method
A scientific law summarizes past observations and makes predictions about natural phenomena without explaining the underlying mechanisms.
All statements about scientific theories are true except
- they explain why nature behaves the way it does.
- they must have the ability to make predictions on future behavior.
- they should use observations to test the theory.
- they are derived from hypothesis.
- they are speculation.
Solution
Answer: E
Concept: Scientific method
Scientific theories are well-substantiated explanations of natural phenomena, based on a large body of evidence from repeated experiments and observations.Which of these represents a hypothesis?
- Sodium reacts with water to form sodium hydroxide and hydrogen gas.
- Nitrogen gas is a fairly inert substance.
- Nickel has a silvery sheen.
- When a substance combusts, it combines with air.
- When wood burns, heat is given off.
Solution
Answer: D
Concept: Scientific method
D is a hypothesis because it is a tentative explanation that is testable and falsifiable.
A is an observed chemical reaction and a statement of fact.
B is a general property of nitrogen and a descriptive statement.
C is a physical property of nickel and a statement of observation.
E is a description of an observed phenomenon and not a testable explanation.Which of the following is a homogeneous mixture?
- wine
- mud
- salad
- salsa
Solution
Answer: A
Concept: Mixtures
Which of the following is a heterogeneous mixture?
- sports drink
- chlorine gas
- black coffee
- bowl of cereal
- carbon (graphite)
Solution
Answer: D
Concept: Mixtures
Which of the following is not an example of a homogeneous mixture?
- air
- rain
- steel
- blood
- coffee
Solution
Answer: D
Concept: Mixtures
Molecules can be described as
- mixtures of two or more pure substances
- mixtures of two or more elements that has a specific ratio between components
- two or more atoms chemically joined together
- heterogeneous mixtures
- homogeneous mixtures
Solution
Answer: C
Concept: Classification of matter
Which of these separations cannot be achieved using a physical process and, therefore, requires a chemical process?
- salt from sea water
- water from honey
- hydrogen from water
- ice from water
- fish from water
Solution
Answer: C
Concept: Separation of mixtures
H2O is a molecule and each hydrogen atom is chemically bonded to an oxygen atom.
A, D, and E can be separated via filtration. B is separated via distillation.
All of the following will lead to a heterogeneous mixture except (select all that apply)
- sugar and coffee
- mud and water
- salt and water
- tea and no-pulp lemonade
- vinegar and water
Solution
Answer: A, C, D, E
Concept: Mixtures
How are gases different from solids and liquids?
- Gases can only be made up of atoms
- Particles in a gas attract each other more strongly than in solids and liquids
- Gases are compressible
- Only gases can take the shape of their container
- Gases are colorless
Solution
Answer: C
Concept: States of matter
Identify the incorrect statement.
- Helium in a balloon is an element
- Paint is a mixture
- Tap water is a compound
- Mercury in a barometer is an element
Solution
Answer: C
Concept: Classification of matter
Tap water is not a compound as it is a mixture/solution that contains many other dissolved particles such as O2(g), Mg, Fe, Ca, etc. Pure water, H2O(l), would be a compound.Which answer includes all the following that are chemical changes and not physical changes?
I. freezing of water
II. rusting of iron
III. dropping a piece of iron into hydrochloric acid where H2(g) is produced
IV. burning of a piece of wood
V. emission of light by a kerosene oil lamp- III and IV
- II and V
- I, II, III, IV, and V
- II, III, and V
- II, III, IV, and V
Solution
Answer: E
Concept: chemical and physical changes
Which answer includes all of the following properties of sulfur that are physical properties and not other properties?
I. It reacts with hydrogen when heated
II. It is a yellow solid at room temperature
III. It is soluble in carbon disulfide
IV. It has a density of 2.97 g cm–3
V. It melts at 112 °C- II, III, IV, and V
- II, IV, and V
- I
- II, III, and IV
- III, IV, and V
Solution
Answer: A
Concept: chemical and physical properties
Which of the following is a chemical change?
- forging of metal
- boiling water
- shattering glass
- rust forming on a bike
- melting ice
Solution
Answer: D
Concept: chemical changes
Which of the following would you expect to have the highest density?
- oxygen
- lead
- water
- helium
Solution
Answer: B
Concept: chemical changes
Oxygen and helium are gasses and water is a liquid at room temperature. States of matter with less order are more likely to be less dense (although not always)
Measurements, Derived Units, Significant Figures
A student is using a graduated cylinder to measure the volume of a liquid. What reading should be recorded?
Solution
Answer: 24.0 mL; the zero in the tenths-place is the uncertain (estimated) number
Concept: uncertainty; measurement
A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.11 g mL–1, 1.81 g mL–1, 1.95 g mL–1, 1.75 g mL–1. The actual value for the density of the sugar solution is 1.75 g mL–1. Determine if the following statements are true or false.
- Her results are precise.
- Her results are accurate.
Solution
Answer: A and B are both false.
Concept: Accuracy and precision
A student goes to the lab and is tasked with measuring the mass of a block of nickel metal that has a known mass of 15.00 g. The student takes three different measurements in which they collect the following masses: 13.29 g, 13.28 g, and 13.29 g. What statement best describes their results?
- Their measurements are both precise and accurate
- Their measurements are precise, but not accurate
- Their measurements are accurate, but not precise
- Their measurements are neither precise or accurate
Solution
Answer: B
Concept: Accuracy and precision
Write the following numbers in normalized scientific notation to three significant figures.
- 0.000 467 01
- 8 973.002
- 0.027 902 0
- 32.451 3
- 0.000 090 999
Solution
Answer:
- 4.67 × 10–4
- 8.97 × 103
- 2.79 × 10–2
- 3.25 × 101
- 9.10 × 10–5
Concept: Scientific notation
How many significant figures are in each number?
- 554 km
- 7 pennies
- 1.01 × 105 m
- 0.00099 s
- 1.4500 km
- 21 000 m
Solution
Answer:
- 3
- ∞ (obtained via counting)
- 3
- 2
- 5
- ambiguous but considered to be 2 for this class
Concept: Significant figures
Determine the number of significant figures for each of the following values.
- 4 700
- 0.470 0
- 4 700 pencils
- 00.470 0
- 4 700.0 g
- 4 700.010 0
Solution
Answer:
- ambiguous but considered to be 2 for this class
- 4
- ∞ (obtained via counting)
- 4
- 5
- 8
Concept: Significant figures
How many significant figures are in the answer to the following calculation?
\[\left ( 29.0025 + 0.2 \right ) \left (6.134 - 6.101 \right ) / 5.196\times 10^{-2}\]
Solution
Answer: 2
Concept: Significant figures
\[\begin{align*} (29.0025 + 0.2)(6.134-6.101) / 5.196\times 10^{-2} &= \\ (29.\bar{2}02)(0.03\bar{3}0) / 5.196\times 10^{-2} &= \\ (0.9\bar{6}36) / 5.196\times 10^{-2} &= 1\bar{8}.546 \\ &= 19 \end{align*}\]
What answer should be reported, with the correct number of significant figures, for the following calculation? Report your answer in scientific notation.
\[\left ( 433.621 - 333.9 \right ) \times 11.900\]
Solution
Answer: 1.19 × 103
Concept: Significant figures
\[\begin{align*} \left ( 433.621 - 333.9 \right ) \times 11.900 &= \\[1.5ex] \left ( 99.\bar{7}21 \right ) \times 11.900 &= 11\bar{8}6.7 \\[1.5ex] &= 1,190 \\[1.5ex] &= 1.19\times 10^3 \end{align*}\]
Perform the following conversions below. Report your answers in normalized scientific notation to the appropriate number of significant figures.
- 145.21 kg to mg
- 0.490 nm to m
- 42.1 cm3 to L
- 1.73 × 105 m3 to cm3
Solution
Answer:
- 1.4521 × 108 mg
- 4.50 × 10–10 m
- 4.21 × 10–2 L
- 1.73 × 1011 cm3
Concept: Significant figures; scientific notation
\[\begin{align*} \mathbf{A.}~ &145.21~\mathrm{kg} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{10^3~\mathrm{mg}}{\mathrm{g}} \right ) = 1.4521\times 10^{8}~\mathrm{mg} \\[1.5ex] \mathbf{B.}~ &0.490~\mathrm{nm} \left ( \dfrac{\mathrm{m}}{10^9~\mathrm{nm}} \right ) = 4.50\times 10^{-10}~\mathrm{m} \\[1.5ex] \mathbf{C.}~ &42.1~\mathrm{cm}^3 \left ( \dfrac{\mathrm{mL}}{\mathrm{cm}^3} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) = 4.21\times 10^{-2}~\mathrm{L}\\[1.5ex] \mathbf{D.}~ &1.73\times 10^5~\mathrm{m}^3 \left ( \dfrac{\mathrm{10^2~cm}}{\mathrm{m}} \right )^3 = 1.73\times 10^{11}~\mathrm{cm^3} \end{align*}\]
Which of the following options is at the highest temperature?
- 71.4 °F
- 22.1 °C
- 294.65 K
Solution
Answer: B
Concept: Temperature
This solution converts all temperatures to °C.
\[\begin{align*} \textbf{A.}~~~ T\left ( ^{\circ}\mathrm{C} \right ) &= \dfrac{ [ T(^{\circ}\mathrm{F}) - 32 ]}{1.8} \\[1.5ex] &= \dfrac{ \left ( 71.4~^{\circ}\mathrm{F} - 32\right ) }{1.8} \\[1.5ex] &= 21.\bar{8}8~^{\circ}\mathrm{C} \\[1.5ex] &= 21.9~^{\circ}\mathrm{C}\\[6ex] \textbf{B.}~~~ T\left ( ^{\circ}\mathrm{C} \right ) &= 22.1~^{\circ}\mathrm{C} \\[6ex] \textbf{C.}~~~ T\left (^{\circ}\mathrm{C} \right ) &= T(\mathrm{K}) - 273.15 \\[1.5ex] &= 294.65~\mathrm{K} - 273.15 \\[1.5ex] &= 21.50~^{\circ}\mathrm{C} \end{align*}\]
The average daytime temperatures on Earth and Jupiter are 72 °F and 313 K, respectively. Which planet is hotter, on average, and by how much (in °C)?
Solution
Answer: Jupiter, 18 °C
Concept: Temperature
\[\begin{align*} T_{\mathrm{Earth}}\left ( ^{\circ}\mathrm{C} \right ) &= \dfrac{ [ T_{\mathrm{Earth}}(^{\circ}\mathrm{F}) - 32 ]}{1.8} \\[1.5ex] &= \dfrac{\left ( 72~^{\circ}\mathrm{F} - 32 \right )}{1.8} \\[1.5ex] &= 2\bar{2}.22~^{\circ}\mathrm{C}\\[6ex] T_{\mathrm{Jupiter}}\left (^{\circ}\mathrm{C} \right ) &= T_{\mathrm{Jupiter}}(\mathrm{K}) - 273.15 \\[1.5ex] &= 313~\mathrm{K} - 273.15 \\[1.5ex] &= 3\bar{9}.85~^{\circ}\mathrm{C} \\[6ex] \Delta T &= T_{\mathrm{Jupiter}}\left (^{\circ}\mathrm{C} \right ) - T_{\mathrm{Earth}}\left (^{\circ}\mathrm{C} \right ) \\[1.5ex] &= 3\bar{9}.85~^{\circ}\mathrm{C} - 2\bar{2}.22~^{\circ}\mathrm{C}\\[1.5ex] &= 1\bar{7}.63~^{\circ}\mathrm{C} \\[1.5ex] &= 18~^{\circ}\mathrm{C} \end{align*}\]
What is the length (in mm) of 3.20 cm?
Solution
Answer: 32.0 mm
Concept: Metric conversions
\[\begin{align*} 3.20~\mathrm{cm} \left ( \dfrac{\mathrm{m}}{10^2~\mathrm{cm}} \right ) \left ( \dfrac{10^3~\mathrm{mm}}{\mathrm{m}} \right ) = 32.0~\mathrm{mm} \end{align*}\]
If the walls in a room are 955 ft2 in area, and a gallon of paint covers 15 yd2, how many gallons of paint are needed for the room? (3 ft = 1 yd)
Solution
Answer: 7.1 gal
Concept: Conversions
\[\begin{align*} 955~\mathrm{ft^2} \left ( \dfrac{\mathrm{gal}}{15~\mathrm{yd^2}} \right ) &= 7.\bar{0}74~\mathrm{gal} \\[1.5ex] &= 7.1~\mathrm{gal} \end{align*}\]
A piece of metal ore has a mass of 9.25 g. When a student places it into a graduated cylinder containing water, the liquid level rises from 21.25 mL to 26.47 mL. What is the density (in g mL–1) of the ore?
Solution
Answer: 1.77 g mL–1
Concept: Density; volume displacement
\[\begin{align*} V(\mathrm{ore}) &= V_{\mathrm{f}} - V_{\mathrm{i}} \\[1.5ex] &= 26.47~\mathrm{mL} - 21.25~\mathrm{mL} \\[1.5ex] &= 5.22~\mathrm{mL}\\[3ex] d(\mathrm{ore}) &= \dfrac{m}{V} \\[1.5ex] &= \dfrac{9.25~\mathrm{g}}{5.22~\mathrm{mL}} \\[1.5ex] &= 1.7\bar{7}20~\mathrm{g~mL^{-1}}\\[1.5ex] &= 1.77~\mathrm{g~mL^{-1}} \end{align*}\]
A sample containing 33.42 g of metal pellets is poured into a graduated cylinder that initially contains 12.7 mL of water. The final water level in the cylinder is 21.6 mL. What is the density (in g mL–1) of the metal?
Solution
Answer: 3.8 g mL–1
Concept: Density; volume displacement
\[\begin{align*} \Delta V &= V_{\mathrm{final}} - V_{\mathrm{initial}} \\[1.5ex] &= 21.6~\mathrm{mL} - 12.7~\mathrm{mL} \\[1.5ex] &= 8.9~\mathrm{mL}\\[3ex] d &= \dfrac{m}{V} \\[1.5ex] &= \dfrac{33.42~\mathrm{g}}{8.9~\mathrm{mL}} \\[1.5ex] &= 3.\bar{7}55~\mathrm{g~mL^{-1}} = 3.8~\mathrm{g~mL^{-1}} \end{align*}\]
Nitric acid is a very important industrial chemical and 1.612 × 1010 lb were produced in 1992. If the density of nitric acid is 12.53 lb gal–1, what is the volume (in L)? 1 gal = 3.7854 L
Solution
Answer: 4.870 × 109 L
Concept: Conversions
\[\begin{align*} 1.612\times 10^{10}~\mathrm{lb} \left ( \dfrac{1~\mathrm{gal}}{12.53~\mathrm{lb}} \right ) \left ( \dfrac{3.7854~\mathrm{L}}{1~\mathrm{gal}} \right ) &= 4.86\bar{9}96\times 10^{9}~\mathrm{L} \\[1.5ex] &= 4.870\times 10^{9}~\mathrm{L} \end{align*}\]
A rectangular block has the following dimensions: 2.9 cm × 3.5 cm × 10.0 cm. The mass of the block is 615.0 g. What is the volume (in cm3) and density (in kg m–3) of the block?
Solution
Answer: 1.0 × 102 cm3; 6.1& times; 103 kg m–3
Concept: volume; density
\[\begin{align*} V &= l \times w \times h \\[1.5ex] &= 2.9~\mathrm{cm} \times 3.5~\mathrm{cm} \times 10.0~\mathrm{cm} \\[1.5ex] &= 1\bar{0}1.5~\mathrm{cm^3} \\[1.5ex] &= 1.0\times 10^2~\mathrm{cm^3} \\[3ex] d &= \dfrac{m}{V} \\[1.5ex] &= \dfrac{615.0~\mathrm{g}}{1\bar{0}1.5~\mathrm{cm^3}} \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \left ( \dfrac{10^2~\mathrm{cm}}{\mathrm{m}} \right )^3 \\[1.5ex] &= 6\bar{0}59~\mathrm{kg~m^{-3}} = 6.1\times 10^{3}~\mathrm{kg~m^{-3}} \end{align*}\]
At room temperature, elemental bromine (Br2) is a liquid with a density of 3.12 g cm–3. What is the mass (in g) of 125 mL of bromine and what volume (in mL) does 85.0 g of bromine occupy?
Solution
Answer: 390. g; 27.2 mL
Concept: volume; density
\[\begin{align*} d &= \dfrac{m}{V} \longrightarrow \\[1.5ex] m &= Vd \\ &= 125~\mathrm{mL} \left ( \dfrac{3.12~\mathrm{g}}{\mathrm{cm}^3} \right ) \left ( \dfrac{\mathrm{cm}^3}{\mathrm{mL}} \right ) \\[1.5ex] &= 390.~\mathrm{g}\\[3ex] d &= \dfrac{m}{V} \longrightarrow \\[1.5ex] V &= \dfrac{m}{d} \\[1.5ex] &= \dfrac{85.0~\mathrm{g}}{3.12~\mathrm{g~cm^{-3}}} \left ( \dfrac{\mathrm{mL}}{\mathrm{cm}^3} \right ) \\[1.5ex] &= 27.\bar{2}43~\mathrm{mL} \\[1.5ex] &= 27.2~\mathrm{mL} \end{align*}\]
The Hindenburg contained 7.062 × 106 ft3 of a gas. If the gas was helium ( d(He) = 1.78 × 10–4 g mL–1 ), how much lighter (in kg) would it be than if the gas was air (d(air) = 1.293 × 10–3 g mL–1)? (1 ft = 0.3048 m)
Solution
Answer: 2.23 × 105 kg
Concept: volume; density
\[\begin{align*} V &= 7.062\times 10^{6}~\mathrm{ft}^3 \left ( \dfrac{0.3048~\mathrm{m}}{\mathrm{ft}} \right )^3 \left ( \dfrac{10^2~\mathrm{cm}}{\mathrm{m}} \right )^3 \left ( \dfrac{\mathrm{mL}}{\mathrm{cm^3}} \right ) \\[1.5ex] &= 2.000\times 10^{11}~\mathrm{mL}\\[3.0ex] d &= \dfrac{m}{V} \longrightarrow \\[1.5ex] m(\textrm{He}) &= Vd(\textrm{He}) \\[1.5ex] &= 2.000\times 10^{11}~\mathrm{mL} \left ( \dfrac{1.78\times 10^{-4}~\mathrm{g}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[1.5ex] &= 3.5\bar{6}0\times 10^{4}~\mathrm{kg}\\[3ex] m(\textrm{air}) &= Vd(\textrm{air}) \\[1.5ex] &= 2.000\times 10^{11}~\mathrm{mL} \left ( \dfrac{1.293\times 10^{-3}~\mathrm{g}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[1.5ex] &= 2.58\bar{6}0\times 10^{5}~\mathrm{kg}\\[3ex] \Delta m &= m(\textrm{air}) - m(\textrm{He}) \\[1.5ex] &= 2.58\bar{6}0\times 10^{5}~\mathrm{kg} - 3.56\times 10^{4}~\mathrm{kg} \\[1.5ex] &= 25.8\bar{6}0\times 10^{4}~\mathrm{kg} - 3.56\times 10^{4}~\mathrm{kg} \\[1.5ex] &= (25.8\bar{6}0 - 3.56) \times 10^4 \\[1.5ex] &= 22.30\times 10^4~\mathrm{kg} \\[1.5ex] &= 2.23\times 10^{5}~\mathrm{kg} \end{align*}\]
Atoms, Atomic Theory, Periodic Table
An unused flashbulb contains magnesium and oxygen. After use, the contents are changed to magnesium oxide but the total mass does not change. This observation can best be explained by the…
- Law of Constant Composition
- Law of Multiple Proportions
- Avogadro’s Law
- Law of Conservation of Mass
Solution
Answer: D
Concept: Laws
Which of the statements about two isotopes is false?
- They will have the same charge on the nucleus.
- They will have different numbers of neutrons.
- They will have essentially the same chemical reactivity.
- They will have the same atomic numbers.
- They will have the same atomic weights.
Solution
Answer: E
Concept: atomic structure; isotopes
What is the mass number for 25Mg?
Solution
Answer: 25
Concept: atomic structure; isotopes
Which of the following represents isotopes?
I. \(^{46}_{21}X\)
II. \(^{46}_{22}X\)
III. \(^{46}_{21}X\)
IV. \(^{46}_{23}X\)- I and II
- I and III
- I and IV
- III and IV
Solution
Answer: B
Concept: atomic structure; isotopes
Isotopes of a given element vary in mass number have the same atomic number.
Complete the table.
Solution
Answer: B
Concept: atomic structure; isotopes
Write the group number for each of the following elements.
- Ba
- Sb
- Zr
- Tl
- Rn
Solution
Answer:
- group 2 (or 2A)
- group 15 (or 5A)
- group 4 (or 4B)
- group 13 (or 3A)
- group 18 (or 8A)
Concept: periodic table
Write the period number for each of the following elements.
- Hf
- Am
- Ga
- Si
- He
Solution
Answer:
- 6
- 7
- 4
- 3
- 1
Concept: periodic table
Which of the following elements is a metal?
- Ge
- Sb
- H
- Sn
- Cl
Solution
Answer: D
Concept: periodic table
Which of the following elements is a nonmetal?
- Ca
- Si
- Mo
- Cs
- Se
Solution
Answer: E
Concept: periodic table
What quantity did J. J. Thompson determine using a cathode ray tube?
- the mass of an electron
- the charge-to-mass ratio of an electron
- the density of a proton
- the charge of a neutron
- the charge-to-mass ratio of a proton
Solution
Answer: B
Concept: famous experiments
Which of the following is found in the nucleus?
- cations
- neutrons
- electrons
- anions
- none of these
Solution
Answer: B
Concept: atomic structure
Which of these is not found in an atom?
- electron
- neutron
- photon
- proton
- all of them are
Solution
Answer: C
Concept: atomic structure
True or False: an atom consists mostly of empty space.
- true
- false, an atom consists primarly of electrons
- false, an atom consists primarily of the nucleus
- false, an atom consists primarly of neutrons
- false, an atom consists primarly of protons
Solution
Answer: A
Concept: atomic structure
Which of the following contribute to the mass of an atom? (select all that apply)
- protons
- neutrons
- electrons
- electrons
- the nucleus
- atomic charge
Solution
Answer: A and B
Concept: atomic structure
An element consists of four naturally occurring isotopes. The percent abundance and isotope masses are given as:
- 1.40 %; 203.973 u
- 24.10 %; 205.9754 u
- 22.10 %; 206.9759 u
- 52.40 %; 207.9766 u
Find the standard atomic weight (in u) of the element
Solution
Answer: 207.2 u
Concept: percent abundance; isotopes
\[\begin{align*} A_{\mathrm{r}}^{\circ}(E) &= \left ( \dfrac{x \! \left ( X_1 \right )\%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left ( X_1 \right ) \right ] + \left ( \dfrac{x \! \left ( X_2 \right )\%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left ( X_2 \right ) \right ] + \\ &\phantom{=}~\, \left ( \dfrac{x \! \left ( X_3 \right )\%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left ( X_3 \right ) \right ] + \left ( \dfrac{x \! \left ( X_4 \right )\%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left ( X_4 \right ) \right ] \\[2ex] A_{\mathrm{r}}^{\circ}(\mathrm{B}) &= \left ( \dfrac{1.40~\%}{100~\%} \right ) \left ( 203.973~\mathrm{u} \right ) + \left ( \dfrac{24.10~\%}{100~\%} \right ) \left ( 205.9754~\mathrm{u} \right ) + \\ &\phantom{=}~\, \left ( \dfrac{22.10~\%}{100~\%} \right ) \left ( 206.9759~\mathrm{u} \right ) + \left ( \dfrac{52.40~\%}{100~\%} \right ) \left ( 207.9766~\mathrm{u} \right ) \\[2ex] &= 20\bar{7}.21~\mathrm{u} \\[0.5ex] &= 207~\mathrm{u} \end{align*}\]
The element rhenium (Re) has a standard atomic weight of 186.207 and exists as two naturally occurring isotopes. Given the following percent abundance and isotope mass
- 185Re
- 187Re: 62.60 %; 186.956
find the relative atomic weight of rhenium-185.
Solution
Answer: 185
Concept: percent abundance; isotopes
\[\begin{align*} A_{\mathrm{r}}^{\circ} \left ( \mathrm{Re} \right ) &= x \left (\mathrm{^{185}Re} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{185}Re} \right ) + x \left (\mathrm{^{187}Re} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{187}Re} \right ) \longrightarrow \\[1.5ex] A_{\mathrm{r}} \left ( \mathrm{^{185}Re} \right ) &= \dfrac{ A_{\mathrm{r}}^{\circ} \left ( \mathrm{Re} \right ) - \Biggr [ x \left (\mathrm{^{187}Re} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{187}Re} \right ) \Biggr] } {x \left (\mathrm{^{185}Re} \right )}\\[0.5ex] &= \dfrac{ 186.207 - \Biggr[ \left ( \dfrac{62.60~\%}{100~\%} \right ) \left ( 186.956 \right ) \Biggr] } { \dfrac{\left ( 100~\%-62.60~\%\right ) }{100~\%} }\\[0.5ex] &= 184.\bar{9}53 \\[0.5ex] &= 185 \end{align*}\]
Ions, Ionic Compounds, Molecules, and Nomenclature
What is the correct charge of a sulfur atom with 16 protons, 16 neutrons, and 18 electrons?
Solution
Answer: 2–
Concept: atomic structure; ions
What charge will most likely be formed by lithium?
Solution
Answer: 1+
Concept: ions
Is iodine more likely to gain or lose an electron?
Solution
Answer: Gain an electron (to increase its stability)
Concept: ions
What is the atomic mass (in amu) of a magnesium ion containing 12 protons, 13 neutrons, and 10 electrons?
Solution
Answer: 25 amu
Concept: isotopes
Fill out the missing information regarding a binary ionic compound.
- Name: sodium chloride
- Cation:
- Anion:
- Formula:
Solution
Answer:
- Name: sodium chloride
- Cation: Na+
- Anion: Cl−
- Formula: NaCl
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation: Mg2+
- Anion: O2–
- Formula:
Solution
Answer:
- Name: magnesium oxide
- Cation: Mg2+
- Anion: O2–
- Formula: MgO
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation:
- Anion:
- Formula: Al2S3
Solution
Answer:
- Name: aluminum sulfide
- Cation: Al3+
- Anion: S2–
- Formula: Al2S3
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name: iron(III) oxide
- Cation:
- Anion:
- Formula:
Solution
Answer:
- Name: iron(III) oxide
- Cation: Fe3+
- Anion: O2–
- Formula: Fe2O3
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation: Sn2+
- Anion: Cl–
- Formula:
Solution
Answer:
- Name: tin(II) chloride
- Cation: Sn2+
- Anion: Cl–
- Formula: SnCl2
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation:
- Anion:
- Formula: PbF4
Solution
Answer:
- Name: lead(IV) fluoride
- Cation: Pb4+
- Anion: F–
- Formula: PbF4
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name: sodium hydroxide
- Cation:
- Anion:
- Formula:
Solution
Answer:
- Name: sodium hydroxide
- Cation: Na+
- Anion: OH–
- Formula: NaOH
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation: Mg2+
- Anion: CO32–
- Formula:
Solution
Answer:
- Name: magnesium carbonate
- Cation: Mg2+
- Anion: CO32–
- Formula: MgCO3
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation:
- Anion:
- Formula: Na2SO3
Solution
Answer:
- Name: sodium sulfite
- Cation: Na+
- Anion: SO32–
- Formula: Na2SO3
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation: Fe2+
- Anion: OH–
- Formula:
Solution
Answer:
- Name: iron(II) hydroxide
- Cation: Fe2+
- Anion: OH–
- Formula: Fe(OH)2
Concept: ionic compounds
Fill out the missing information regarding a binary ionic compound.
- Name:
- Cation:
- Anion:
- Formula: TiPO3
Solution
Answer:
- Name: titanium(III) phosphate
- Cation: Ti3+
- Anion: PO33–
- Formula: TiPO3
Concept: ionic compounds
Provide the names for the following compounds
- N2O
- P2O5
- SF2
- S2O
- P2I4
Solution
Answer:
- dinitrogen monoxide
- diphosphorous pentoxide
- sulfur difluoride
- disulfur monoxide
- diphosphorous tetraiodide
Concept: ionic compounds
What is the proper chemical formula for lithium hypoiodite?
- LiIO
- LiO2
- LiIO3
- LiIO3
- LiIO4
Solution
Answer: A
Concept: ionic compounds
Write the chemical formula for a compound formed between aluminum and sulfur.
Solution
Answer: Al2S3
Concept: ionic compounds
In which set do all elements tend to form cations in binary ionic compounds?
- Na, Al, S
- Ca, Mn, Sn
- P, As, Sb
- S, Cl, Br
Solution
Answer: B
Concept: ions
Which of the following contains both ionic and covalent bonds?
- CaBr2
- COS
- BaSO4
- SF6
- none of these
Solution
Answer: C
Concept: ionic compounds
What is the name of the C3H8?
- hexane
- propane
- decane
- butane
- ethane
Solution
Answer: B
Concept: nomenclature; organic molecule
What is the name of this compound?
Solution
Answer: octanol
Concept: nomenclature; organic molecule
What is the chemical formula for magnesium carbide?
Solution
Answer: Mg2C3
Concept: nomenclature
What is the name of PH3?
Solution
Answer: phosphine
Concept: nomenclature
Write the empirical formulas for the following compounds
- iron(II) sulfide
- iron(III) sulfide
- iron(II) sulfate
- iron(III) sulfate
- iron (II) sulfite
Solution
Answer:
- FeS
- Fe2S3
- FeSO4
- Fe2(SO4)3
- FeSO3
Concept: nomenclature
Provide the name for each of the following and identify it as a molecular or ionic compound
Solution
Answer: B
Concept: nomenclature
Provide the name for each of the following and identify it as a molecular or ionic compound
Solution
Answer: B
Concept: nomenclature
Provide the name for each of the following and identify it as an alkane, alcohol, or carboxylic acid.
Solution
Answer: B
Concept: nomenclature
Provide the chemical formulas and identify each as molecular or ionic compound. Include phase labels where necessary.
Solution
Answer: B
Concept: nomenclature
Provide the chemical formulas and identify each as molecular or ionic compound. Include phase labels where necessary.
Solution
Answer: B
Concept: nomenclature
Questions are written by UGA Chemistry unless otherwise indicated and translated with minor tweaks into an online format by Eric Van Dornshuld.