Part B-2 and Part C Questions 2011 Exam

Part B-2 Questions 2011 Exam

51 Explain, in terms of protons and neutrons, why U-235 and U-238 are different isotopes of uranium. [1]

Isotopes of an element have the same number of protons but different number of neutrons.

(Recall atomic number = # of protons; mass number = ( # of protons + neutrons)

U-235 versus U-238

Base your answers to questions 52 through 54 on the information below.The bright-line spectra for three elements and a mixture of elements are shown below.

52 Explain, in terms of both electrons and energy, how the bright-line spectrum of an element is produced. [1]53 Identify all the elements in the mixture. [1]54 State the total number of valence electrons in a cadmium atom in the ground state. [1]

Base your answers to questions 52 through 54 on the information below.The bright-line spectra for three elements and a mixture of elements are shown below.

52 Explain, in terms of both electrons and energy, how the bright-line spectrum of an element is produced. [1]

Possible Answers:

Bright line spectra are produced when electrons in an excited state return to a lower energy state; specific amounts of energy areemitted and these energies are associated with specific wavelengths of light that are characteristic of the bright-line spectrum of an element.

Or

Energy is emitted when excited electrons fall back to lower shells.

53 Identify all the elements in the mixture. [1] Answer: lithium and strontium (Sr and Li) Match spectral lines of unknown with the reference spectra.

54 State the total number of valence electrons in a cadmium atom in the ground state. [1] Answer: 2 valence electrons. (Cd has a 2-8-18-18-2 electron configuration so 2 valence electrons).

Base your answers to questions 55 through 59 on the information below.

The ionic radii of some Group 2 elements are given in the table below.

Base your answers to questions 55 through 59 on the information below.

The ionic radii of some Group 2 elements are given in the table below.

55 and 56 On the grid in your answer booklet, mark an appropriate scale on the axis labeled “Ionic Radius (pm).” [1] On the same grid, plot the data from the data table. Circle and connect the points. [1]

(57) Estimate the ionic radius of strontium. [1] Answer: 117 pm ± 2 pmUse P table to find Atomic # of Sr (38) and then use your graph

58 State the trend in ionic radius as the elements in Group 2 are considered in order of increasing atomic number. [1]

Answer: As the atomic number of elements in Group 2 increases, the ionic radius increases (Recall periodic trends.)

59 Explain, in terms of electrons, why the ionic radius of a Group 2 element is smaller than its atomic radius. [1]

Answer: A Group 2 ion has two fewer electrons than the atom from which it was formed.

Base your answers to questions 60 and 61 on the information below.The balanced equation below represents the decomposition of

potassium chlorate.2KClO3(s) → 2KCl(s) + 3O2(g)

60 Determine the oxidation number of chlorine in the reactant in the equation. [1]

Answer: Oxidation number is +5 . (Recall: use the sum rule to determine the oxidation number of chlorine in KClO3 )

61 State why the entropy of the reactant is less than the entropy of the products.

Answer: The gaseous product is more disordered than the solid reactant or The solid reactant is more ordered than the products (which are gas and solid).

Base your answers to questions 62 and 63 on the information below.At 550°C, 1.00 mole of CO2(g) and 1.00 mole of H2(g) are placed in a 1.00-liter reaction vessel. The substances react to form CO(g) and H2O(g). Changes in the concentrations of the reactants and the concentrations of the products are shown in the graph below.

62 Determine the change in the concentration of CO2(g) between time t0 and time t1. [1]

63 What can be concluded from the graph about the concentrations of the reactants and the concentrations of the products between time t1 and time t2? [1]

Base your answers to questions 62 and 63 on the information below.At 550°C, 1.00 mole of CO2(g) and 1.00 mole of H2(g) are placed in a 1.00-liter reaction vessel. The substances react to form CO(g) and H2O(g). Changes in the concentrations of the reactants and the concentrations of the products are shown in the graph below.

62 Determine the change in the concentration of CO2(g) between time t0 and time t1.

Answer: the concentration decreases by 0.27 mol/L

63 What can be concluded from the graph about the concentrations of the reactants and the concentrations of the products between time t1 and time t2?

Answer: The concentrations of reactants and products do not change ( the concentrations become constant)

Base your answers to questions 64 and 65 on the information below.A reaction between bromine and a hydrocarbon is represented by the

balanced equation below.

64 Identify the type of organic reaction. Br2 + C3H6 C3H6Br2

Answer: This is an addition reaction. (This is one of the types of organic reactions that you need to know)

65 Write the name of the homologous series to which the hydrocarbon belongs.

Answer: Use Table Q to help you identify the answer as alkenes.

• Q 66-68 are based on the following Unbalanced Equation

O2 (g) O3 (g)

66. Balance the equation using the smallest whole-number coefficients

67. Identify the type of Bonding between the atoms of oxygen

68.Identify , in terms of electron configuration, why an oxygen molecule is more stable than an oxygen atom.

electricity

• Q 66-68 are based on the following Unbalanced EquationO2 (g) O3 (g)

(“electricity” written over the arrow)

66. Balance the equation using the smallest whole-number coefficients

3 O2 2 O3 (Balance for Mass; 6 O atoms on each side)

67. Identify the type of Bonding between the atoms of oxygenAns: Oxygen has 6 valence electrons and is a nonmetal; bonding will be covalent

68.Identify , in terms of electron configuration, why an oxygen molecule is more stable than an oxygen atom.

Ans. A single O atom has six valence electrons and so does not have a stable octet

Q 69-70 are based on data for four components of natural gas

Q69. Identify a process used to separate the components of natural gas.

Q70. List the four components of natural gas in order of increasing strength of intermolecular forces.

Components of Natural Gas Boiling Point at Standard Pressure ( oC )

Butane -0.5

Ethane -88.6

Methane -161.6

Propane -42.1

69. Ans: Distillation(You want a separation technique based on differences in boiling points/condensation points). Distillation or selective condensation are possible processes).

Q70. List in order of increasing strength of intermolecular forces. ( recall: the weaker the forces of attraction between molecules, the lower the

boiling pts) Answer: methane ethane propane butane

Photos from POGIL on Distillation and Wikipedia

Components of Natural Gas

Boiling Point at Standard Pressure ( oC )

Butane -0.5

Ethane -88.6

Methane -161.6

Propane -42.1

Typical lab Distillation Apparatus

www.chemheritage.org

Q 71-73 are based on the following reaction

NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl

71. Write the chemical formula for one compound in the equation that contains both ionic bonds and covalent bonds.

72. Explain, in terms of electronegativity difference, why the bond between hydrogen and oxygen in a water molecule is more polar than the bond between hydrogen and nitrogen in an ammonia molecule.

73. In the space in your answer booklet, draw a Lewis dot diagram for the reactant containing nitrogen in the equation.

NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl71. Either NaHCO3 or NH4Cl are acceptable answers

Compounds that contain polyatomic ions (HCO3 - or NH4 +will have both ionic bonds and covalent bonds (Table E: Selected Polyatomic Ions)

72. ANS: The electronegativity difference between oxygen and hydrogen (3.4-2.2 = 1.2) is greater than the electronegativity difference between nitrogen and hydrogen (3.0-2.2 = 0.8) and so the bond between oxygen and hydrogen in water is more polar than the bond between nitrogen and hydrogen in ammonia. (Use table S for Electronegativity values)

73. Ans: Draw a dot diagram for NH3

(N has 5 valence electrons and each H atom has 1 valence electron; atoms will bond in order to achieve the same electron configuration as a noble gas)

73. Ans: Draw a dot diagram for NH3

• N has 5 valence electrons (2-5 is the electron configuration listed in P. Table) • H atoms have 1 valence electron • Atoms will bond to achieve the same electron configuration as a noble gas. • Start by drawing the dot structures of each atom• Make shared pairs where needed• Tidy up dot diagram

74. Classify each mixture using the term “homogeneous” or the term “heterogeneous”.

75. Determine the volume of the Fe filings used to produce mixture 2.

76. Describe a procedure to physically remove the water from mixture 1.

Mixture 1 Mixture 2

composition NaCl in H2O Fe filings in H2O

Student observations •Colorless liquid•No visible solid in bottom of beaker

•Colorless liquid•Black solid on bottom of beaker

Other data Mass of dissolved NaCl (s) = 2.9 g •Mass of Fe (s) = 15.9 g•Density of Fe(s) = 7.87 g/cm3

Questions 74 – 76 are based on the following Data Table

74. Mixture 1 is homogeneous, Mixture 2 is heterogeneous.

75. Density = 7.87 g/cm3 so 7.87 g in 1 cm3

15.9 g x ________ = ______ cm 3

(You are given mass and density and need to find the volume. Solve with dimensional analysis or use density formula in table T to solve for missing variable)

74. Water can be removed by…………..

Mixture 1 Mixture 2

composition NaCl in H2O Fe filings in H2O

Student observations •Colorless liquid•No visible solid in bottom of beaker

•Colorless liquid•Black solid on bottom of beaker

Other data Mass of dissolved NaCl (s) = 2.9 g •Mass of Fe (s) = 15.9 g•Density of Fe(s) = 7.87 g/cm3

1 cm3

7.87 g

2.02

Heating Salt Water removes the H2O from the Salt Water Mixture leaving NaCl behind.(Pictures from Wikipedia (http://upload.wikimedia.org/wikipedia/commons/4/43/Image-Zigong_Salt.jpg and http://upload.wikimedia.org/wikipedia/commons/9/9f/SaltCrystaUSGOV.jpg )

Other ways of removing water include” :(1)Just allowing the water to evaporate at room temperature(2) Removing water by distillation since Water and NaCl will have very different boiling points

Salt Crystal NaCl (s)

Q 77-79 involve a lab experiment with Aluminum foil and Copper(II) chloride solution. The reaction equation is given

and student observations are listed in a table.

2 Al (s) + 3 CuCl2 (aq) 3 Cu (s) + 2 AlCl3 (aq)

Procedure Observation

In a beaker, completely dissolve 5.0 g of CuCl2 in 80.0 mL of water.

•The solution is blue green

Cut 1.5 g of Al (s) foil into small pieces. Add all the foil to the mixture in the beaker. Stir the contents for 1 min.

•The surface of the Al(s) foil appears partially black.•The beaker feels warm to the touch.

Observe the beaker and contents after 10 minutes

•The liquid in the beaker appears colorless•A reddish-brown solid is seen at the bottom of the beaker•Some pieces of Al(s) with a partially black coating remain in the beaker

http://www.youtube.com/watch?v=zg-gsLEGk2A&feature=related for youtube video for this experiment

77. State one observation that indicates Cu+2 ions became Cu atoms.

Possible Answers: The solution is no longer blue or blue green (characteristic of

Cu+2 ions in aqueous solution) . Or A reddish-brown solid is formed (corresponds to color of copper)

Procedure Observation

In a beaker, completely dissolve 5.0 g of CuCl2 in 80.0 mL of water.

•The solution is blue green

Cut 1.5 g of Al (s) foil into small pieces. Add all the foil to the mixture in the beaker. Stir the contents for 1 min.

•The surface of the Al(s) foil appears partially black.•The beaker feels warm to the touch.

Observe the beaker and contents after 10 minutes

•The liquid in the beaker appears colorless•A reddish-brown solid is seen at the bottom of the beaker•Some pieces of Al(s) with a partially black coating remain in the beaker

78. Describe one change in the procedure that would cause the reaction to occur at a faster rate.

Possible Answers: • Increase the concentrations of the CuCl2 solution

• Cut the aluminum into even smaller pieces• Use a catalyst• Use heated solutions

79. State one safety procedure the student should perform after completing the laboratory activity.

Possible answers: wash hands before leaving the lab or dispose of chemicals as teacher instructs or wash and return equipment to cabinets, etc.

Q 77-79 continued 2 Al (s) + 3 CuCl2 (aq) 3 Cu (s) + 2 AlCl3 (aq)

Base your answers to questions 80 through 82 on the information below.

Some carbonated beverages are made by forcing carbon dioxide gas into a beverage solution. When a bottle of one kind of carbonated beverage is first opened, the beverage has a pH value of 3.

80. State, in terms of the pH scale, why this beverage is classified as acidic.

Ans: beverage is acidic because its pH is less than than 7

What happens when CO2 (g) is bubbled through water ? (Hint; See table K Common Acids

81 Using Table M, identify one indicator that is yellow in a solution that has the same pH value as this beverage.

Use Table L, Possible Answers = bromthymol blue, bromcresol green or thymol blue

82 After the beverage bottle is left open for several hours, the hydronium ion concentration in the beverage solution decreases to 1/1000 of the original concentration. Determine the new pH of the beverage solution.

Started at pH = 3 ( so H + ion concentration = 10 –pH = 1 x 10-3 M (mol/L)

New H+ concentration = 1 x 10-3 ÷ 1000 = 1 x 10-6 M Remember pH = -log (H+) so the new pH = 6

• Base your answers to questions 83 through 85 on the information below.

Polonium-210 occurs naturally, but is scarce. Polonium-210 is primarily used in devices designed to eliminate static electricity in machinery. It is also used in brushes to remove dust from camera lenses.

Polonium-210 can be created in the laboratory by bombarding bismuth-209 with neutrons to create bismuth-210. The bismuth-210 undergoes beta decay to produce polonium-210. Polonium-210 has a half-life of 138 days and undergoes alpha decay.

83.State one beneficial use of Po-210. (Extract answer from reading passage above)

• Base your answers to questions 83 through 85 on the information below.

Polonium-210 occurs naturally, but is scarce. Polonium-210 is primarily used in devices designed to eliminate static electricity in machinery. It is also used in brushes to remove dust from camera lenses.

Polonium-210 can be created in the laboratory by bombarding bismuth-209 with neutrons to create bismuth-210. The bismuth-210 undergoes beta decay to produce polonium-210. Polonium-210 has a half-life of 138 days and undergoes alpha decay.

84.Complete the nuclear equation in your answer booklet for the decay of Po-210, by writing a notation for the missing product.

(Use ref table O, N and periodic table)

Ans booklet: 210 Po 4He + ________ 284

• Base your answers to questions 83 through 85 on the information below.

Polonium-210 occurs naturally, but is scarce. Polonium-210 is primarily used in devices designed to eliminate static electricity in machinery. It is also used in brushes to remove dust from camera lenses.

Polonium-210 can be created in the laboratory by bombarding bismuth-209 with neutrons to create bismuth-210. The bismuth-210 undergoes beta decay to produce polonium-210. Polonium-210 has a half-life of 138 days and undergoes alpha decay.

84.Complete the nuclear equation in your answer booklet for the decay of Po-210, by writing a notation for the missing product. (Use ref table O, N and periodic table)

Ans booklet: 210Po 4He + 206Pb84 2 82

85. Determine the total mass of an original 28.0-milligram sample of Po-210 that remains unchanged after 414 days.

Steps: (1) determine the number of half-lives during the elapsed time period# of half-lives = total elapsed time ÷ half-life of isotope

414 days ÷ 138 = 3 half-lives

(2) Setup figure to predict the remaining mass

28.0 mg

85. Determine the total mass of an original 28.0-milligram sample of Po-210 that remains unchanged after 414 days.

Steps: (1) determine the number of half-lives during the elapsed time period# of half-lives = total elapsed time ÷ half-life of isotope

414 days ÷ 138 = 3 half-lives

(2) Setup figure to predict the remaining mass

28.0 mg 14.0 mg 7.0 mg 3.5 mg