Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Similar presentations


Presentation on theme: "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."— Presentation transcript:

1 Part B-2 and Part C Questions 2011 Exam

2 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

3 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]

4 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 are emitted 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.

5 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 electron configuration so 2 valence electrons).

6 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.

7

8 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 pm Use P table to find Atomic # of Sr (38) and then use your graph

9 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.

10 Base your answers to questions 60 and 61 on the information below. The balanced equation below represents the decomposition of potassium chlorate. 2KClO 3 (s) → 2KCl(s) + 3O 2 (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 KClO 3 ) 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).

11 Base your answers to questions 62 and 63 on the information below. At 550°C, 1.00 mole of CO 2 (g) and 1.00 mole of H 2 (g) are placed in a 1.00-liter reaction vessel. The substances react to form CO(g) and H 2 O(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 CO 2 (g) between time t 0 and time t 1. [1] 63 What can be concluded from the graph about the concentrations of the reactants and the concentrations of the products between time t 1 and time t 2 ? [1]

12 Base your answers to questions 62 and 63 on the information below. At 550°C, 1.00 mole of CO 2 (g) and 1.00 mole of H 2 (g) are placed in a 1.00-liter reaction vessel. The substances react to form CO(g) and H 2 O(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 CO 2 (g) between time t 0 and time t 1. 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 t 1 and time t 2 ? Answer: The concentrations of reactants and products do not change ( the concentrations become constant)

13 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. Br 2 + C 3 H 6 C 3 H 6 Br 2 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.

14 Q are based on the following Unbalanced Equation O 2 (g) O 3 (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

15 Q are based on the following Unbalanced Equation O 2 (g) O 3 (g) (“electricity” written over the arrow) 66. Balance the equation using the smallest whole-number coefficients 3 O 2 2 O 3 (Balance for Mass; 6 O atoms on each side) 67. Identify the type of Bonding between the atoms of oxygen Ans: 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

16 Q 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 GasBoiling Point at Standard Pressure ( o C ) Butane-0.5 Ethane-88.6 Methane Propane-42.1

17 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 ethanepropanebutane Photos from POGIL on Distillation and Wikipedia Components of Natural Gas Boiling Point at Standard Pressure ( o C ) Butane-0.5 Ethane-88.6 Methane Propane-42.1

18 Typical lab Distillation Apparatus

19 Q are based on the following reaction NaCl + NH 3 + CO 2 + H 2 O NaHCO 3 + NH 4 Cl 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.

20 NaCl + NH 3 + CO 2 + H 2 O NaHCO 3 + NH 4 Cl 71. Either NaHCO 3 or NH 4 Cl are acceptable answers Compounds that contain polyatomic ions (HCO 3 - or NH 4 + will have both ionic bonds and covalent bonds (Table E: Selected Polyatomic Ions) 72.ANS: The electronegativity difference between oxygen and hydrogen ( = 1.2) is greater than the electronegativity difference between nitrogen and hydrogen ( = 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 NH 3 (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)

21 73. Ans: Draw a dot diagram for NH 3 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

22 74. Classify each mixture using the term “homogeneous” or the term “heterogeneous”. 75. Determine the volume of the Fe filings used to produce mixture Describe a procedure to physically remove the water from mixture 1. Mixture 1Mixture 2 compositionNaCl in H 2 OFe filings in H 2 O Student observations Colorless liquid No visible solid in bottom of beaker Colorless liquid Black solid on bottom of beaker Other dataMass of dissolved NaCl (s) = 2.9 g Mass of Fe (s) = 15.9 g Density of Fe(s) = 7.87 g/cm 3 Questions 74 – 76 are based on the following Data Table

23 74. Mixture 1 is homogeneous, Mixture 2 is heterogeneous. 75. Density = 7.87 g/cm 3 so 7.87 g in 1 cm 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 1Mixture 2 compositionNaCl in H 2 OFe filings in H 2 O Student observations Colorless liquid No visible solid in bottom of beaker Colorless liquid Black solid on bottom of beaker Other dataMass of dissolved NaCl (s) = 2.9 g Mass of Fe (s) = 15.9 g Density of Fe(s) = 7.87 g/cm 3 1 cm g 2.02

24 Heating Salt Water removes the H 2 O 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/4/43/Image-Zigong_Salt.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)

25 Q 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 CuCl 2 (aq) 3 Cu (s) + 2 AlCl 3 (aq) ProcedureObservation In a beaker, completely dissolve 5.0 g of CuCl 2 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 for youtube video for this experiment

26 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) ProcedureObservation In a beaker, completely dissolve 5.0 g of CuCl 2 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

27 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 CuCl 2 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 continued 2 Al (s) + 3 CuCl 2 (aq) 3 Cu (s) + 2 AlCl 3 (aq)

28 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 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

29 What happens when CO 2 (g) is bubbled through water ? (Hint; See table K Common Acids

30 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 M (mol/L) New H + concentration = 1 x ÷ 1000 = 1 x M Remember pH = -log (H + ) so the new pH = 6

31 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)

32 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 4 He + ________ 2 84

33 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 4 He Pb

34 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

35 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

36


Download ppt "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."

Similar presentations


Ads by Google