CHEM 142 - Exam 1 Version B!!! DO NOT OPEN THIS EXAM BOOK UNTIL TOLD TO DO SO BY THE INSTRUCTOR!!! Instructor: Dr. Kari Pederson NO GRAPHING CALCULATORS ALLOWED Date: Friday, July 15 ONLY CALCULATORS MAY BE USED AS CALCULATORS (you may not use your cellular phone as a calculator) Time: 9:40-10:40am NO HEADPHONES ALLOWED (EARPLUGS ARE OK) Location: BAG 154 NO HATS WITH BRIMS ALLOWED!!! PLEASE READ THIS!!! Indicate all of the following on your scantron form or five points will be deducted from your exam score: First Name, Last Name, Student Number, Section, Exam Version YOUR FULL NAME: first name last name YOUR SECTION/SEAT: discussion section seat number EXAM 1 VERSION B Page 1 of 10
Question Points Possible Score 1-8 24 9-16 40 17-20 16 21 20 Scantron Info? -5 TOTAL 100 EXAM 1 VERSION B Page 2 of 10
MULTIPLE CHOICE: CONCEPTS. 8 @ 3 pts each 24 POINTS TOTAL Please mark the one correct answer for each of the following questions on your scantron. 1. Which of the following is a physical property of the element chlorine? A) Chlorine is a gas that is poisonous if inhaled. B) Chlorine will react with sodium to make salt. C) Chlorine is a gas that will become a liquid at 34.4 ºC. D) Chlorine is used in water treatment as a disinfectant. E) Chlorine exists as a diatomic molecule. 2. Which metric prefix symbol means 1 x 10-3? A) p B) m C) M D) k E) n 3. As chemists and other scientists seek to understand the principles that govern nature they employ what is called the scientific method. The first step in this method is. A) the proposal of a hypothesis B) the testing of a scientific law C) the making of observations D) the application for a research grant E) the testing of a hypothesis 4. Reaction of 10.0 grams of calcium with of sulfur gave 22.0 grams of calcium sulfide. All the calcium was used, but there were 20.0 grams of sulfur left over. A) 12.0 grams B) 22.0 grams C) 32.0 grams D) 42.0 grams Conservation of Mass: total mass reactants = total mass products mass sulfur = ( 22.0g calcium sulfide + 20.0g excess sulfur)! 10.0g calcium = 32.0g 5. Of the following, only would not be classified as a pure substance. A) sucrose B) acetic acid C) oxygen D) a granite rock E) water EXAM 1 VERSION B Page 3 of 10
6. Which statement best describes isotopes? A) atoms in the same vertical group B) atoms with the same number of neutrons and different numbers of protons C) atoms with the same number of protons and different numbers of neutrons D) atoms with the same numbers of protons and neutrons E) atoms in the same horizontal period 7. The element is classified as a metal whereas is classified as a metalloid. A) Sr, Ge B) Ni, Ba C) As, Sb D) Br, Si E) Rb, I 8. In an atom of 65 30 Zn. A) the number of protons equals the number of neutrons B) 30 electrons are found outside the nucleus C) 35 protons are found in the nucleus D) the total number of electrons and protons equals 65 E) 30 neutrons are found in the nucleus EXAM 1 VERSION B Page 4 of 10
MULTIPLE CHOICE: SHORT CALCULATIONS. 8 @ 5 pts each 40 TOTAL POINTS Please mark the one correct answer for each of the following questions on your scantron. 9. The volume of a ball bearing is 1.37 mm 3. What is its volume in cubic centimeters (cm 3 )? A) 1.37 x 10 3 cm 3 B) 1.37 x 10 1 cm 3 C) 1.37 x 10-1 cm 3 D) 1.37 x 10-3 cm 3 3 1cm 1.37mm 3 " %! # $ 10mm & ' = 1.37! 10 (3 cm 3 10. A sugar contains 42.14% C, 6.48% H and 51.46% O. What is the molecular formula of the sugar if its molar mass is approximately 330 g/mol? A) C 11 H 22 O 11 B) C 13 H 2 O 16 C) C 20 H 20 O 17 D) CH 2 O 42.14g C! 1mol C 12.001g C = 3.511mol C 6.48g H! 1mol H 1.008g H = 6.43mol H 1mol O 51.46g C! 15.9994g O = 3.216mol O empirical formula: CH 2 O empirical mass = 1! 12.001amu C empirical mass molecular mass = 30.0amu 330.1amu = 11 molecular formula: C 11 H 22 O 11 ( ) + ( 2! 1.008amu H ) + ( 1! 15.9994amu O) = 30.0amu 11. Calculate the moles of calcium ions present in a milliliter of 0.265 M calcium sulfate. A) 2.65 x 10-1 moles B) 5.30 x 10-1 moles C) 2.65 x 10-4 moles D) 5.30 x 10-4 moles Formula: CaSO 4 1L 1mL! 1000mL! 0.265mol CaSO 4 1mol Ca2+! = 2.65! 10 "4 mol Ca 2+ 1L 1mol CaSO 4 EXAM 1 VERSION B Page 5 of 10
12. A stock solution of potassium dichromate, K 2 CrO 4, is made by dissolving 76.2 g of the compound in 1.00 L of solution. How many milliliters of this solution are required to prepare 1.00 L of 0.100 M K 2 CrO 4? A) 392 ml B) 255 ml C) 76.2 ml D) 39.2 ml M 1 V 1 = M 2 V 2 76.2g mol K 2 CrO 4 = 2! 39.0983g mol K = ( ) + ( 51.9961g mol Cr) + ( 4! 15.9994 g mol O) 76.2g 194.1903g mol = 0.392mol K 2CrO 4 M 1 = 0.392mol K 2CrO 4 = 0.392M K 2 CrO 4 1.00L 1.00L! 0.100 mol L V 1 = = 0.255L! 1000mL 0.392 mol L 1L = 255mL 13. Calculate the number of lithium ions in 21.4 mg of lithium nitride. A) 3.70 x 10 26 ions B) 4.62 x 10 20 ions C) 6.15 x 10 20 ions D) 3.70 x 10 20 ions Formula: Li 3 N 1g 21.4mg Li 3 N! 1000mg! 1mol Li N 3 3mol Li+ 6.02! 10 23 Li + ions! = 1.1! 10 21 Li + ions 34.83g Li 3 N 1mol Li 3 N 1mol Li + 14. What is the mass % of H in ammonium hydrogen sulfate (NH 4 HSO 4 )? A) 0.9% B) 3.5% C) 4.4% D) 12% ( ( )) + 1.008 g mol H + 32.066 g mol S + ( 4! 15.9994 g molo) molar mass = 14.00674 g mol N + 4! 1.008 g mol H = 115.11g mol NH 4 HSO 4 5! 1.008 g mol H mass%h =! 100% = 4.4% 115.11g mol NH 4 HSO 4 EXAM 1 VERSION B Page 6 of 10
15. Naturally occurring copper (63.546 amu) is composed of two isotopes, copper-63 (62.916 amu) and copper-65 (64.955 amu). From this, one can conclude that naturally occurring copper is composed of. A) approximately 20% copper-63 and 80% copper-65 B) approximately 80% copper-63 and 20% copper-65 C) approximately 30% copper-63 and 70% copper-65 D) approximately 70% copper-63 and 30% copper-65 63.546amu = ( x)62.916amu + ( 1! x)64.955amu x = 0.69 16. What is the correct name of Pb(ClO 4 ) 2? A) lead(ii) perchlorate B) lead(iv) perchlorate C) lead(ii) chlorate D) lead(iv) chlorate E) lead(ii) chlorite EXAM 1 VERSION B Page 7 of 10
MATCHING. 8 @ 2 pts each 16 POINTS TOTAL Please indicate the letter of the one correct answer for each of the following questions in the blank. 17. The conversion of CO (g) into CO 2 (g) is an example of a(n) C change/reaction. 18. Give the chemical symbol/name (whichever is missing) for each of the following elements: A) S O B) Boron J C) Na M D) Ni Q E) Iron L 19. Reaction of 5.80 grams of aluminum with 4.80 grams of oxygen gives 10.20 grams of aluminum oxide with 0.40 grams of aluminum left over. This illustrates the Law of Z. 20. The number of I is equal to the number of electrons in a neutral atom. A) Physical B) K C) Chemical D) Os E) Calcium F) Electrons G) Nitrogen H) I I) Protons J) B K) Decomposition L) Fe M) Sodium N) Ir O) Sulfur P) In Q) Nickel R) P S) Carbon T) O U) Neutrons V) Lead W) Br X) Multiple Proportions Y) Definite Proportions Z) Conservation of Mass EXAM 1 VERSION B Page 8 of 10
LONG ANSWER. 20 POINTS TOTAL 21. In lab, a student mixes 8.0 ml aqueous solutions of sodium sulfide (4.0 M) and copper(iii) chloride (2.5 M) are mixed together. A) What is the net ionic equation for this reaction? (3 pts) 2Cu 3+ (aq) + 3S 2 (aq) Cu 2 S 3 (s) If you are unable to write the balanced net ionic equation or no precipitate would form in this reaction (i.e., no reaction would occur), use the following information for the remainder of the problem: 60. ml of 0.50 M iron(iii) chloride solution and 40.0 ml of 7.5 M ammonium sulfide solution are mixed in order to precipitate iron(iii) sulfide according to the following balanced net ionic equation: 2Fe 3+ (aq) + 3S 2- (aq) Fe 2 S 3 (s) B) Assuming the reaction goes to completion, what mass of precipitate will form? (6 pts) 0.008L solution! 4.0mol Na 2 S 1L solution 0.008L solution! 2.5mol CuCl 3 1L solution 1mol S2"! = 0.032mol S2" 1mol Na 2 S! 1mol Cu 3+ 1mol CuCl 3 = 0.020mol Cu 3+ Need 3 S 2- ions for every 2 Cu 2+ ions, so Cu 2+ is limiting reactant. 0.020mol Cu 3+! 1mol Cu 2 S 3 2mol Cu 3+! 223.29g Cu 2 S 3 1mol Cu 2 S 3 = 2.2g Cu 2 S 3 Alternate problem: 0.50mol FeCl 3 L solution! 0.060L solution = 0.030mol FeCl 3! 7.5mol Na 2 S L solution! 0.0400L solution = 0.30mol ( NH 4 ) 2 S! 1mol Fe3+ 1mol FeCl 3 = 0.030mol Fe 3+ 1mol S 2" ( ) 2 S 1mol NH 4 2:3 Fe 3+ :S 2- mole ratio in balanced net ionic equation, so 0.015 equivalents Fe 3+ and 0.10 equivalents S 2-. Fe 3+ is limiting reagent. 0.030mol S 2! " 1mol Fe 2 S 3 2mol Fe 3+ = 0.015mol Fe 2 S 3 " 207.888g Fe 2 S 3 1mol Fe 2 S 3 = 3.1g Fe 2 S 3 = 0.30mol S2" EXAM 1 VERSION B Page 9 of 10
What mass of each excess ion will remain? (9 pts) 0.032mol S 2! " 2mol Na+ 1mol S 2! 0.064mol Na +! = 0.064mol Na+ initially! 0mol Na + used = 0.064mol Na + excess 22.989770g Na+ 1mol Na + = 1.5g Na + excess " 3mol Cl 0.020mol Cu 3+! 1mol Cu = 0.060mol Cl " initially " 0mol Cl " used = 0.060mol Cl " excess 3+! 35.4527g Cl 0.060mol Cl! " = 2.1g Cl! excess 1mol Cl! # 3mol 0.032mol S 2! initially! 0.020mol Cu 3+ S2! & " $ % 2mol Cu 3+ ' ( used = 0.002mol S2! excess 0.002mol S 2! " 32.066g S2! 1mol S 2! = 0.064g S 2! excess Alternate Problem: 0.30mol S 2! " 2mol NH + 4 = 0.60mol NH + 1mol S 2! 4 initially! 0mol NH + 4 used = 0.60mol NH + 4 excess 0.60mol NH + 4! 18.03874g NH 4 = 11g NH + + 4 excess 1mol NH 4 " 3mol Cl 0.030mol Fe 3+! 1mol Fe = 0.090mol Cl " initially " 0mol Cl " used = 0.090mol Cl " excess 3+! 35.4527g Cl 0.090mol Cl! " = 3.2g Cl! excess 1mol Cl! # 3mol 0.30mol S 2! initially! 0.030mol Fe 3+ S2! & " $ % 2mol Fe 3+ ' ( used = 0.255mol S2! excess + 0.255mol S 2! " 32.066g S2! 1mol S 2! = 8.2g S 2! excess D) If the student recovers 2.00 g of precipitate, what is the percent yield? (If you did not calculate a theoretical yield in part B, use 3.53 g here) (2 pts) 2.00g PbCl 2 2.2g PbCl 2! 100% = 91% yield Alt. 2.00g Fe 2 S 3 3.1g Fe 2 S 3! 100% = 65% yield EXAM 1 VERSION B Page 10 of 10