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- Question : (TCO 2) The atomic number of an atom is equal to the number of Student Answer: electrons plus protons. nuclei. neutrons plus protons. protons. neutrons. Instructor Explanation: Chapter 3; The atomic number = number of protons
- Question : (TCO 2) Consider a neutral atom with 30 protons and 34 neutrons. The atomic number of the element is Student Answer: 94. 30. 64. 32. 34. Instructor Explanation: Chapter^ 3;^ Atomic^ number^ =^ number^ of^ protons. Therefore, in this case, the atomic number = 30.
- Question : (TCO 2) No matter how much extra oxygen is available, 12 grams of carbon always combines with 32 grams of oxygen. This best illustrates the law of Student Answer: conservation of mass. multiple proportions. definite proportions.
All of the above
- Question : (TCO 3) The maximum number of electrons that may reside in the n= energy level is Student Answer: 2.
Instructor Explanation: Chapter^3
- Question : (TCO 3) How many electrons are in a phosphorus atom? Student Answer: 15 19 23 39 84 Instructor Explanation: Chapter^3
- Question : (^) (TCO 3) Which is an impossible electron configuration? Student Answer: 1s^2 2s^2 1s^2 2s^2 2p^4 1s^2 2s^2 2p^4 1s^3 2s^2 2p^4 3s^1 1s^2 2s^2 2p^4 3s^2 Instructor Explanation: Chapter^3
- Question : (TCO 5) A reaction that releases energy as it occurs is classified as a(n) . Student Answer: catalyzed reaction exothermic reaction decomposition reaction endothermic reaction oxidation-reduction reaction Instructor Explanation: Week^3 Lecture;^ Reactions^ that^ release^ energy are exothermic.
- Question : (^) (TCO 5) What should be the coefficient of hydrogen, H 2 , in the following equation in order to make it balanced? 2 Al + 3 H 2 SO 4 ---> Al 2 (SO4) 3 +? H 2 Student Answer: 5 1 4 2 3 Instructor Explanation: Chapter 5; There are 6 hydrogens on the reactant side of the equation from the 3 H 2 SO 4 molecules. There must be 6 hydrogens on the product side, as well. Therefore, there are 3 H 2.
- Question : (^) (TCO 5) Magnesium reacts with oxygen to form magnesium oxide and has the following balanced chemical equation: 2 Mg + O 2 --> 2 MgO. How many mole(s) of MgO are produced when 0.20 mole O 2 reacts completely?
Student Answer: (1) mass/mass = [3.5 g/ (3.5.g + 50 g)] x 100 = 6.5%; (2) Molarity = mol/L, moles KCl = 3.5 g x (1 mol/74.6 g KCl) = 0.0469 mol; volume in Liters = 50 ml x ( L/1000 mL) = 0.050 L; mol/L = 0.0469 mol/0. L = 0.94 M Instructor Explanation: Chapter 5; (1) mass/mass = [3.5 g/(3.5.g + 50 g)] x 100 = 6.5%; (2) Molarity = mol/L, moles KCl = 3. g x (1 mol/74.6 g KCl) = 0.0469 mol; volume in Liters = 50 ml x (1 L/1000 mL) = 0.050 L; mol/L = 0.0469 mol/0.05 L = 0.94 M
- Question : (TCO 7) What volume (L) of a 3M KOH solution can be prepared by diluting 0.5 L of a 5M KOH solution? Student Answer: (5M)(0.5L)=(3M)(X)= 2.5M/L=3x 2.5/3=x x=.83L Instructor Explanation: Week 3 Lecture; C 1 V 1 =^ C 2 V 2.^ C 1 = 5 M,^ V 1 =0.5L, C 2 = 3M; V 2 = [(5M)(0.5L)]/(3M) = 0.83L Grade Details - All Questions
- Question : (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the acid? CO 32 -^ + H 2 O - > OH-^ + HCO 3 - Student Answer: OH-
HCO 3 2 - CO 3 H 2 O None of the above Instructor Explanation: Chapter 7
- Question : (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the base? H 3 PO 4 + NH 3 - > NH 4 + H 2 PO 4 Student Answer: NH 3 NH 4 +
H 2 PO 4 H 3 PO 4 None of the above Instructor Explanation: Chapter 7
- Question : (TCO 8) What is reduced in the following reaction? 2 Bi3+^ + 3 Mg - > 2 Bi
- 3 Mg2+ Student Answer: Mg Bi Mg2+ Bi3+
- Question : (^) (TCO 6) A sample of helium gas occupies 1245 mL at 705 mmHg. For a gas sample at constant temperature, determine the volume of helium at 745 mmHg. Student Answer: Using^ Boyle’s^ law,^ P 1 V 1 = P 2 V 2. We have V (1245 mL), P1 ( mmHg) and P2 ( mmHg). We want to determine V2, so V2 = (P1 x V1)/P2 = ( mL * 705 mmHg)/ mmHg) = 1178 mL. Instructor Explanation: Chapter^ 6;^ Using^ Boyle’s^ law,^ P 1 V 1 =^ P 2 V 2.^ We have V1 (1245 mL), P1 (705 mmHg) and P2 ( mmHg). We want to determine V2, so V2 = (P1 x V1)/P2 = (1245 mL * 705 mmHg)/745 mmHg) = 1178 mL.
- Question : (TCO 6) A gas at a temperature of 85 degrees C occupies a volume of 175 mL. Assuming constant pressure, determine the volume at 15 degrees C. Student Answer: T1=85+273=358K T2=15+273=288K 175288/358=141mL Instructor Explanation: Chapter 6; Using Charles’ Law, (V1/T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t +273) Thus, T1 = 85 + 273 = 358. We have V (175 mL) & T2 = (15 + 273) = 288. V2 = (V1T2)/T1 = (175 mL*288)/358 = 141 mL.
- Question : (^) (TCO 6) For a gas at standard temperature and pressure with a density of 2.75 g/L, determine its molar mass. Student Answer: 1 mol of a gas at STP has a volume of 22.4L. x molar mass = 2.75 g/L x (22.4 L/1 mol) = 61. g/mol. Instructor Explanation: Chapter 6; 1 mol of a gas at STP has a volume of 22.4L. x molar mass = 2.75 g/L x (22.4 L/1 mol) = 61.6 g/mol.
- Question : (^) (TCO 6) Calculate the pressure, in atmospheres, of 3.94 mol CO(g) in a 4.25 L tank at 31 degrees C. Student Answer: Given that n = 4.64 mol; V = 3.96 L; and temperature is 31 degrees C, which is 304 K; we want to calculate pressure in atm. First, convert your
Instructor Explanation: Chapter 9; CH3-CH2-CH2-CH3 is an alkane with 4 carbons and is called butane.
- Question : (^) (TCO 9) Name the following organic compound: CH 3 CH 2 CH 2 CH 2 CH 2 CH3. Student Answer: Hexane Instructor Explanation: Chapter^ 9.^ This^ molecule^ has^6 carbons,^ and^ thus, is hexane.
- Question : (TCO 9) Name the following organic compound: CH 3 CH 2 NH2. Student Answer: This molecule has an amine functional group (-NH 2 ); it has a 2- carbon-long chain; thus, it is ethylamine. Instructor Explanation: Chapter 9. This molecule has an amine functional group (-NH 2 ); it has a 2 - carbon-long chain; thus, it is ethylamine.
- Question : (^) (TCO 10) Match the organic compound with its use or characteristic. Student Answer: : Methyl salicylate » 5 : wintergreen flavor : Amine » 3 : derived from ammonia
: Morphine » 4 : alkaloid : Acetic acid » 2 : found in vinegar : Ethylene oxide » 1 : sterilize medical instruments Instructor Explanation: Chapter 9
- Question : (^) (TCO 9) Match the organic compound with the functional group found in that molecule. Student Answer: : » 2 : Amine CH 3 CH 2 CH 2 CH 2 NH 2 : CH 3 CH 2 CHO » 3 : Aldehyde : CH 3 CH 2 CH 2 OH » 1 : Alcohol : HCOOCH 2 CH 3 » 4 : Ester : » 5 : Carboxylic CH 3 CH 2 CH 2 COOH acid Instructor Explanation: Chapter 9
- Question : (TCO 9) Which of the following compounds is an alkyne? Student Answer: C3H CH3-CH2-CCH H2C=CH-CH=CH CH3-CH2-CH 2 - pentene Instructor Explanation: Chapter 9
- Question : (TCO 9) What is the name of this compound? CH3-CH2-CH2-CH2- CH3? Student Answer: Pentane Methylbutane Octane Hexane Heptane Instructor Explanation: Chapter 9; CH3-CH2-CH2-CH2-CH3 is an alkane with 5 carbons and is called pentane.
- Question : (^) (TCO 9) Name the following organic compound: CH 3 CH 2 CH 2 CH 2 CH 2 CH3. Student Answer: Hexane Instructor Explanation: Chapter^ 9.^ This^ molecule^ has^6 carbons,^ and^ thus, is hexane.
- Question : (^) (TCO 9) Name the following organic compound: CH 3 CH 2 CH 2 CH 2 CH 2 COOH. Student Answer: It is hexanoic acid and the common name is caproic acid Instructor Explanation: Chapter 9. This molecule has a carboxylic acid functional group (COOH) and has a 6 - carbon-long chain; thus, it is hexanoic acid.
- Question : (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the acid? CO 32 -^ + H 2 O - > OH-^ + HCO 3 - Student Answer: OH-
HCO 3 2 - CO 3 H 2 O None of the above Instructor Explanation: Chapter^7
- Question : (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the base? H 3 PO 4 + H 2 O - > H 3 O+^ + H 2 PO 4 - Student Answer: H 2 O H 3 O+ H 3 PO 4
H 2 PO 4 None of the above Instructor Explanation: Chapter^7
- Question : (^) (TCO 8) What is reduced in the following reaction? Cl 2 + 2 NaBr - > 2 NaCl + Br 2 Student Answer: NaCl Br 2 NaBr Cl 2 None of the above Instructor Explanation: Chapter^8
- Question : (TCO 8) What is the pH of a solution with a [H 3 O+] of 1 x 10 -^12 M? Student Answer: - 12. 2. - 2. 12. None of the above Instructor Explanation: Chapter^7
