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Devillez (ld2653) – Test 1 Review – Devillez – (99998) This print-out should have 133 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 1 3. 4 4. 3 correct 5. 1 001 10.0 points A chemist investigates the I) melting point II) boiling point III) flammability of acetone, a component of fingernail polish remover. Which is/are physical? 1. III only 2. I, II and III 3. II and III only 4. I and II only correct 5. I only 6. I and III only 7. II only 8. None of these Explanation: Flammability is a chemical change; melting and boiling (change of state) are physical changes. 002 10.0 points Consider the statement: “The temperature of the land is an important factor for the ripening of oranges, because it affects the evaporation of water and the humidity of the surrounding air.” How many of these factors, temperature ripening evaporation humidity are physical properties or changes? Explanation: The temperature and humidity are physical properties and the evaporation of water is a physical change. 003 10.0 points All of the following are characteristic properties of phosphorus. Which one is a chemical property? 1. The white form is soluble in liquid carbon disulfide, but is insoluble in water. 2. The red form melts at about 600◦ C and the white form melts at 44◦ C. 3. Both red phosphorus and white phosphorus exist in solid allotropic forms. 4. When exposed to air, white phosphorus will burn spontaneously, but red phosphorus will not. correct 5. The red form of phosphorus is insoluble in both water and carbon disulfide. Explanation: Chemical properties are exhibited as matter undergoes chemical changes. The burning of white or red phosphorus explains its interaction with chemical reactions and changes in composition which justifies its being a chemical property. 004 10.0 points Which of the following is an extensive physical property? 1. reactivity 1. 2 2. color 2. None 3. density Devillez (ld2653) – Test 1 Review – Devillez – (99998) 2 4. mass correct 5. refractive index Explanation: Extensive properties are those which depend on the amount of substance present. Mass depends on and is directly proportional to the amount of matter present. 005 10.0 points Which of the following is an intensive property? 1. number of moles of molecules 2. volume I II III IV best characterize(s) a mixture? 3. weight 1. I, II and III only 4. mass 2. I and III only 5. density correct 3. III and IV only correct 4. IV only Explanation: Intensive properties are independent of the amount of matter present. The density of a substance will be the same regardless of the size (large or small) of the sample. 006 10.0 points Which is NOT a pure substance? 1. water 2. chlorine 3. copper 4. brass correct Explanation: The dark and light spheres represent atoms of different elements. A mixture consists of different substances combined together physically. Although some atoms are chemically bonded in the figures, the samples do not consist of all of the same type of bonded atoms. 008 10.0 points Which diagram(s) best characterize(s) an element? Explanation: Brass is an alloy (a physical mixture of the metals copper and zinc). 007 10.0 points Which diagram(s) I II Devillez (ld2653) – Test 1 Review – Devillez – (99998) 3 4. I only correct III IV best characterize(s) an element? Explanation: The dark and light spheres represent atoms of different elements. A compound is a combination of two or more elements, chemically bound together, with the same composition throughout. 010 10.0 points Which diagram(s) 1. III and IV only 2. II only correct 3. I only 4. I and II only Explanation: The dark and light spheres represent atoms of different elements. An element consists of only one kind of atom. In II the atoms happen to form diatomic molecules, but only one type of atom is present in II. 009 10.0 points Which diagram(s) best characterize(s) a compound? I II III IV best represent(s) molecules? 1. II and IV only 2. I only I II 3. I and II only correct 4. I and III only 5. III only 6. II and III only III 1. II and III only 2. II only 3. I and IV only IV 7. Another combination 8. II only 9. IV only Explanation: The dark and light spheres represent atoms Devillez (ld2653) – Test 1 Review – Devillez – (99998) of different elements. Molecules consist of two or more atoms chemically bonded together. 011 10.0 points Which diagram(s) 4 neous solution. correct 2. A pure substance cannot be heterogeneous. 3. Heterogeneous solutions cannot be pure substances. 4. Mixtures can be homogeneous. I II Explanation: A solution, although homogeneous, cannot be a pure substance since it consists of two components: the solute and the solvent. 013 10.0 points Which diagram(s) III IV best represent(s) atoms? 1. I and III only I II 2. I and II only 3. I only 4. II only 5. IV only III 6. III only correct 7. Another combination 8. II and IV only 9. II and III only Explanation: The dark and light spheres represent atoms of different elements. Atoms are single and unattached to each other. 012 10.0 points Which of the following statements concerning substances is NOT true? 1. A pure substance can only be a homoge- best characterize(s) a pure substance that is an element? 1. I and III only 2. I and II only 3. II and III only 4. None of these 5. II only correct 6. All of these 7. III only Devillez (ld2653) – Test 1 Review – Devillez – (99998) 5 8. I only Explanation: 014 10.0 points Which of the following is a pure substance? 1. a puddle of milk 2. a rod of steel 3. a block of wood 4. a cube of sugar correct 5. a strap of leather Explanation: A cube of sugar is an example of a pure compound. 015 10.0 points Which of these substances I) carbon dioxide gas II) distilled water III) jello is a pure substance? 1. I and II only correct 2. I only 3. II only 4. II and III only 5. All of these 6. III only 7. None of these 8. I and III only Explanation: 016 10.0 points What is the proper solution to the following equation? (150 + 2.5 + 36.75) ÷ (6.60 + 0.173) = 1. 30 2. 27.942 3. 27.94183 4. 28 correct 5. 27.9 6. 27.94 7. 27.9418 Explanation: The last position retained for the first addition problem will be the tens position giving it 2 sigfigs. The last position retained for the second addition problem will be the hundredths position giving it 3 sigfigs. Therefore the final answer 27.94 needs to be rounded to 2 sigfigs: 28. 017 10.0 points Which of the following numbers has the greatest number of significant figures? 1. 130 2. 0.690 correct 3. 1.2 4. 33,000 5. 0.0032 6. 4.3 × 102 Explanation: Zeros to the left of the decimal are not significant. Zeros used to place the decimal are not significant. 0.690 has 3 significant figures. All the other choices have 2 significant figures. 018 10.0 points What is the length of the line segment indicated on the scale provided using proper Devillez (ld2653) – Test 1 Review – Devillez – (99998) 6 significant figures? 0 1 2 3 3. 3.4308 g 4. 3.431 g correct 1. 3.00 units correct 5. 3.43 g 2. 3.000 units 6. 3 g Explanation: sum of given values = ? 3. 3.0 units 4. 3 units 2.636 g + 0.7948 g = 3.431 g Explanation: The scale is in units of tenths, so hundreths is the uncertain digit. 019 10.0 points What is the proper measurement of the volume represented? 5 mL 4 3 021 10.0 points Keeping in mind the rules for significant figures, divide 92.80 g by 3.91111 mL. 1. 23.73 g/mL correct 2. 23.7 g/mL 3. 23.727 g/mL 4. 23.7273 g/mL 2 1 5. 24 g/mL Explanation: 1. 2.5 mL correct 92.80 g (4 sf) = 23.73 g/mL 3.91111 mL (6 sf) 2. 2.0 mL 3. 2.50 mL 4. 2.500 mL Explanation: The scale is in units of 1, so tenths is the uncertain digit. 020 10.0 points Keeping in mind the rules for significant digits, calculate the sum of 2.636 g and 0.7948 g. (round to 4 sig. figs) 022 10.0 points For the conversion of 0.0003140 kilograms to ounces, the conversion factors are 103 grams per kilogram, 1 pound per 453.6 grams and 16 ounces per pound. The correctly expressed answer is 1. 1.108 × 10−5 oz 2. 4.329 × 10−5 oz 3. 1.108 × 10−2 oz correct 1. 3.4 g 4. 1.108 × 102 oz 2. 3.43080 g 5. 8.902 × 10−1 oz Devillez (ld2653) – Test 1 Review – Devillez – (99998) 7 Explanation: 4. 2.69 miles/min2 ? ounces = 0.0003140 kg 1000 g 1 pound × × kg 453.58 g 16 ounces × 1 pound = 1.108 × 10−2 ounces Zeroes at the end of numbers containing a decimal point are considered to be significant, so 0.0003140 has 4 significant figures. The 1 kg to 1000 g and 1 pound to 16 ounces conversions are exact definitions, so these numbers do not affect how many significant figures the answer should have. 023 10.0 points How many milliliters are in a 67.8 ft3 box? 1. 1.92×106 mL correct 2. 13300 mL 3. 0.00239 mL 4. 2070 mL 5. 34.7 miles/min2 correct 6. 1.15 miles/min2 7. 1.49 miles/min2 Explanation: Acceleration rate = 15.5 m/s2 . 1 km 1 mile m × 15.5 2 × s 1000 m 1.609 km miles (60 s)2 = 34.7 × 2 (1 min) min2 10.0 points 025 Given: 1 pound (lb) = 453.6 grams 1 pound = 16 ounces (oz) How many ounces are equal to 0.0003140 kilograms? Taking significant figures into account, the correctly expressed answer is 1. 1.1076 × 10−2 oz 2. 1 × 10−2 oz 3. 1.108 × 10−2 oz correct 4. 1.1 × 10−2 oz 5. 7150 mL Explanation: (12 in)3 1 ft3 1 mL (2.54 cm)3 × × 3 1 in 1 cm3 6 = 1.92 × 10 mL ? mL = 67.8 ft3 × 024 10.0 points An Olympic hurdler accelerates at a rate of 15.5 m/s2 . What is the rate in miles/min2 ? 1. 31.8 miles/min2 2. 6.89 miles/min2 3. 0.581 miles/min2 5. 1.11 × 10−2 oz Explanation: ? ounces = 0.0003140 kg × 1000 g 1 pound × kg 453.58 g × 16 ounces 1 pound = 1.108 × 10−2 ounces Zeroes at the end of numbers containing a decimal point are considered to be significant, so 0.0003140 has 4 significant figures. The 1 kg to 1000 g and 1 pound to 16 ounces conversions are exact definitions, so these numbers are not included when deciding how many significant figures the answer should have. Devillez (ld2653) – Test 1 Review – Devillez – (99998) 026 10.0 points An oxygen molecule at 25◦ C has a speed of about 927 meters per second. Express this speed in miles/hour. (1 mi = 5280 ft) 1. 13400 miles/hour d= 8 4.2 mol CH3OH 5.67 L × 1L 32 g CH3 OH × 1 mol CH3 OH 1000 mL = 0.0237 g/mL 2. 0.000482 miles/hour 028 10.0 points Polycarbonate plastic has a density of 1.3 g/cm3 . A photo frame is constructed from two 6.2 mm sheets of polycarbonate. Each sheet measures 28 cm by 24 cm. What is the mass of the photo frame? 3. 20700 miles/hour 4. 2070 miles/hour correct 5. 0.000160 miles/hour 6. 0.576 miles/hour 1. 730 g 7. 3340000 miles/hour 2. 1100 g correct 3. 1000 g Explanation: m T = 25◦ C v = 927 s 1 in 1 ft m 100 cm × × × 927 s 1m 2.54 cm 12 in 1 mi 3600 s × = 2073.64 mi/h 5280 ft 1 hr 027 10.0 points Calculate the density of methanol (CH3 OH) if you know that there are 4.2 moles of methanol in 5.67 L. 4. 850 g 5. 1200 g Explanation: d = 1.3 g/cm3 w = 24 cm ℓ = 28 cm h = 2(6.2 mm) = 1.24 cm V = length × width × height m V m = d V = d (ℓ w h) = (1.3 g/cm3 ) (28 cm) (24 cm) (1.24 cm) = 1083.26 g d= 1. 0.741 g/mL 2. 0.0237 g/mL correct 3. 1334 g/mL 4. 0.0431 g/mL 029 10.0 points The density of gold is 19.3 g/mL. What is the mass of a gold nugget which has a volume of 3.28 mL? 5. 0.131 g/mL 6. 11.34 g/mL 1. 5.88 g 2. 0.170 g Explanation: n = 4.2 mol V = 5.67 L 3. 63.3 g correct Devillez (ld2653) – Test 1 Review – Devillez – (99998) 9 5. 797 ◦ F 4. 30.4 g Explanation: Explanation: 030 10.0 points A solid has a mass of 95 grams. A graduated cylinder contains 35 mL of water. When the solid is completely submerged in the water, the new water level reads 84 mL. What is the density of the solid? 1. 2.7 g/mL 2. 0.80 g/mL 3. 1.9 g/mL correct 4. -1.9 g/mL 5. 1.1 g/mL 6. 190 g/mL 7. 0.52 g/mL Explanation: 031 10.0 points On the Kelvin scale, a temperature of 43◦ C is 1. 43 K. 2. 316 K. correct 3. 273 K. 4. 230 K. Explanation: 032 10.0 points Copper melts at 1083 ◦ C. What is its melting temperature in ◦ F? 1. 583 ◦ F 2. 1981 ◦ F correct 3. 619 ◦ F ? ◦ F = 1083◦ C × 1.8◦ F + 32◦ F = 1981◦ F 1.0◦ C 033 10.0 points Identify the incorrect statement regarding Dalton’s Atomic Theory. 1. A given compound always has the same relative numbers and types of atoms. 2. All atoms of a given element are identical. 3. All matter is composed of atoms. 4. All matter is composed of protons, electrons, and neutrons. correct Explanation: In the early 1800s when Dalton formulated his atomic theory, it was believed that all atoms of the same element were the same. The fundamental particles of the atom – protons, neutrons, and electrons – were not known at this time. 034 10.0 points The following are all proposals of Dalton’s atomic theory. Which part of Dalton’s atomic theory was shown to be wrong by the discovery of natural radioactivity? 1. Atoms combine in simple numerical ratios to form compounds. 2. Atoms are permanent, unchanging, indivisible bodies. correct 3. Two or more kinds of atoms may combine in different ways to form more than one kind of chemical compound. 4. Atoms of a given element all weigh the same. 4. 1324 ◦ F 5. Compounds consist of collections of Devillez (ld2653) – Test 1 Review – Devillez – (99998) molecules made up of atoms bonded together. Explanation: The discovery of natural radioactivity provided evidence that atoms can change. They can undergo radioactive decay, emitting electrons, protons, or neutrons. They can also undergo nuclear fission, in which a heavy nucleus splits into two lighter nuclei. 035 10.0 points The discovery and characterization of cathode rays was important in the development of the atomic theory because 1. it led to the suggestion of the existence of the neutron. 2. it indicated that all matter contained electrons. correct 3. it indicated that all matter contains protons. 4. None of the these is correct. 5. it indicated that all matter contains alpha particles. Explanation: The cathode ray experiment results were the same, even when different types of gas, composition of electrodes, and power sources were used. 036 10.0 points Rutherford’s α-scattering experiment showed that most α particles directed toward a thin metallic foil passed through with only slight deviations. From this evidence we can conclude that 1. the mass of the atom is concentrated in a very small area. correct 2. an α particle is a type of light which should not be deflected. 3. the foil was too thin. 10 4. α particles are uncharged. 5. α particles are too small to hit anything. Explanation: In Rutherford’s gold foil experiment in 1910, α (alpha) particles were fired at gold foil, and the resulting deflection of the particles were observed. Most of the α particles went through the sample undeflected, suggesting that much of the atom was empty space. But of the few α particles that were deflected, they were deflected at all angles, including some very wide angles! The wide deflections suggested a very hard (dense) positively charged core in the atom. However, this core, or nucleus, must be small in relation to the overall size of the atom, since so few of the α particles were deflected in the first place. It also suggests that the nucleus is surrounded by a cloud of electrons at relatively great distances from the nuclei. 037 10.0 points What experimental evidence led Thomson to conclude that cathode rays were negatively charged? 1. Cathode rays were deflected towards a negatively charged plate. 2. Cathode rays pushed a paddle wheel counterclockwise. 3. Cathode rays pushed a paddle wheel clockwise. 4. False; cathode rays are positively charged, not negatively charged. 5. Cathode rays were deflected towards a positively charged plate. correct Explanation: Cathode rays are repelled by negative plates and attracted to positive plates. Since opposites attract, cathode rays must be negatively charged. Devillez (ld2653) – Test 1 Review – Devillez – (99998) 11 038 10.0 points Identify the experimental evidence that indicates an atom has a positively charged nucleus. about the same. correct 1. Electrons can be ejected from a metallic surface with high energy light. Explanation: A proton weighs 1 amu, as does a neutron. 1 An electron weighs amu. Hydrogen 2000 has one proton, one electron, and usually no neutrons, so H weighs about 1 amu. 2. α particles are deflected at large angles when projected toward a thin sheet of metal. correct 3. Cathode rays produce diffraction patterns. 4. Line spectra are produced from elemental gases in a gaseous discharge tube. 4. The mass of a hydrogen atom is about the same as a proton. 040 10.0 points How many electrons are present in one neon atom? 1. 10 correct 2. 9 5. Cathode rays are attracted to the positive plate of an applied electrical field. 3. 11 Explanation: In Rutherford’s gold foil experiment in 1910, α (alpha) particles were fired at gold foil, and the resulting deflection of the particles were observed. Most of the α particles went through the sample undeflected, suggesting that much of the atom was empty space. But of the few α particles that were deflected, they were deflected at all angles, including some very wide angles! The wide deflections suggested a very hard (dense) positively charged core in the atom. However, this core, or nucleus, must be small in relation to the overall size of the atom, since so few of the α particles were deflected in the first place. It also suggests that the nucleus is surrounded by a cloud of electrons at relatively great distances from the nuclei. 4. 8 039 10.0 points Which of the following is FALSE? 1. The mass of a proton is much greater than the mass of an electron. 2. The mass of protons and neutrons are about the same. 3. The mass of neutrons and electrons are 5. 12 Explanation: The atomic number of an element is equal to the number of protons in an atom. In a neutral atoms the number of protons is equal to the number of electrons. Neon has an atomic number of 10. 041 10.0 points How many protons are present in one iron(III) ion? 1. 25 2. 28 3. 26 correct 4. 27 5. 29 Explanation: The atomic number of an element is equal to the number of protons in that element’s nucleus. The number of protons defines an element and cannot be changed without changing the element. Devillez (ld2653) – Test 1 Review – Devillez – (99998) Iron has an atomic number of 26. 042 10.0 points How many protons are present in one Ca2+ ion? 1. 19 2. 22 3. 20 correct 4. 21 5. 18 Explanation: The atomic number of an element is equal to the number of protons in that element’s nucleus. The number of protons defines an element and cannot be changed without changing the element. Calcium has an atomic number of 20. 043 10.0 points What is the symbol for the element which contains 5 electrons in the neutral state? 1. C 2. B correct 3. Be 4. Li Explanation: Boron has 5 protons and in its neutral state, has 5 electrons as well. 044 10.0 points A magnesium (Mg) atom with a mass number of 25 contains 1. 13 protons, 12 neutrons and 12 electrons. 2. 12 protons, 13 neutrons and 12 electrons. correct 3. 25 protons, 25 neutrons and 25 electrons. 12 4. None of these 5. 25 protons, 13 neutrons and 25 electrons. Explanation: 045 10.0 points Identify the isotope that has atoms with 12 neutrons, 10 protons, and 10 electrons. 1. 22 Na 2. 10 S 3. None of these 4. 22 Ne correct 5. 10 P 6. 12 Mg 7. 12 Al Explanation: The element Ne is identified by the number of protons (atomic number), and the syntax is n+pp Nep−e . 046 10.0 points Write the hyphen notation for the element that contains 15 electrons and 15 neutrons. 1. copper-60 2. phosphorus-30 correct 3. silicon-30 4. fluorine-15 5. sulfur-31 6. phosphorus-31 7. oxygen-15 8. zinc-60 Explanation: Devillez (ld2653) – Test 1 Review – Devillez – (99998) #e− = 15 #n = 15 hyphen notation for given element = ? atomic number = number of protons = number of electrons mass number = number of protons + number of neutrons atomic number = 15 (element is phosphorus) mass number = 15 protons + 15 neutrons = 30 nuclide is phosphorus-30 047 10.0 points Choose the correct nuclear symbol and hyphen notation for the isotope which has a mass number of 28 and atomic number of 14. 1. 14 28 S, sulfur-14 2. 14 14 P, phosphorus-14 3. 14 28 P, phosphorus-14 4. 14 14 Si, silicon-14 5. 28 14 P, phosphorus-28 6. 28 14 Si, silicon-28 correct 7. 14 28 Si, silicon-14 8. 14 14 S, sulfur-14 9. 28 14 S, with 57.25% abundance. The other has an atomic weight of 122.8831 amu. What is the atomic weight of the element? 1. 122.38 amu 2. 123.45 amu 3. 121.17 amu 4. 122.15 amu 5. 121.54 amu 6. 121.75 amu correct Explanation: Average Atomic Mass = 0.5725(120.9038) + 0.4275(122.8831) = 121.75 amu 049 10.0 points Calculate the molar mass of the noble gas krypton in a natural sample, which is 0.81% 78 Kr (molar mass 77.92 g/mol), 1.82% 80 Kr (molar mass 79.91 g/mol), 10.92% 82 Kr (molar mass 81.91 g/mol), 12.42% 83 Kr (molar mass 82.92 g/mol), 56.65% 84 Kr (molar mass 83.91 g/mol), and 17.38% 86 Kr (molar mass 85.91 g/mol). Correct answer: 83.7949 g/mol. Explanation: sulfur-28 Explanation: mass number = 28 atomic number = 14 nuclear symbol and hyphen notation for given nuclide = ? atomic number = number of protons element with 14 protons: silicon (Si) nuclear symbol −→ 28 14 Si hyphen notation −→ silicon-28 048 10.0 points A newly discovered element has two isotopes. One has an atomic weight of 120.9038 amu 13 MMkrypton = 0.0081 (77.92 g/mol) + 0.0182 (79.91 g/mol) + 0.1092 (81.91 g/mol) + 0.1242 (82.92 g/mol) + 0.5665 (83.91 g/mol) + 0.1738 (85.91 g/mol) = 83.7949 g/mol 050 10.0 points Devillez (ld2653) – Test 1 Review – Devillez – (99998) Which of the following statements about a nonmetal is true? 14 5. C only 6. B only 1. A nonmetal forms covalent bonds with metals. 2. A nonmetal has metallic bonding. 3. A nonmetal tends to form cations. 7. None of these 8. A and C only Explanation: The transition metals lie in the shaded area 4. A nonmetal has more than 3 electrons in its outer shell. correct Explanation: Nonmetals start in group VI of the periodic table, so they have more than 3 electrons in their outer shells. 051 10.0 points Which characteristic describes a metal? 053 10.0 points Classify the elements Si, Sr, Sn, Sb. 1. brittle 1. metal; nonmetal; metal; nonmetal 2. malleable correct 2. metalloid; metal; metal; nonmetal 3. good insulator 3. nonmetal; nonmetal; metal; metalloid 4. low conductivity 4. None of these 5. tend to be reduced Explanation: Metals are lustrous, malleable, ductile, good conductors, and tend to be oxidized. 052 10.0 points Identify the transition metals. A. hafnium B. radium C. radon 5. metalloid; metal; metal; metalloid correct Explanation: Silicon (Si) is a group 14 metalloid, strontium (Sr) is a group 2 metal, tin (Sn) is a group 14 metal and antimony (Sb) is a group 15 metalloid. 054 10.0 points List the elements that are members of the alkali metals. 1. B and C only 1. C, Si, Ge, Sn, Pb 2. A only correct 2. Li, Na, K, Rb, Cs, Fr correct 3. A, B, and C 3. B, Al, Ga, In, Tl 4. A and B only 4. None of these Devillez (ld2653) – Test 1 Review – Devillez – (99998) 5. Be, Mg, Ca, Sr, Ba, Ra 3. those on opposite sides 6. N, P, As, Sb, Bi 4. those in the same rows Explanation: Elements in the same group have the same number of valance (outer) electrons and often react in similar ways. 7. F, Cl, Br, I, At 8. He, Ne, Ar, Kr, Xe, Rn 9. O, S, Se, Te, Po Explanation: The name for the group I metal is alkali metals 055 15 057 10.0 points The letters in the periodic table are used to identify four different elements. 10.0 points D C Consider the periodic table B A X Y Z Based on the organization within the periodic table, which element is expected to have 2+ charge when it forms an ion? 1. C Identify the nonmetal. 1. Z correct 2. D 2. X 3. B 3. Y 4. A correct Explanation: The nonmetals lie in the shaded area: X Explanation: Group II elements form ions with a 2+ charge. Y Z 058 10.0 points Of the following elements, the one which is NOT a metal is 1. potassium. 2. iron. 056 10.0 points In the periodic table, which elements typically have similar properties? 1. those in the same columns correct 2. those related diagonally 3. cobalt. 4. arsenic. correct 5. silver. Explanation: Iron, cobalt, silver, and potassium are all Devillez (ld2653) – Test 1 Review – Devillez – (99998) metals since they are to the left of the line dividing metals and non-metals. Arsenic is the only element to the right of that line. 3. a transition. 059 10.0 points Of the following elements, the one which is NOT a nonmetal is 5. an orbital. 1. phosphorous. 2. bismuth. correct 3. sulfur. 4. silicon. 5. iodine. Explanation: Only bismuth is a metal in the list of elements because it is to the left of the line separating the metals and non-metals. 060 10.0 points Which of the following elements would be expected to resemble strontium (Sr) most closely in chemical properties? 1. Si 2. Ba correct 4. a lanthanide. Explanation: Horizontal rows on the periodic table are called periods. The vertical columns are groups or families. 062 10.0 points In the periodic table, a vertical column of elements is called 1. a lanthanide. 2. a group. correct 3. a period. 4. an orbital. 5. a transition. Explanation: Horizontal rows on the periodic table are called periods. The vertical columns are groups or families. 063 10.0 points Arsenic (As) is classified as 3. Li 1. a transition element. 4. Cs 2. an actinide. 5. Sc 3. a lanthanide. Explanation: Elements with similar chemical properties are arranged in the same group or vertical column. Thus strontium (Sr) would be expected to have similar properties to barium (Ba). 061 10.0 points In the periodic table, a horizontal row of elements is called 1. a group. 2. a period. correct 16 4. a noble gas. 5. a representative element. correct Explanation: Representative elements are found in the A Groups of the periodic table. As is a Group VA element. 064 10.0 points What is the formula for the ionic compound formed from sodium and oxide ions? Devillez (ld2653) – Test 1 Review – Devillez – (99998) 17 1. NaO2 7. Ca3 N 2. None of these 3. Na3 O2 4. NaO 5. Na2 O correct 6. Na2 O3 Explanation: Ca is in Group 2 so it forms a +2 ion; N is in Group 5 so it forms a −3 ion. To balance charges you need three Ca2+ for each two N3− . 067 10.0 points What is the formula of the compound made by combining calcium (Ca) with sulfur (S)? Explanation: The sodium ion is Na+ and the oxide ion is O2− , so the formula is Na2 O. 1. SCa2 065 10.0 points What is the formula for the ionic compound formed from lithium and sulfide ions? 3. SCa 1. LiS 2. None of these 2. CaS correct 4. CaS2 Explanation: Ca has two valence electrons and forms a 2+ ion. S has six valence electrons and forms a 2− ion. The ratio will be 1 : 1, so the formula is CaS. 3. Li4 S 4. Li2 S correct 5. Li2 SO4 6. Li2 S3 Explanation: The lithium ion is Li+ and the sulfide ion is S2− , so the formula is Li2 S. 066 10.0 points Write the formula of calcium nitride. 1. Ca2 N 2. Ca3 N2 correct 3. Ca2 N3 068 10.0 points What is the formula for the salt which contains both Ca and NO3 ions? (Note that the charges of the ions are not indicated.) 1. Ca(NO3 )2 correct 2. Ca3 (NO3 )2 3. Ca2 (NO3 )2 4. Ca2 NO3 5. CaNO3 Explanation: The calcium ion is Ca2+ and the nitrate − ion is NO− 3 . Two NO3 are needed to balance the charge on each Ca2+ . The formula is Ca(NO3 )2 . 4. None of these 5. CaN2 6. CaN3 069 10.0 points Which of the following ionic compounds has the wrong chemical formula? Devillez (ld2653) – Test 1 Review – Devillez – (99998) 18 1. NaO2 correct 2. KBr 3. K2 S 4. CaO 072 10.0 points An element (call it X) in Group IA of the periodic table forms a compound with an element (call it Y) of Group VIA. What would be the most likely formula and nature of the compound? 1. X2 Y; covalent 5. MgCl2 Explanation: The sodium ion is Na1+ and the oxide ion is O2− . The compound should thus be Na2 O. 070 10.0 points Which of the following species has the greatest number of electrons? 1. Cl− 2. F− 3. O2− 4. S2− 5. I− correct Explanation: The atomic number is the number of protons in an atom. In a neutral atom this also equals the number of electrons. For ions you must account for the lost or gained electrons to find the total number. O2− has 10 electrons, F− has 10 electrons, S2− has 18 electrons, Cl− has 18 electrons, and I− has 54 electrons. 071 10.0 points Which of the following species has the greatest number of electrons? 1. Al 2. Na+ 3. F− 4. Fe2+ correct 5. Ne Explanation: 2. XY2 ; ionic 3. XY6 ; ionic 4. X6 Y; ionic 5. X2 Y; ionic correct 6. XY2 ; covalent 7. XY6 ; covalent Explanation: Metals reacts with non-metals to form ionic compounds, which involve the transfer of electrons. A metal X in group IA would have one valence electron. It would lose one electron to become isoelectronic with the preceding noble gas. This metal would now have one fewer electron than protons and would have a +1 charge. An element Y in group VIA would have six valence electrons. It would gain two electrons to become isoelectronic with the nearest noble gas. This element would now have two more electrons than protons and would have a −2 charge. We need twice as many X+ ions as Y2− ions to equalize the overall charge, so the ratio of X to Y ions is 2:1, and the formula would be X2 Y. 073 10.0 points Choose that pair of names and formulas that do not match. 1. Fe2 (SO4 )3 : iron(III) sulfate 2. SnCl4 : tin(IV) chloride Devillez (ld2653) – Test 1 Review – Devillez – (99998) 3. MnCO3 : manganese(II) carbonate 4. Cu2 O : copper(II) oxide correct 5. SnO : tin(II) oxide Explanation: The copper(II) ion is Cu2+ ; the oxide ion is O2− . One O2− is needed to balance the charge of each Cu2+ , so the correct formula for copper(II) oxide is CuO. In Cu2 O, the total charge from copper ions must be +2 to balance the −2 charge from the oxide ion. Each copper ion would have a charge of +1. This compound is copper(I) oxide. The manganese(II) ion is Mn2+ ; the carbonate ion is CO3 2− . One CO3 2− is needed to balance the charge of each Mn2+ , so the formula is MnCO3 . The tin(IV) ion is Sn4+ ; the chloride ion is Cl− . Four Cl− are needed to balance the charge of each Sn4+ , so the formula is SnCl4 . The tin(II) ion is Sn2+ ; the oxide ion is O2− . One O2− is needed to balance the charge of each Sn2+ , so the formula is SnO. The iron(III) ion is Fe3+ ; the sulfate ion is SO4 2− . Three SO4 2− are needed to balance the charge of every two Fe3+ . (This gives a total anion charge of −6 and a total cation charge of +6.) The formula is Fe2 (SO4 )3 . 074 10.0 points Which of the following pairs of names and formulas are correct? Z1) dinitrogen tetroxide : N2 O4 Z2) calcium hydroxide : CaOH Z3) tin : Sn 19 The calcium ion is Ca2+ ; the hydroxide ion is OH− . Two OH− are needed to balance the charge on each Ca2+ , so the formula should be Ca(OH)2 . N2 O4 is a covalent compound and is correctly named using prefixes to indicate the number of atoms of each element in the compound. Sn is the symbol for tin. 075 10.0 points The binary compound PCl3 is called 1. phosphorus chloride. 2. None of these. 3. monophosphorus trichloride. 4. triphosphorus chloride. 5. phosphorus trichloride. correct Explanation: PCl3 is a covalent compound and is therefore named using prefixes to indicate the number of each type of atom. The prefix for three is “tri” so this molecule is named phosphorous trichloride. The “mono” prefix is commonly left off of the first element name in a binary compound. 076 10.0 points What is the name of the compound with the formula CCl4 ? 1. carbon tetrachloride correct 2. carbon chloride 1. Z1) and Z3) correct 3. chlorine carbonide 2. Z1) only 4. carbon(IV) chloride 3. Z1) and Z2) 4. Z1), Z2), and Z3) 5. Z3) only Explanation: Explanation: This is a covalent compound and should be named using prefixes to indicate the number of atoms of each element in the compound. Binary covalent compounds are named with the name of the first element followed by the Devillez (ld2653) – Test 1 Review – Devillez – (99998) name of the second element with the suffix “-ide” added. CCl4 is carbon tetrachloride. 077 10.0 points Write the formula of perchloric acid. 20 1. H2 SO4 : sulfurous acid correct 2. HIO (or HOI) : hypoiodous acid 3. HNO3 : nitric acid 1. HClO2 2. HClO 3. HClO5 4. HClO4 correct 5. HClO3 Explanation: Change ‘-ic’ to ‘-ate’. The perchlorate an+ ion is ClO− 4 . Now add enough H to make the overall species neutral. 078 10.0 points What is the chemical formula for hypopophosphorous acid? 4. HMnO4 : permanganic acid Explanation: Sulfuric acid is H2 SO4 . Sulfurous acid has one fewer oxygen atom, so its formula is H2 SO3 . Iodic acid is HIO3 . Hypoiodous acid has two fewer oxygen atoms, so its formula is HIO. MnO− 4 is the permanganate ion. HMnO4 is permanganic acid. NO− 3 is the nitrate ion. HNO3 is nitric acid. 080 10.0 points Pick the name and formula that do NOT match. 1. HCl : hydrochloric acid 1. H2 PO 2. H2 CO3 : carbonic acid 2. H2 PO4 3. H2 SO4 : sulfuric acid 3. H2 PO5 4. HClO4 : chloric acid correct 4. HPO5 5. HPO8 6. HPO 7. H2 PO2 8. H2 PO3 9. H2 PO6 10. H3 PO2 correct Explanation: Cloric acid is HClO3 . P erchloric acid has one more oxygen atom, so its formula is HClO4 . HCl is hydrogen chloride. In aqueous solution, HCl is known as hydochloric acid. SO2− 4 is the sulfate ion so H2 SO4 is sulfuric acid. CO2− is the carbonate ion so H2 CO3 is 3 carbonic acid. 081 10.0 points Which one has the greatest number of atoms? Explanation: 1. 3.05 moles of CH4 correct 079 10.0 points Pick the name and formula that do NOT match. 2. 3.05 moles of water 3. 3.05 moles of helium Devillez (ld2653) – Test 1 Review – Devillez – (99998) 4. All have the same number of atoms 5. 3.05 moles of argon How many moles of H2 O are in 1.8 g of AuCl3 · 2 H2 O? Correct answer: 0.0106085 mol. Explanation: mAuCl3 ·2 H2 O = 1.8 g Explanation: For 3.05 moles of water: ? atoms = 3.05 mol H2O × 21 6.02 × 1023 molec 1 mol 3 atoms 1 molecule = 5.51 × 1024 atoms × MWAuCl3 ·2 H2 O = 196.97 g/mol + 3 (35.45 g/mol) + 4 (1.00079 g/mol) + 2 (16.0 g/mol) = 339.352 g/mol For 3.05 moles of CH4 : 6.02 × 1023 molec ? atoms = 3.05 mol CH4 × 1 mol 5 atoms × 1 molecule = 9.18 × 1024 atoms nH 2 O 1.8 g AuCl3 · 2 H2 O = 339.352 g/mol 2 mol H2 O × 1 mol AuCl3 · 2 H2 O = 0.0106085 mol H2 O For 3.05 moles of helium: 6.02 × 1023 atoms 1 mol atoms ? atoms = 3.05 mol He × = 1.84 × 1024 For 3.5 moles of argon: 1. 1.5 × 1023 atoms 6.02 × 1023 atoms 1 mol atoms ? atoms = 3.05 mol Ar × = 1.84 × 1024 082 (part 1 of 2) 10.0 points How many moles of Ag+ are in 3 g of AgCl? Correct answer: 0.0209322 mol. Explanation: mAgCl = 3 g MWAgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol nAgCl = 084 10.0 points How many fluorine atoms are in 4.0 moles of fluorine molecules? 3g = 0.0209322 mol 143.32 g/mol The number of moles of Ag+ ions equals the number of moles of AgCl. 083 (part 2 of 2) 10.0 points 2. 6.6 × 10−24 atoms 3. 2.4 × 1024 atoms 4. 4.8 × 1024 atoms correct Explanation: nF = 4.0 mol Fluorine is diatomic. Each F2 molecule contains two fluorine atoms. We can use Avogadro’s number and the ratio of F atoms to F2 molecules to find the number of fluorine atoms: ? atoms F = 4.0 mol F2 6.022 × 1023 F2 × 1 mol F2 2 atoms F × 1 molec F2 = 4.8 × 1024 atoms F Devillez (ld2653) – Test 1 Review – Devillez – (99998) 22 = 2.41 × 1024 H atoms 085 10.0 points Each mole of Al(NO3 )3 contains how many moles of oxygen atoms? 1. 12 mol 087 10.0 points How many moles of carbon are in 8.00 moles of C12 H22 O11 ? 2. 6 mol 1. 8.00 mol 3. 1 mol 2. 88.0 mol 4. 3 mol 3. 96.0 mol correct 5. 9 mol correct 4. 10.0 mol Explanation: There are 9 O atoms in each molecule of Al(NO3)3 . This same ratio holds for the number of moles of O in one mole of Al(NO3)3 . 9O ? mol O = mol Al(NO3 )3 1 mol Al(NO3 )3 Therefore there are 9 mol O in one mole of Al(NO3)3 . 086 10.0 points How many atoms of hydrogen are contained in 1 mole of methane (CH4 )? 1. 6.02 × 10 23 atoms 2. The correct answer is not given. 3. 2.41 × 1024 atoms correct 4. 4 atoms 5. 3.01 × 1024 atoms Explanation: n = 1 mol Each methane molecule contains 4 hydrogen atoms. There are Avogadro’s number of methane molecules in one mole of methane molecules: nH = 1 mol CH4 6.02 × 1023 molec CH4 × 1 mol CH4 4 H atoms × 1 molec CH4 Explanation: n = 8.00 mol In 1 mole of C12 H12 O11 , there are 12 moles of C, 22 moles of H, and 11 moles of O. If you have 8 moles of C12 H12 O11 , then 12 mol C x = 1 mol C12 H22 O11 8 mol C12 H22 O11 12 mol C x= 1 mol C12 H22 O11 × 8 mol C12 H22 O11 = 96 mol C 088 10.0 points How many sulfur atoms are in 78.4 g of sulfur? Correct answer: 1.47235 × 1024 atoms. Explanation: mSu = 78.4 g To solve this problem, we will need both the atomic mass of sulfur (S, 32.066 g/mol) and Avogadro’s number (6.022×1023 atoms/mol). First convert grams to moles of sulfur using the atomic mass: 1 mol S ? mol S = 78.4 g S × 32.066 g S = 2.44496 mol S Now convert moles to the number of atoms using Avogadro’s number: ? atoms S = 2.44496 mol S 6.022 × 1023 S atoms × 1 mol S = 1.47235 × 1024 atoms S Devillez (ld2653) – Test 1 Review – Devillez – (99998) 23 Explanation: 089 10.0 points How many moles are in 418.8 g of Ba(OH)2? Correct answer: 2.44412 mol. Explanation: mBa(OH)2 = 418.8 g 137.33 g Ba 1 mol Ba × mol Ba 16.00 g O 2 mol O × mol O 1.01 g H 2 mol H × mol H Molar mass n=? = 137.33 g = 32.00 g 1. 1.68% correct = 2.02 g 2. 0.413% = 171.35 g /mol 3. 42.8% 418.8 g 171.35 g/mol Ba(OH)2 = 2.44412 mol Ba(OH)2 nBa(OH)2 = 090 10.0 points How many moles are in 158.55 g of Fe3 (PO4 )2 ? Correct answer: 0.443447 mol. Explanation: mFe3 (PO4 )2 = 158.55 g 55.85 g Fe 3 mol Fe × mol Fe 30.97 g P 2 mol P × mol P 16.00 g O 8 mol O × mol O Molar mass 092 10.0 points Although coal is not a pure chemical compound, its elemental composition may be approximated by the formula C135 H96 NO9 S. What is the approximate percentage by mass of sulfur in coal? n=? = 167.6 g = 61.94 g = 128.00 g 4. 3.11% Explanation: FWcoal = 135(12.0107 g/mol) + 96(1.00794 g/mol) + 1(14.0067 g/mol) + 9(15.9994 g/mol) + 1(32.065 g/mol) = 1908.27 g/mol 1(32.065 g/mol) % S in coal = × 100% 1908.27 g/mol = 1.68032 % 093 (part 1 of 2) 10.0 points Zinc chloride (ZnCl2 ) is 52.02% chlorine by mass. What mass of chlorine is contained in 61.7 g of ZnCl2 ? = 357.54 g /mol Correct answer: 32.0963 g. 158.55 g 357.54 g/mol Fe3 (PO4 )2 = 0.443447 mol nFe3 (PO4 )2 = 091 10.0 points Aspirin has a formula C9 H8 O4 . What is the molar mass of aspirin? 1. 95 g 2. 220 g Explanation: % mass of chlorine in ZnCl2 = 52.02% mCl in 61.7 g of ZnCl2 = ? mCl = mass % × mass ZnCl2 = 0.5202 × 61.7 g = 32.0963 g Cl 3. 325 g 094 (part 2 of 2) 10.0 points How many moles of Cl is this? Correct answer: 0.905397 mol. 4. 180 g correct Explanation: Devillez (ld2653) – Test 1 Review – Devillez – (99998) mass Cl molar mass of Cl 32.0963 g Cl = 35.45 g Cl = 0.905397 mol Cl nCl = 095 10.0 points What is the percentage of hydrogen in C7 H8 O? The molecular weight of C is 12.011 g/mol, of H 1.00794 g/mol, and of O 15.9994 g/mol. Correct answer: 7.45656%. Explanation: MWC = 12.011 g/mol MWH = 1.00794 g/mol MWO = 15.9994 g/mol 24 097 (part 1 of 3) 10.0 points a) How many CaH2 formula units are present in 6.341 g of CaH2 ? Correct answer: 9.08837 × 1022 formula units. Explanation: mCaH2 = 6.341 g MWCaH2 = 40.0 g/mol + 2 (1.0079 g/mol) = 42.0158 g/mol n= 6.341 g 42.0158 g/mol × 6.022 × 1023 formula units/mol = 9.08837 × 1022 formula units MWtotal = #C · MWC + #H · MWH + #O · MWO = 7(12.011 g/mol) + 8(1.00794 g/mol) + 1(15.9994 g/mol) = 108.14 g/mol , so the percentage of hydrogen in 0-cresol is #H · MWH 100.0% MWtotal 8 (1.00794 g/mol) = 100.0% 108.14 g/mol = 7.45656% %= 096 10.0 points What mass of rhodium contains as many atoms as there are iron atoms in 28 g of iron? 098 (part 2 of 3) 10.0 points b) Find the mass of 6.22 × 1024 formula units of NaBF4 (sodium tetrafluroborate). Correct answer: 1134.1 g. Explanation: nNaBF4 = 6.22 × 1024 formula units MWNaBF4 = 22.99 g/mol + 10.81 g/mol + 4 (19.0 g/mol) = 109.8 g/mol 6.22 × 1024 formula units units 6.022 × 1023 formula mol × (109.8 g/mol) = 1134.1 g mNaBF4 = Correct answer: 51.596 g. Explanation: mFe = 28 g mRh 28 g Fe = 55.847 g/mol N × (102.91 g/mol Rh) = 51.596 g Rh 099 (part 3 of 3) 10.0 points c) How many moles of CeI3 are in 9.66 × 1021 formula units of CeI3 (cerium(III) iodide; a bright, water, soluble solid). Correct answer: 0.0160412 mol. Explanation: nCeI3 = 9.66 × 1021 formula units Devillez (ld2653) – Test 1 Review – Devillez – (99998) 9.66 × 1021 formula units 6.022 × 1023 formula units/mol = 0.0160412 mol nCeI3 = 100 (part 1 of 3) 10.0 points Find the mass of 2.72 mol Cl2 . Correct answer: 193.12 g. Explanation: nCl2 = 2.72 mol 71 g Cl2 m = (2.72 mol Cl2 ) mol Cl2 = 193.12 g Cl2 101 (part 2 of 3) 10.0 points Find the mass of 2.65 × 1023 molecules H2 S. Correct answer: 14.9618 g. Explanation: nH2 S = 2.65 × 1023 molec m=? m = 2.65 × 1023 molec 34 g H2 S mol · 6.022 × 1023 molec mol = 14.9618 g H2 S 102 (part 3 of 3) 10.0 points Find the mass of 17 molecules SO2 . Correct answer: 1.80671 × 10−21 g. Explanation: nSO2 = 17 molec m=? mol m = (17 molec) 6.022 × 1023 molec 64 g SO2 · mol = 1.80671 × 10−21 g SO2 103 10.0 points How many molecules are in 1.20 g of SO2 ? Correct answer: 1.12789 × 1022 molec. Explanation: mSO2 = 1.20 g 32.07 g S 1 mol S × = 32.07 g mol S 16.00 g O 2 mol O × = 32.00 g mol O Molar mass = 64.07 g /mol n=? 1.20 g 64.07 g/mol SO2 6.022 ×1023 molec × 1 mole = 1.12789 × 1022 molec SO2 nSO2 = m=? 25 104 10.0 points A can of mineral water contains 345 mL of water. How many molecules of water does the can contain? (water = 1.00 g/mL) 1. 57.6 molec 2. 3.74 × 1027 molec 3. 3.46 × 1025 molec 4. 1.15 × 1025 molec correct 5. 19.2 molec Explanation: VHCl = 345 mL We can use the density of water (H2 O) to convert from mL of water to grams of water: ? g H2 O = 345 mL H2 O × 1 g H2 O 1 mL H2 O = 345 g H2 O Before we can use Avogadro’s number to convert to the number of water molecules, we must first convert from grams water to moles of water. We do this using the formula weight, which can be determined from the atomic weights on the periodic table: 1.01 g H mol H 16.00 g O + 1 mol O × mol O ? FWH2 O = 2 mol H × Devillez (ld2653) – Test 1 Review – Devillez – (99998) = 18.02 g H2 O mol H2 O × 1 mol H2 O 18.02 g H2 O = 19.1 mol H2 O Now we can use Avogadro’s number to find the number of water molecules: ? molec H2 O = 19.1 mol H2 O 6.02 × 1023 H2 O × 1 mol H2 O = 1.15 × 1025 molec H2 O 1.01 g H mol H 14.01 g N + 1 mol N × mol N 17.04 g NH3 = mol NH3 FW = 3 mol H × in Explanation: m = 198.273 g Each Al2 O3 molecule contains two aluminum atoms. There are Avogadro’s number of Al2 O3 molecules in one mole of Al2 O3 . We need the formula mass of Al2 O3 so we can convert grams of Al2 O3 to moles Al2 O3 . Molecular mass of Al2 O3 : ? g/mol = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol We can use this formula mass to convert g Al2O3 to mol Al2 O3 : ? mol Al2 O3 = 198.273 g Al2 O3 × 106 10.0 points A 46 gram sample of NH3 contains how many grams of H? Correct answer: 8.17958 g. Explanation: m = 46 g To solve this problem we need the formula weight of NH3 : 10.0 points How many aluminum atoms are 198.273 g of Al2O3 ? Correct answer: 2.34209 × 1024 atoms. 2 atoms Al 1 molec Al2 O3 = 2.34209 × 1024 atoms Al ? mol H2 O = 345 g H2 O × 105 26 1 mol Al2 O3 101.96 g Al2 O3 This formula weight is used to convert grams NH3 to moles NH3 : 1 mol NH3 17.04 g NH3 = 2.69953 mol NH3 ? mol NH3 = 46 g NH3 × Each mole of NH3 contains 3 mol of H. We can use this ratio to convert from moles NH3 to moles H: ? mol H = 2.69953 mol NH3 × 3 mol H 1 mol NH3 = 8.09859 mol H Finally, the atomic weight of H can be used to convert from moles H to grams H: = 1.94462 mol Al2 O3 We can now use Avogadro’s number and the ratio of Al atoms to Al2 O3 molecules to find the number of aluminum atoms: ? g H = 8.09859 mol H × 1.01 g H 1 mol H = 8.17958 g H ? atoms Al = 1.94462 mol Al2 O3 × 6.022 × 1023 Al2 O3 1 mol Al2 O3 107 10.0 points Iron forms a compound called ferrocene. It Devillez (ld2653) – Test 1 Review – Devillez – (99998) has the composition 64.56% C, 5.42% H, and 30.02% Fe. What is the empirical formula of ferrocene? 1. FeC6 H5 2. Fe2 C10 H11 27 21.18 g = 1.324 mol 16.0 g/mol 62.93 g = 0.3309 mol moles of Os = 190.2 g/mol Dividing each number by 0.3309 mol gives a ratio of 4.0 C : 4.0 O : 1.0 Os . The empirical formula is OsC4 O4 . moles of O = 3. FeC10 H10 correct 4. FeC11 H9 5. Fe3 C11 H12 109 (part 2 of 2) 10.0 points From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g/mol. What is its molecular formula? 6. FeC7 H8 Explanation: For 100 g of compound, 64.56 g moles of C = = 5.376 mol 12.01 g/mol 5.42 g = 5.38 mol moles of H = 1.0079 g/mol 30.02 g moles of Fe = = 0.5375 mol 55.85 g/mol Dividing by 0.5375 mol gives 10.0 C : 10.0 H : 1.0 Fe. The empirical formula is FeC10 H10 . 108 (part 1 of 2) 10.0 points Osmium forms a number of molecular compounds with carbon monoxide. One lightyellow compound was analyzed to give the following elemental composition: 15.89% C, 21.18% O, and 62.93% Os. What is the empirical formula of this compound? 1. OsCO 2. Os2 C4 O5 3. OsC3 O5 4. OsCO4 5. OsC4 O4 correct 6. Os2 CO4 Explanation: For 100 g of osmium carbonyl compound, 15.89 g moles of C = = 1.323 mol 12.01 g/mol 1. Os3 C12 O12 correct 2. Os3 C3 O12 3. Os3 C9 O15 4. Os4 C8 O11 5. Os4 C2 O8 6. Os4 C8 O10 Explanation: The formula mass of OsC4 O4 is 302.24 g/mol. The molar mass is 907 g/mol which is 3 times the formula mass, so the molecular formula is Os3 C12 O12 . 110 10.0 points Paclitaxel, which is extracted from the Pacific yew tree Taxus brevifolia, has antitumor activity for ovarian and breast cancer. It is sold under the trade name Taxol. On analysis, its mass percentage composition is 66.11% C, 6.02% H, and 1.64% N, with the balance being oxygen. What is the empirical formula of paclitaxel? 1. C47 H51 NO14 correct 2. C45 H50 NO13 3. C3 H5 NO4 4. C4 H5 NO9 Devillez (ld2653) – Test 1 Review – Devillez – (99998) 28 5. C35 H41 NO8 6. C49 H53 NO16 Explanation: For 100 g of Paclitaxel, 66.11 g moles of C = = 5.504 mol 12.01 g/mol 6.02 g moles of H = = 5.97 mol 1.0079 g/mol 1.64 g = 0.117 mol moles of N = 14.01 g/mol mass of O = 100 − 66.11 − 6.02 − 1.64 = 26.23 g 26.32 g = 1.639 mol 16.0 g/mol Dividing by 0.117 mol gives 47.0 C : 51.0 H : 1.0 N : 14.0 O. The empirical formula is C47 H51 NO14 . 112 10.0 points An analysis of nicotine (MW = 162 g/mol) gives 74.0% carbon, 8.65% hydrogen, and 17.3% nitrogen. What is the TRUE (MOLECULAR) formula for nicotine? 1. C12 H4 N 2. C4 H5 N2 3. C10 H14 N2 correct 4. C9 H12 N3 5. C6 H8 N5 moles of O = 111 10.0 points What is the molecular formula of the molecule that has an empirical formula of CH2 O and a molar mass of 120.12 g/mol? 1. CH2 O 2. C3 H6 O3 3. C4 H8 O4 correct 4. C5 H10 O5 6. C5 H7 N 7. C8 H10 N4 8. C4 H5 N4 9. C11 H16 N Explanation: % C = 74.0% % H = 8.65% % N = 17.3% MW = 162 g/mol Assume we have 100 g of nicotine. (In fact, we can assume any mass, but choosing 100 g makes the math easier.) If we have 100 g of nicotine, 74.0 g (74.0%) of it is carbon, 8.65 g (8.65%) of it is hydrogen, and 17.3 (17.3%) of it is nitrogen. This means that, in this 100 g sample: 5. C2 H4 O2 6. C6 H12 O6 Explanation: empirical formula: CH2 O molar mass = 120.12 g/mol molecular formula mass x= empirical formula mass Empirical formula mass of CH2 O = 12.01 amu + 2.02 + 16.00 amu = 30.03 amu 120.12 g/mol = 4.000 → 4 x= 30.03 g/mol molecular formula: Cx H2x Ox = C4 H8 O4 . ? mol of C = 74.0 g C × 1 mol C 12.011 g C = 6.16 mol C ? mol of H = 8.65 g H × 1 mol H 1.0079 g H = 8.58 mol H ? mol of N = 17.3 g N × 1 mol N 14.0067 g N = 1.24 mol N The smallest number of moles here is then 1.24, and we can use this number to detemine the empirical (simplest) formula for this compound. For each element, we should now Devillez (ld2653) – Test 1 Review – Devillez – (99998) divide the number of moles we just calculated by this smallest number of moles to force the mole ratio to integers. So for C, we have 6.16 mol C = 5 mol C 1.24 For H, we have 8.58 mol H = 7 mol H 1.24 And for N, we have 1.24 mol N = 1 mol N 1.24 This means this compound’s empirical formula is C5 H7 N1 , better written as C5 H7 N. However, the question asked what the compound’s TRUE molecular formula was. For this, we need to compare the empirical formula’s apparent molecular weight to the known molecular weight of the compound. The apparent molecular weight of the empirical formula is 81 g/mol, exactly half of the known molecular weight. This means that the TRUE formula is double the empirical formula, i.e., C10 H14 N2 . 113 10.0 points Calculate the nuclear binding energy of a 32 16 S S atom. The measured atomic mass of 32 16 is 31.972 070 amu. 1. 2.62 × 10−11 J nuclear binding energy of 32 16 S = ? 16 protons: (16×1.007 276 amu) = 16.116 416 amu 16 neutrons: (16×1.008 665 amu) = 16.138 640 amu 16 electrons: (16×0.000 548 6 amu) = 0.008 777 6 amu total combined mass: 32.263 834 amu mass defect = 32.263 834 amu − 31.972 070 amu = 0.291 764 amu 1.6605 × 10−27 kg = (0.291 764 amu) 1 amu −28 = 4.8447 × 10 kg E = m c2 = (4.8447 × 10−28 kg)(3.00 × 108 m/s)2 = 4.36 × 10−11 kg·m2 /s2 = 4.36 × 10−11 J 114 10.0 points The mass of a 73 Li atom is 7.016 00 amu. Calculate its mass defect. 1. None of these 2. 7 amu 3. 0.294 91 amu 4. 0.042 13 amu correct 5. 3 amu 2. 4.36 × 10−11 J correct 6. 1 amu 3. None of these 7. 0.126 39amu 4. 4.36 × 1011 J 5. 2.73 × 10−12 J 6. 1.36 × 10−12 J 7. 2.62 × 1013 J Explanation: measured atomic mass of 29 32 16 S = 31.972 070 amu Explanation: measured atomic mass of 73 Li = 7.016 00 amu mass defect = ? 3 protons: (3×1.007 276 amu) = 3.021 828 amu 4 neutrons: (4×1.008 665 amu) = 4.034 66 amu 3 electrons: (3×0.000 5486 amu) = 0.001 646 amu Devillez (ld2653) – Test 1 Review – Devillez – (99998) 30 total combined mass: 7.058 134 amu mass defect = 7.058 134 amu − 7.016 00 amu = 0.042 13 amu 118 10.0 points What type of particle is emitted in the transformation 201 115 10.0 points What is the mass defect of 28 14 Si which has an actual mass of 27.97693 g/mol? 1. 0.0102 g 2. 0.52416 g 3. 0.1004 g 4. 0.25406 g correct Explanation: 116 10.0 points When I emits a β particle, what nuclide is produced? 131 1. 131 Xe correct 2. 131 Te 3. 130 I 4. 130 Te 5. 127 Sb Explanation: 1. positron 2. β particle correct 3. γ particle 4. No particle is emitted because electron capture occurs. 5. α particle Explanation: 119 10.0 points What nuclide is formed when 24 Na undergoes β decay? 1. 24 Ne 2. 28 P 3. 28 S 4. 24 Mg correct Explanation: The nuclear reaction is 24 11 Na 117 10.0 points A nuclide undergoes proton emission to form 52 Fe. What is the nuclide? 3. 53 Co correct 4. 52 Co 5. 53 Fe Explanation: 0 →A Z ? + −1 e A = 24 − 0 = 24 Z = 11 − (−1) = 12 1. 56 Ni 2. 52 Mn Pt → 201 Au ? 24 12 ? is magnesium: 24 11 Na 0 → 24 12 Mg + −1 e 120 10.0 points What is the least massive particle? 1. proton Devillez (ld2653) – Test 1 Review – Devillez – (99998) 2. α particle 31 123 10.0 points For the nuclear reaction 239 235 94 Pu → 92 U + ?, the missing particle is 3. electron correct 4. neutron 1. 1 n. Explanation: 121 10.0 points Uranium-238 decays by emission of an alpha particle. The product of this decay is 2. e− . 3. 42 He. correct 4. 1 p. 1. 234 88 Ra . Explanation: 2. 234 91 Pa . 239 94 Pu 3. 234 90 Th . correct 4 → 235 92 U +2 He 124 10.0 points The half-life of carbon-14 is 5,730 years. If you started with 100.0 g of carbon-14, how much would remain after 4 half-lives? 4. 234 92 U . Explanation: 1. 12.5 g 122 10.0 points Balance the nuclear equation 2. 57.3 g ? −→ 33 15 P 3. 25.0 g 32 15 P + 4. 6.25 g correct 1. None of these Explanation: After 4 half lives the amount of carbon has been halved 4 times, so 2. ? = −10 β A= 3. ? = 34 15 P 4. ? = 42 He A0 100 g = = 6.25 g n 2 24 125 10.0 points If you have 200.0 g of radioisotope with a halflife of 5 days, how much isotope would remain after 15 days? 5. ? = 10 n correct 6. ? = 151 P 1. 40.0 g 7. ? = 11 p 2. 13.3 g Explanation: mass number: 33 − 32 = 1 atomic number: 15 − 15 = 0 32 1 15 P + 0 n −→ 33 15 P 3. 12.5 g 4. 25.0 g correct Explanation: Devillez (ld2653) – Test 1 Review – Devillez – (99998) The half life is 5 days, so after 15 days, three half lives have elapsed. The amount of isotope remaining is A= A0 200.0 g = = 25 g n 2 23 126 10.0 points The half-life of P-32 is 14 days. How long after a sample is delivered can a laboratory wait to use a sample in an experiment if they need at least 10 percent of the original radioactivity? 1. 42 days correct 2. 56 days 3. 14 days 4. 28 days 5. 70 days Explanation: After one half life 50% remains; after two half lives 25% remains; after three half lives 12.5% remains. So, after three half-lives, over 10% of the 32 P remains. 3(14 days) = 42 days, so the laboratory can wait about 42 days but shouldn’t wait longer than that if at least 10% of the original radioactive substance 32 P is needed. 127 10.0 points The isotope 90 Sr has a half-life of 28 years. If you have 16 g of 90 Sr today, how much will be left in 112 years? 1. 1 g correct 2. 8 g 3. 2 g 4 g to 2 g is the third half-life; 2 g to 1 g is the fourth half-life. 1 g of 90 Sr will be left after four half-lives (112 years). 128 10.0 points Calculate the energy change when one nucleus undergoes the fission reaction 235 235 U U + n → 142 Ba + 92 Kr + 2 n . The masses needed are 235 U : 235.04 u; Ba : 141.92 u; 92 Kr : 91.92 u; n : 1.0087 u. 142 1. −2.8 × 10−11 J correct 2. −1.7 × 10−10 J 3. +1.8 × 10−10 J 4. +2.9 × 10−11 J 5. −3.2 × 10−28 J Explanation: 129 10.0 points What nuclear reaction probably does not occur in a conventional fission nuclear reactor designed to produce energy? 1. 235 U + n →239 Pu correct 2. 235 U + n →103 Mo + 131 Sn + 2 n 3. 235 U + n →90 Sr + 143 Xe + 3 n 4. 235 U + n →141 Ba + 92 Kr + 3 n Explanation: If 235 U captures a neutron, it forms an unstable species that breaks into barium and krypton. To make plutonium, 238 U is needed as the starting material. 4. 4 g Explanation: 112 = 4, so 112 years is 4 half-lives. 18 16 g to 8 g is the first half-life; 8 g to 4 g is the second half-life; 32 130 10.0 points In the fission process 239 143 86 92 U → 56 Ba + 36 Kr + X n, what is X? 1. 5 Devillez (ld2653) – Test 1 Review – Devillez – (99998) 2. 20 3. 15 4. 10 correct 5. 25 Explanation: 239 143 86 1 92 U → 56 Ba + 36 Kr + 10 0 n 131 10.0 points In nuclear fission, which is true? 1. One large atom splits into two or more smaller atoms. correct 2. Two or more smaller atoms combine to form one larger atom. 3. Individual protons split into smaller pieces. 4. Electrons combine to form larger, negatively charged particles. Explanation: 132 10.0 points The nuclear equation 2 3 4 1 1 H + 1 H → 2 He + 0 n is an example of 1. a branching chain. 2. nuclear fission. 3. nuclear fusion. correct 4. a chain reaction. Explanation: Two individual atoms combined to form one atom. This is nuclear fusion. 133 10.0 points Which reaction best describes nuclear fusion? 1. 2 C2 H6 + 3 O2 → 4 C + 6 H2 O + energy 33 90 143 2. 10 n + 235 92 U → 38 Sr + 54 Xe + energy 3. 21 H + 31 H → 42 He + 10 n + energy correct 238 4. 42 He + 235 92 U → 94 Pu + energy Explanation: Fusion starts with multiple atoms which form a single atom.