NAME:______Student Number:______

Fall 2019 Chemistry 1000 Midterm #1B ____/ 65 marks

INSTRUCTIONS: 1) Please read over the test carefully before beginning. You should have 5 pages of questions and a formula/periodic table sheet. 2) If your work is not legible, it will be given a mark of zero. 3) Marks will be deducted for incorrect information added to an otherwise correct answer. 4) Marks will be deducted for improper use of significant figures and for missing or incorrect units. 5) Show your work for all calculations. Answers without supporting calculations will not be given full credit. 6) You may use a calculator. 7) You have 90 minutes to complete this test.

Confidentiality Agreement: I agree not to discuss (or in any other way divulge) the contents of this test until after 8:00pm Mountain Time on Thursday, October 10th, 2019. I understand that breaking this agreement would constitute academic misconduct, a serious offense with serious consequences. The minimum punishment would be a mark of 0/65 on this exam and removal of the “overwrite midterm mark with final exam mark” option for my grade in this course; the maximum punishment would include expulsion from this university.

Signature: ______Date: ______Course: CHEM 1000 (General Chemistry I) Semester: Fall 2019 The University of Lethbridge

Question Breakdown Spelling matters! Q1 / 4 Fluorine = F Fluorene = C13H10 Q2 / 4 Q3 / 6 Q4 / 4 Q5 / 6 Flourine = Q6 / 3 Q7 / 2 Q8 / 10 Q9 / 10 Q10 / 3 Q11 / 8 Q12 / 5

Total / 65 NAME:______Student Number:______

1. Complete the following table. [4 marks] Symbol # protons # neutrons # electrons

11 12 11 23 11 𝑁𝑁𝑁𝑁 29 34 27 63 2+ 29 𝐶𝐶𝐶𝐶 9 10 10 19 − 9𝐹𝐹

2. For each of the images below, give the angular momentum quantum number ( ) and indicate whether the orbital is s, p, d or f. [4 marks] (a) (b) ℓ

= ____2____; orbital is __d___ = ____2____; orbital is __d___

(c) ℓ (d) ℓ

= ____0____; orbital is __s___ = ____1____; orbital is __p___

ℓ ℓ 3. List all orbitals in an atom for which = 3. Include the subscript part of each name (where appropriate). Do not draw the orbitals. [6 marks] 𝑛𝑛 3 1 mark 3 3 3 2 marks 3𝑠𝑠 3 3 3 3 3 marks 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 2 2 2 𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 𝑥𝑥 −𝑦𝑦 𝑧𝑧 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑑𝑑 NAME:______Student Number:______

4. [4 marks] (a) Draw an orbital occupancy diagram (“orbital box diagram”) for a neutral ground state nitrogen atom (N). Include all electrons and label all subshells. 2 marks

1s 2s 2p

(b) How many valence electrons does a neutral ground state nitrogen atom have? 5 1 mark

(c) Is a neutral ground state nitrogen atom paramagnetic or diamagnetic? paramagnetic 1 mark

5. Write electron configurations for each of the following species in the ground state. Use the noble gas notation, but always show all electrons in the valence shell explicitly. [6 marks] (a) [ ]5 4 5 2 10 4 𝑇𝑇𝑇𝑇 𝐾𝐾𝐾𝐾 𝑠𝑠 𝑑𝑑 𝑝𝑝 (b) [ ]3 3 − 2 6 𝐶𝐶𝐶𝐶 𝑁𝑁𝑁𝑁 𝑠𝑠 𝑝𝑝 (c) [ ]3 3+ 2 𝑉𝑉 𝐴𝐴𝐴𝐴 𝑑𝑑

6. There are two naturally occurring isotopes of silver ( ). If one of the isotopes is and its abundance is 52%, what must the other isotope be? [3 marks]107 For this question only, it is not necessary to show your𝐴𝐴𝐴𝐴 work; however, partial credit may𝐴𝐴𝐴𝐴 be awarded for correct work in the absence of a correct final answer.

109 𝐴𝐴𝐴𝐴 According to the periodic table, the average atomic mass of Ag is 107.868 u. This rounds to 108 u. The isotopic mass of is 107 u. Therefore, if the two107 isotopes of silver exist in approximately equal amounts, the other isotope must have an isotopic𝐴𝐴𝐴𝐴 mass of 109 u.

It was acceptable to show the calculation to find the exact isotopic mass of , but the logic above was sufficient for full credit. 109 𝐴𝐴𝐴𝐴 NAME:______Student Number:______

7. Write a balanced nuclear equation for the fusion of and . In addition to the main product, a neutron is produced.22 4 [2 marks] 10𝑁𝑁𝑁𝑁 2𝐻𝐻𝐻𝐻 + + 22 4 25 1 10𝑁𝑁𝑁𝑁 2𝐻𝐻𝐻𝐻 → 12𝑀𝑀𝑀𝑀 0𝑛𝑛

8. Potassium-40 is the greatest source of natural radioactivity in humans. It is also useful in the dating40 of minerals. [10 marks] (a) Nine times out� of 𝐾𝐾ten,� decays to . What type(s) of decay could this be? 40 40 beta emission 𝐾𝐾 𝐶𝐶𝐶𝐶 + 40 40 0 (b) One time out of ten, decays to . What19𝐾𝐾 type(s)→ 20𝐶𝐶 of𝐶𝐶 decay−1𝛽𝛽 could this be? 40 40 positron emission or electron𝐾𝐾 capture 𝐴𝐴 𝐴𝐴 + or + 40 40 0 40 0 40 (c) A human with a mass of 60 kg contains about19 120𝐾𝐾 →g potassium,18𝐴𝐴𝐴𝐴 +1 𝛽𝛽of which19𝐾𝐾 0.0123−1𝑒𝑒 g→ are18 𝐴𝐴𝐴𝐴 . The resulting activity is 3.2 × 10 . Calculate the decay constant for . 40 Report your answer in . 3 40 𝐾𝐾 −1 𝐵𝐵𝐵𝐵 𝐾𝐾 𝑠𝑠 = In this question, activity (A) is provided along with enough data to calculate the number of 𝐴𝐴 𝑘𝑘𝑘𝑘 radioactive atoms (N). Step 1: Calculate the number of atoms. The isotopic mass of is on the data sheet. 𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 . × = 0.0123 × × = 1.85 × 10 3 sig. fig. . 𝑲𝑲 23 𝑲𝑲 1 𝑚𝑚𝑚𝑚𝑚𝑚 6 022 141 10 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 20 𝑁𝑁Step 2: Calculate𝑔𝑔 39 the963 decay998 166 constant𝑔𝑔 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = . × = = × = 1.7 × 10 2 sig. fig. 𝐴𝐴 𝑘𝑘𝑘𝑘 . × 3 −1 𝐴𝐴 3 2 10 𝐵𝐵𝐵𝐵 1 𝑠𝑠 −17 −1 20 𝑘𝑘 𝑁𝑁 1 85 10 1 𝐵𝐵𝐵𝐵 𝑠𝑠

(d) Calculate the half-life for . Report your answer in years. There are 31,557,600 seconds40 in a year (factoring in leap years). 𝐾𝐾 (2) = 1 ( ) ( ) 𝑙𝑙𝑙𝑙 = 𝑘𝑘=∙ 𝑡𝑡2 = 4.0 × 10 × = 1.3 × 10 . × 2 sig. fig. 1 𝑙𝑙𝑙𝑙 2 𝑙𝑙𝑙𝑙 2 16 1 𝑦𝑦 9 −17 −1 𝑡𝑡2 𝑘𝑘 1 7 10 𝑠𝑠 𝑠𝑠 31 557 600 𝑠𝑠 𝑦𝑦

NAME:______Student Number:______

9. Calculate the energy change for the reaction in which decays to . [10 marks] (a) Report your answer in J/mol. 40 40 𝐾𝐾 𝐴𝐴𝐴𝐴 + or + You40 can40 use either0 equation40 because0 40the masses of positrons and electrons are not included in 19𝐾𝐾 → 18𝐴𝐴𝐴𝐴 +1𝛽𝛽 19𝐾𝐾 −1𝑒𝑒 → 18𝐴𝐴𝐴𝐴 mass defect calculations. See Exercise 2.4 if you need a reminder of why. Step 1: Calculate the mass change for this reaction (products minus reactants; no β particles) =

∆𝑚𝑚 = 𝑀𝑀39𝐴𝐴𝐴𝐴.−96240 − 383𝑀𝑀𝐾𝐾 −12440 39.963 998 166 𝑔𝑔 𝑔𝑔 ∆𝑚𝑚 = � 0.001 615 042 𝑚𝑚𝑚𝑚𝑚𝑚 � − � 𝑚𝑚𝑚𝑚𝑚𝑚� 9 decimal places therefore 7 sig. fig. 𝑔𝑔 ∆Step𝑚𝑚 2:− Calculate the energy𝑚𝑚𝑚𝑚𝑚𝑚 change for this reaction = 2

∆𝐸𝐸 = ∆𝑚𝑚0𝑐𝑐.001 615 042 2.997 925 × 10 = 1.451 528 × 10 2 2 𝑔𝑔 1 𝑘𝑘𝑘𝑘 8 𝑚𝑚 11 𝑘𝑘𝑘𝑘∙𝑚𝑚 2 ∆𝐸𝐸 = �−1.451 528 × 10 𝑚𝑚𝑚𝑚𝑚𝑚� �1000× 𝑔𝑔� � = 1.451 528 ×𝑠𝑠 �10 − 𝑠𝑠 ∙7𝑚𝑚𝑚𝑚𝑚𝑚 sig. fig. 2 11 𝑘𝑘𝑘𝑘∙𝑚𝑚 1 𝐽𝐽 11 𝐽𝐽 2 𝑘𝑘𝑘𝑘∙𝑚𝑚2 𝑠𝑠 ∙𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 ∆ 𝐸𝐸 − 1 𝑠𝑠2 −

(b) Report your answer in J (for a single decay).

= 1.451 528 × 10 × = 2.410 318 × 10 . × 7 sig. fig. 11 𝐽𝐽 1 𝑚𝑚𝑚𝑚𝑚𝑚 −13 𝐽𝐽 23 ∆𝐸𝐸 − 𝑚𝑚𝑚𝑚𝑚𝑚 6 022 141 10 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

(c) Some of this energy is carried by a neutrino and some is carried by a photon. If the energy of the photon is 2.339 × 10 , what is its wavelength? Report your answer in pm. −13 Combining the steps below via = is𝐽𝐽 an equally valid approach. ℎ𝑐𝑐 Step 1: Calculate frequency of𝐸𝐸 electromagnetic𝜆𝜆 radiation = . × = = = 3.530 × 10 4 sig. fig. 𝐸𝐸 ℎ𝜈𝜈 . × −13 𝐸𝐸 2 339 10 𝐽𝐽 20 −34 𝐽𝐽 Step𝜈𝜈 ℎ 2: Calculate6 626 070 10wavelength𝐻𝐻𝐻𝐻 of electromagnetic𝐻𝐻𝐻𝐻 radiation =

. × 𝑐𝑐 = 𝜈𝜈𝜈𝜈= 8 𝑚𝑚 × = 8.493 × 10 4 sig. fig. 2 997. 925× 10 𝑐𝑐 𝑠𝑠 1 𝐻𝐻𝐻𝐻 −13 20 −1 𝜆𝜆Step𝜈𝜈 3: C3onvert530 10 units𝐻𝐻𝐻𝐻 from1 𝑠𝑠 meters to picometers𝑚𝑚

= 8.493 × 10 × = 0.8493 4 sig. fig. 12 −13 10 𝑝𝑝 𝑝𝑝 𝜆𝜆 𝑚𝑚 1 𝑚𝑚 𝑝𝑝𝑝𝑝 NAME:______Student Number:______

10. Fill in the blanks in the following paragraph. [3 marks] Exposure to radiation can be measured using different units to communicate different information. The ______absorbed______dose is measured in Grays where

1 Gy = 1 ____J/kg____. Multiplying this value by the radiation weighting factor (WR) gives the ______equivalent______dose which is measured in Sieverts.

11. Fill in the blanks in the following paragraphs. [8 marks] It is possible to calculate the exact energies for the orbitals of a cation because 2+ it only has _one electron (or “a single electron”)_. To calculate𝐿𝐿𝐿𝐿 the energy of one = of these orbitals, the formula _ 2_ is used. 𝑍𝑍 𝑛𝑛 𝐻𝐻 2 If the cation is in the _ground𝐸𝐸 _− state𝑅𝑅 𝑛𝑛 then = 1. We also need to know the 2+ value for𝐿𝐿𝐿𝐿 – which represents the number of _protons𝑛𝑛 in the nucleus (or “protons”). The value𝑍𝑍 of for the cation is _3_, and the energy for an electron with = 1 2+ in a cation𝑍𝑍 is _ 1𝐿𝐿𝐿𝐿.961 885 × 10 _ J. 𝑛𝑛 2+ −17 If light𝐿𝐿𝐿𝐿 is absorbed −by the cation, the electron will be _excited_ into another 2+ orbital. The energy of the 𝐿𝐿𝐿𝐿absorbed light will be equal to: (circle one choice) the energy of the = 1 orbital / the energy of the second orbital / the sum of the energies

of the two orbitals𝑛𝑛 / the difference between the energies of the two orbitals.

12. Fill in the blanks in the following paragraphs. [5 marks] The photoelectric effect experiment demonstrated the _particle_ nature of light. In this experiment, light shines on a metal surface. In some cases, this causes a current to flow. If a current flows, that means that _electrons_ have been ejected from atoms in the metal.

Whether or not a current flows depends on the frequency of the incident light. If the frequency of the light is _greater (circling the word “greater” is fine)_ (greater/less) than or equal to the _threshold_ frequency for the metal, current will flow. This experiment therefore demonstrated a relationship between the energy and frequency of light as described by Planck’s equation: _ = _.

𝐸𝐸 ℎ𝜈𝜈 NAME:______Student Number:______

Some Useful Constants and Formulae

Fundamental Constants and Conversion Factors Atomic mass unit (u) 1.660 539 × 10-27 kg Planck's constant 6.626 070 × 10-34 J·Hz-1 Avogadro's number 6.022 141 × 1023 mol–1 Proton mass 1.007 277 u -11 Bohr radius (a0) 5.291 772 × 10 m Neutron mass 1.008 665 u -19 -18 Electron charge (e) 1.602 177 × 10 C Rydberg Constant (RH) 2.179 872 x 10 J Electron mass 5.485 799 × 10-4 u Speed of light in vacuum 2.997 925 x 108 m·s-1

Formulae

h h c = λυ E = hυ p = mv λ = ∆x ⋅ ∆p > p 4π

n2 Z 2 1 r = a E = −R E = mv 2 ∆E = ∆mc2 n 0 Z n H n2 k 2

∆N  N  2 = ⋅ = A = − A = kN ln  = −k(t2 − t1 ) ln(2) k t1/ 2 ∆𝑡𝑡 1   1 �𝑡𝑡 �2 ∆t  N1  𝑁𝑁2 𝑁𝑁1 �2�

1 Chem 1000 Standard Periodic Table 18 1.0079 4.0026 H He 1 2 13 14 15 16 17 2 6.941 9.0122 10.811 12.011 14.0067 15.9994 18.9984 20.1797 Li Be B C N O F Ne 3 4 5 6 7 8 9 10 22.9898 24.3050 26.9815 28.0855 30.9738 32.066 35.4527 39.948 Na Mg Al Si P S Cl Ar 11 12 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 39.0983 40.078 44.9559 47.88 50.9415 51.9961 54.9380 55.847 58.9332 58.693 63.546 65.39 69.723 72.61 74.9216 78.96 79.904 83.80 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 85.4678 87.62 88.9059 91.224 92.9064 95.94 (98) 101.07 102.906 106.42 107.868 112.411 114.82 118.710 121.757 127.60 126.905 131.29 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 132.905 137.327 178.49 180.948 183.85 186.207 190.2 192.22 195.08 196.967 200.59 204.383 207.19 208.980 (210) (210) (222) Cs Ba La-Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 55 56 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 (223) 226.025 (265) (268) (271) (270) (277) (276) (281) (280) (285) (284) (289) (288) (293) (294) (294) Fr Ra Ac-Lr Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og 87 88 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118

138.906 140.115 140.908 144.24 (145) 150.36 151.965 157.25 158.925 162.50 164.930 167.26 168.934 173.04 174.967 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 227.028 232.038 231.036 238.029 237.048 (240) (243) (247) (247) (251) (252) (257) (258) (259) (262) Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103

Developed by Prof. R. T. Boeré (updated 2016)

NAME:______Student Number:______

Some Useful Masses 4 4.002 603 254 u 2α 1 1.007 276 467 u 1 p 1 1.008 664 916 u 0 n 0.000 548 579 9 u 0 0.000 548 579 9 u +1 0𝛽𝛽 39.962 383 124 u −1 40 𝛽𝛽 39.962 590 863 u 𝐴𝐴𝐴𝐴 40 39.963 998 166 u 40𝐶𝐶𝐶𝐶 𝐾𝐾

Band of Stability Graph The graph below shows the band of stability. The black dots represent all known stable isotopes. Stable nuclei 140

130

120

110

100

90

80 N = Z

N 70

60

50

40

30

20

10

0 0 10 20 30 40 50 60 70 80 90 Z