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Chemistry 2100 In-Class Test 1(A)

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Chemistry 2100 In-Class Test 1(A) 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 29 34 27 63 2+ 29 9 10 10 19 − 9 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] 3 1 mark 3 3 3 2 marks 3 3 3 3 3 3 marks 2 2 2 − 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 (b) [ ]3 3 − 2 6 (c) [ ]3 3+ 2 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 your work; however, partial credit may be awarded for correct work in the absence of a correct final answer. 109 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 isotopic 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 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] 10 2 + + 22 4 25 1 10 2 12 0 → 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 ten,� decays to . What type(s) of decay could this be? 40 40 beta emission + 40 40 0 (b) One time out of ten, decays to . What19 type(s)→ 20 of decay−1 could this be? 40 40 positron emission or electron capture + or + 40 40 0 40 0 40 (c) A human with a mass of 60 kg contains about19 120 →g potassium,18 +1 of which19 0.0123−1 g→ are18 . The resulting activity is 3.2 × 10 . Calculate the decay constant for . 40 Report your answer in . 3 40 −1 = In this question, activity (A) is provided along with enough data to calculate the number of radioactive atoms (N). Step 1: Calculate the number of atoms. The isotopic mass of is on the data sheet. . × = 0.0123 × × = 1.85 × 10 3 sig. fig. 23 1 6 022 141 10 20 Step 2: Calculate 39 the963 decay998 166 constant 1 = . × = = × = 1.7 × 10 2 sig. fig. . × 3 −1 3 2 10 1 −17 −1 20 1 85 10 1 (d) Calculate the half-life for . Report your answer in years. There are 31,557,600 seconds40 in a year (factoring in leap years). (2) = 1 ( ) ( ) = =∙ 2 = 4.0 × 10 × = 1.3 × 10 . × 2 sig. fig. 1 2 2 16 1 9 −17 −1 2 1 7 10 31 557 600 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 + or + You40 can40 use either0 equation40 because0 40the masses of positrons and electrons are not included in 19 → 18 +1 19 −1 → 18 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) = ∆ = 39.−96240 − 383 −12440 39.963 998 166 ∆ = � 0.001 615 042 � − � � 9 decimal places therefore 7 sig. fig. ∆Step 2:− Calculate the energy change for this reaction = 2 ∆ = ∆0.001 615 042 2.997 925 × 10 = 1.451 528 × 10 2 2 1 8 11 ∙ 2 1000 ∙ ∆ = �−1.451 528 × 10 � � × � � = 1.451 528 ×�10 − 7 sig. fig. 2 11 ∙ 1 11 2 ∙2 ∙ ∆ − 1 2 − (b) Report your answer in J (for a single decay). = 1.451 528 × 10 × = 2.410 318 × 10 . × 7 sig. fig. 11 1 −13 23 ∆ − 6 022 141 10 − (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 = is an equally valid approach. ℎ Step 1: Calculate frequency of electromagnetic radiation = . × = = = 3.530 × 10 4 sig. fig. ℎ . × −13 2 339 10 20 −34 Step ℎ 2: Calculate6 626 070 10wavelength of electromagnetic radiation = . × = = 8 × = 8.493 × 10 4 sig. fig. 2 997. 925× 10 1 −13 20 −1 Step 3: C3onvert530 10 units from1 meters to picometers = 8.493 × 10 × = 0.8493 4 sig. fig. 12 −13 10 1 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 calculate the energy of one = of these orbitals, the formula _ 2_ is used. 2 If the cation is in the _ground _− state then = 1. We also need to know the 2+ value for – which represents the number of _protons in the nucleus (or “protons”). The value of for the cation is _3_, and the energy for an electron with = 1 2+ in a cation is _ 1.961 885 × 10 _ J. 2+ −17 If light is absorbed −by the cation, the electron will be _excited_ into another 2+ orbital. The energy of the absorbed 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 orbitals / 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: _ = _. ℎ 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 ∆ 1 1
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