M3 Problem Set

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Chemistry

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Jan 9, 2024

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M3: Problem Set Due No due date Points 5 Questions 13 Time Limit None Instructions You may find the following resources helpful while working through these problem sets: | e [ —— ] ————————————— ] ——————————— ) S—— ] e ———— e || S ] | Equation Table =_(https://previous.nursingabc.com/upload/images/Help_file_picture/ CHEM121%20- %20Equation%20Bank%5B20719%5D.pdf) How to Draw Lewis Structures (https://portagelearning.instructure.com/courses/1952/files/637411/download ?wrap=1)_ (https://portagelearning.instructure.com/courses/1952/files/637411/download?download_frd=1) Table of Water Vapor Pressures & —_—f_—_—_—_—_e—_——-- - —— s re—_——eeeeeee e e - ——_—,_—_—_—_—,——— e —1 Attempt History Attempt Time Score LATEST Attempt 1 2,498 minutes S5outof 5 Score for this quiz: 5 out of 5 Submitted Dec 9 at 10:28am This attempt took 2,498 minutes. Question 1 0/0 pts Show your work and be sure to give answer with the correct units.
A gas sample has an original volume of 680 ml when collected at 720 mm and 28°C. What will be the volume of the gas sample if the pressure increases to 820 mm and the temperature increases to 55°C? Your Answer: Pi x Vi = Pf x Vf Ti Tf 680 ml/1000 = 0.680 liters = Vi 720 mm/760 = 0.947 atm = Pi 820 mm/760 = 1.08 atm = Pf 28 C + 273 = 301 K = Ti 55 C + 273 = 328 K = Tf Put in the data: (0.947) x (0.680) = (1.08) x Vf (301) (328) Solve for Vf: 0.002139402 = 0.003292683 x Vf Vf = 0.650 liters P, x V, P,xV, T T, 680 mI/1000 = 0.680 liters =V, 720 mm/760 = 0.947 atm = P; 28°C + 273 = 301K =T, 820 mm/760 = 1.08 atm = P; 55°C + 273 = 328K =T Put in the data: (0.947) x (0.680) = (1.08) x V; (301) (328) Solve for V. 0.002139402 = 0.003292683 x Vs: Vs = 0.002139402 = 0.650 liter 0.003292683 Question 2 0/0 pts
Show your work and be sure to give answer with the correct units. A sample of CO, gas which weighs 1.62 grams has a volume of 1.02 liters when collected at 20°C. What would be the pressure of the gas sample? Your Answer: CO2 C 12.011 O 15.999 x 2 44.009 1.62 / 44.009 = 0.0368 mol = n V= 1.02LR=0.0821T=20+ 273 =293 K? x 1.02 =0.0368 x 0.0821 x 293 Px1.02=0.8850.885/1.02 =P = 0.868 atm PxV=nxRxT CO> molecular weight = 44 1.62 grams/44 = 0.03682 mole = n R =0.0821 1.02 liters =V 20°C + 273 =293K=T P x (1.02) = (0.03682) x (0.0821) x (293) P x (1.02) = 0.8857 P =0.8857 = 0.868 atm 1.02 Question 3 0/0 pts Show your work and be sure to give answer with the correct units.
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The combustion of benzene (CgHg) takes place by the following reaction equation. 2 CeHg (9) + 15 02 (g) 12 CO2 (g) + 6 H20 (9) What is the volume of CO, gas formed by the combustion of 18.5 grams of CgHg at 30°C and 1.50 atm? Your Answer: (MW =78) (MW = 32) (MW =44) (MW =18) 2 C6H6 (g) + 15 02 (g) 12 CO2 (g) + 6 H20 (g) 18.5 grams/78 = 0.2372 mol C6H6 0.2372 mol C6H6 X (12 mol CO2/2 mol C6HB) = 1.4232 mol CO2 PV=nRT V =nRT/P V = (1.4232)(0.0821)(303) / 1.50 V = 23.6 liters (MW =78) (MW=32) (MW=44) (MW =18)2 CgHg(g) + 15 O5 (g) = 12 CO5(s) + 6 H,O (I) 18.5 grams 23.6 liters | 1 0.2372 mol —» - 12/2 x 0.2372 mol V = nRT /P = (1.4232)(0.0821)(303) / 1.50 = 23.6 liters Question 4 0/0 pts Show your work and be sure to give answer with the correct units. A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25°C. Determine the partial pressures of He and Ne in the flask. What is the total pressure? Your Answer: n(He) = g(He) / [MW(He)] = (0.52 mg x 1 g/ 1000 mg) / 4.002 = 0.000129935 mol n(Ne) = g(Ne) / [MW(Ne)] = (2.05 mgx 1 g/ 1000 mg) / 20.18 = 0.000101586 mol PT=n R T/V = (0.000129935 + 0.000101586)
(0.0821) (298K) / 1.00 = 0.005664 atm PT = 0.005664 atm x 760 mm/1 atm = 4.305 mm X(He) = 0.000129935 / (0.000129935 + 0.000101586) = 0.5612 X(Ne) = 0.000101586 / (0.000129935 + 0.000101586) = 0.4388 P(He) = X(He) (4.305 mm) = 0.5612 (4.305 mm) = 2.416 mm P(Ne) = X(Ne) (4.305 mm) = 0.4388 (4.305 mm) = 1.889 mm n(He) = g(He) / [MW(He)] = (0.52 mg x 1 g/ 1000 mqg) / 4.002 = © 0.000129935 mol n(Ne) = g(Ne) / [MW(Ne)] = (2.05 mgx 1 g/ 1000 mg)/ 20.18 = ~ 0.000101586 mol Pr=nRT/V=(0.000129935 + 0.000101586) (0.0821) (298K) / 1.00 = 0.005664 atm Pt = 0.005664 atm x 760 mm/1 atm = 4.305 mm X(He) = 0.000129935 / (0.000129935 + 0.000101586) = 0.5612 X(Ne) = 0.000101586 / (0.000129935 + 0.000101586) = 0.4388 P(He) = X(He) (4.305 mm) = 0.5612 (4.305 mm) = 2.416 mm P(Ne) = X(Ne) (4.305 mm) = 0.4388 (4.305 mm) = 1.889 mm Question 5 0/0 pts Show your work and be sure to give answer with the correct units. A sample of hydrogen (H»>) gas is collected over water at 35°C and 745 mm. The volume of the gas collected is 72.0 ml. How many moles of gas has been collected? How many grams of H, gas has been collected? (Click here (https://portagelearning.instructure.com/courses/1952/files/637992?wrap=1) J N—
(https://portagelearning.instructure.com/courses/1952/files/637992/download download_frd=1) for Water Vapor Pressure Table). Your Answer: PH2 =725 - P H20 (from table) = 725 -42.2 =682.8 x 1 atm / 760 mm = 0.898 atm H2 from Ideal Gas Law: n H2 = PV / RT = (0.898 atm) (72.0 ml X 1 liter / 1000 ml) / (0.0821) (3080K) n H2 = 0.002557 moles grams H2 = moles x MW = 0.002557 moles x 2.016 grams / 1 mole = 0.00515 grams P(Hz) = 745 - PHo0) (from table) = 745 - 42.2 = 702.8 x 1 atm / 760 mm = 0.925 atm H, from ldeal Gas Law: n(H5) = PV / RT = (0.925 atm) (72.0 ml x 1 liter / 1000 ml) / (0.0821) (308K) n(H2) = 0.002634 moles grams(H»>) = moles x MW = 0.002634 moles x 2.016 grams / 1 mole = 0.0053 grams Question 6 0/0 pts Show your work and be sure to give answer with the correct units. The rate of effusion of nitrogen gas (N») is 1.253 times faster than that of an unknown gas. What is the molecular weight of the unknown gas®? Your Answer: 1.253)"2 = MW(unknown) / 28.014 MW(unknown)= 28.014 x (1.253)2 28.014 x 1.570009 = 43.99
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(rn2) / runknown)2 = MW nknown / MWp2 (1.253/1)? = MW nknown / 28.02 MW .nknown = (1.253)? x 28.02 = 43.99 Question 7 0/0 pts Write electron (subshell) configurations for the following elements: 19K 35Br Your Answer: 19K = 1s22s522p63s23p64s1 35Br = 1522s22p63s23p64s23d104p5 19K = 15?2 252 2p® 3s? 3p° 45 35Br = 12 2s? 2p® 3s? 3p® 452 3d'° 4p° Question 8 0/0 pts
The following document will be helpful in understanding how to insert arrows as part of your answers. CHEM 121 - Module 3.7 - How to Insert Arrows (https://portagelearning.instructure.com/courses/1952/files/637394/download wrap=1)_ {, (https://portagelearning.instructure.com/courses/1952/files/637394/download download_frd=1) Write orbital diagrams for the following atoms: 15P 2gNIi G > Your Answer: 15P = 1s22s22p63s23p3 1| 11 TITITL T4- 11T 1 1s2s. 2p 3s. 3p 28Ni = 1522522p63s23p64s23d8 1| 1| TITITL T NI UL LT T 1s 2s 2p 3s 3p 4s 3d
15P = 182 252 2p® 352 3p3 a5 W o MO 21 o s O I W o . 1s 25 2p 3 3p 2sNi = 152 252 2p® 352 3p°® 452 37° 5 e 4 A o 6 3 A N s e s o o G A s 1s 23 2p 3s 3p 4s 3d Question 9 0/0 pts Write the values for the four quantum numbers for a 4d® electron. Your Answer: NNty Ttn=41=2Mi=0Ms =-1/2 A 4d3 electron is in the 4th shell so n = 4. Its "d" orbital has 2 nodal planes so |1 =2. The electron is in the 3rd of the orbitals of the d subshell so m; = 0. And since this is the 2nd electron to occupy this "d" orbital, mg = -1/2. In summary, the values for the four quantum numbers for the 4d® electron are: n =4, | =2, m=0and mg =-1/2. Question 10 0/0pts Write the values for the four quantum numbers for the last electron to fill 2gNi.
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Your Answer: 28Ni = 1822522p63s23p64s23d8 1] 1. TITITL- TL- NI 1s 2s 2p 3s 3p 4s 3d n=31=2Mi=0Ms=-1/2 The last electron to fill ogNi is a 3d® electron which is in the 3rd shell so n = 3. Its "d" orbital has 2 nodal planes so | =2. The electron is in the 3rd of the orbitals of the d subshell som; = 0. And since this is the 2nd electron to occupy this "d" orbital, mg = -1/2. In summary, the values for the four quantum numbers for the 3d® electron are:n=3,1=2, m;=0 and mg = -1/2. Question 11 0/0 pts Choose the atom that matches the identified atomic property and explain your answer based on the periodic trend and its position within a group or period in the periodic table & (http://www.nursingabc.com/upload/images/Help_file_ picture/Periodic_Table. 1. Highest lonization Energy: Li, Be, B 2. Lowest Electronegativity: Li, Be, B 3. Largest Atomic Size: Li, Be, B « B Your Answer: 1. Boron (B) has the highest ionization energy. lonization energy iIncreases across a period from left to right.
2. Lithium (Li) has the lowest electronegativity. Since Lithium is the leftmost element among the three in the same period, it has the lowest electronegativity. 3. Lithium (Li) has the largest atomic size. Lithium is the leftmost among these elements, it has the largest atomic size in this group. 1. B has the highest ionization energy since it is farthest to right of these elements and ionization energy increases as you go from left to right in a period. 2. Li has the lowest electronegativity since it is farthest to left of these elements and electronegativity increases as you go from left to right in a period. 3. Li has the largest atomic size since it is farthest to left of these elements and atomic size decreases as you go from left to right in a period. Question 12 0/0pts Create an orbital diagram and Lewis dot structure for 4¢S. Your Answer: 16S = 1522522p63s23p4 1. 1L TITITL TL 14 11 1s2s 2p 3s 3p S 16S = 152 2522pd 3s23p4 s $0tL MLTLTL H oMTt . S ) 1s 2s 2p 35 3p o0
Question 13 515 pts As a reminder, the questions in this review quiz are a requirement of the course and the best way to prepare for the module exam. Did you complete all questions in their entirety and show your work? Your Answer: Yes Quiz Score: 5 out of 5
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