HOLLINSED CHEM131 PRACTICE EXAM 1 Spring 2014 Name Section Circle 1115 M 12 00 pm 1116 M 1 00 pm 1135 W 12 00 pm 1136 W 1 00 pm 1118 M 3 00 pm 1138 W 3 00 pm Please show all work in the space provided below the problems This will particularly help if you are seeking partial credit You may use a pencil on this exam You are also permitted encouraged to use a calculator You may not use any other electronic device in lieu of a calculator This exam is closed note and closed book When you provide an answer provide the appropriate unit e g g Hz m mL J etc if the answer is numerical Here is some information you may find useful N0 6 022 x 1023 c 3 00 x 108 m sec h 6 626 x 10 34 J sec E 1 2mv2 E h c E hc E 2 18 x 10 18 1 2 n final 1 2 ninitial En 2 18 x 10 18 1 n 2 mass of electron 9 11 x 10 31kg mass of proton 1 67 x 10 27kg Group I Alkali metals form ions with 1 charge Group II Alkaline Earth metals form ions with 2 charge Group VII Halogens form ions with 1 charge Group VI non metals form ions with 2 charge Aluminum forms a cation with a 3 charge Silver forms a cation with a 1 charge hydrogen sulfate bisulfate hydrogen carbonate bicarbonate Some polyatomic ions 2 sulfate SO4 HSO4 2 SO3 HSO3 OH NH4 sulfite hydroxide ammonium hydrogen sulfite bisulfite 2 CO3 HCO3 N3 NO3 CN 3 PO4 carbonate nitride nitrate cyanide phosphate C Hollinsed Chem 131 Spring 2014 Exam 1 PRACTICE February 21 2014 page 1 DO NOT WRITE ON THIS PAGE 1 2 3a 3b 3c 3d 3e 4 5 6 7 8 9 10 Total 10 10 10 10 10 10 5 5 5 5 5 5 5 5 100 C Hollinsed Chem 131 Spring 2014 Exam 1 PRACTICE February 21 2014 page 2 1 Name the following compounds 2 Provide the formula for the 10 points a CuSO4 5H2O Copper II sulfate pentahydrate Cupric sulfate pentahydrate Copper sulfate pentahydrate b Na2CO3 Sodium Carbonate c N2O4 Dinitrogen tetroxide d LiHSO4 e PCl5 Phosphorous pentachloride 3 following compounds 10 points a Magnesium Nitride b Sodium sulfate decahydrate Mg3N2 Na2SO4 Al2 CO3 3 c Aluminum carbonate d Ammonium phosphate e Ammonium bicarbonate NH4HCO3 Lithium hydrogen sulfate Lithium bisulfate NH4 3PO4 a Write the full nuclear symbol elemental symbol with atomic number and mass number for an isotope of Vanadium having 27 neutrons 5 points Since vanadium is element number 23 it has 23 protons If it has 27 neutrons as well then the mass number will be 23 27 51 51 V23 b This element forms an oxide of having the formula V2O5 What is the charge on Vanadium in this compound 5 points Since oxygen has a 2 charge 5 oxygens will equal 10 Each Vanadium must be 5 c What is the electron configuration orbital diagram and noble gas configuration for neutral vanadium 10 points Electron configuration 1s22s22p63s23p64s23d3 or 1s22s22p63s23p64s23d34s2 Noble gas configuration Ar 4s23d3 Orbital diagram 1s 2s 2p 2p 2p 3s 3p 3p 3p 4s 3d 3d 3d 3d 3d C Hollinsed Chem 131 Spring 2014 Exam 1 PRACTICE February 21 2014 page 3 d Is Vanadium diamagnetic 5 points No with 3 unpaired electrons it will be paramagnetic e Is the ion in part b paramagnetic 5 points The ion in part b will have the closed shell configuration and will not be paramagnetic 4 What is the standard atomic weight of Chromium which occurs naturally as 4 345 50Cr mass 49 9460 83 789 52Cr mass 51 9405 9 501 53Cr mass 52 9107 and 2 365 54Cr mass 53 9389 10 points 50Cr 0 04345 x 49 9460 2 1702 52Cr 0 83789 x 51 9405 43 5204 53Cr 0 09501 x 52 9107 5 0270 54Cr 0 02365 x 53 9389 1 2757 total 51 99 4 significant figures 5 The light emitted by an electron falling from the n 5 level to the n 2 level in hydrogen has a wavelength of 434 nm What is the energy difference between the n 2 level and the n 5 level 10 Points Method 1 Method 2 E hc E 6 626 10 34 J sec 3 00 108 m sec 434 10 9 m 1 2 ninitial 2 n final E 2 18 x 10 18 J 1 E 2 18 x 10 18 J 1 22 5 2 E 2 18 x 10 18 J 1 4 1 1 25 E 1 987 10 25 434 10 9 m 4 58 10 19 J E 2 18 x 10 18 J 0 25 0 04 6 It takes 4 184 J of energy to raise 1 gram of water by 1 degree C Assuming the water actually absorbs all of the energy how many photons from the 434 nm wavelength light in problem 4 will it take to raise 3 08 grams of water by 4 70 degrees 10 points E 2 18 x 10 18 J 0 21 4 58 10 19 J 4 70 deg C 3 08 g H 2O 60 6 J 1 32 1020 photons 4 184 J g deg C 60 6 Joules 4 58 10 19 Joule photon points C 22 x 12 01 264 22 g C mol Sildenafil H 30 x 1 01 30 30 g H mol Sildenafil N 6 x 14 01 84 06 g N mol Sildenafil 7 a What is the molecular weight molar mass of Sildenafil Viagra C22H30N6O4S 5 C Hollinsed Chem 131 Spring 2014 Exam 1 PRACTICE February 21 2014 page 4 O 4 x 16 00 64 00 g O mol Sildenafil S 1 x 32 06 32 06 g S mol Sildenafil Total 474 64 g mol Sildenafil 7 b What is the weight nitrogen in Sildenafil 5 points N 17 71 84 06 474 64 8 Minoxidil Rogaine is 51 66 Carbon 7 23 Hydrogen 33 47 Nitrogen and 7 65 Oxygen What is its molecular formula 10 points C 51 66 12 01 4 301 0 4781 8 995 H 7 23 1 01 7 158 0 4781 14 97 N 33 47 14 01 2 389 0 4781 4 997 O 7 65 16 00 0 4781 0 4781 1 000 C9H15N5O C Hollinsed Chem 131 Spring 2014 Exam 1 PRACTICE February 21 2014 page 5