gen chem II LG iv - exam 1 review (key)

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Daniela Sanchez Giannakis Gen Chem II LG Session 4 ds6547@nyu.edu Exam 1 Review – Chapters 11, 13, and 16 Question 1: Write the Lewis structure for the following molecules/polyatomic ions, indicating all non-zero formal charges and showing all important resonance structures . Also, write the electron group arrangement, molecular shape , and hybridization scheme. Molecule or Ion Lewis Structures EGA, Shape, Hybridization CLF 4 + AlCl 4 - ClF 4 - SF 2 ClO 2 - BrO 3 - trigonal bipyramidal - · seesaw je spd tetrahedral oi tetrahedral 1 - see sp3 .. octahedral - square - e sp5c W tetrahedral S · bent W it F : sp3 · Tetrahedral .. si -> : = it .. - . . sp3 11 tetrahedral - : - · je Br - P trigonal pyramidal - 8 · -" e = - : ...:: 0 = .. sp5 B g ..
Daniela Sanchez Giannakis Gen Chem II LG Session 4 ds6547@nyu.edu Question 2: Use the two types of molecular orbital diagrams used in Ch. 11 to answer the following questions – if a particular species can’t exist, don’t include it! (a.) Write out the valence electron configuration for N 2 + (b.) Which of: N 2 2- , N 2 - , N 2 , N 2 + , N 2 2+ has/have the smallest bond energy? (c.) Place the species from the previous question in order of increasing bond length. * heavy-weight * light-weight (light-weight) Ga Os Ta Ga / o Band order= 18-7 e -> Bond order= + (6-2) 2 Na < Nat <Na-c Nat > Na
Daniela Sanchez Giannakis Gen Chem II LG Session 4 ds6547@nyu.edu (d.) Which of the species from part b is/are diamagnetic? (e.) Write out the valence electron configuration for F 2 - (f.) Which of F 2 2- , F 2 - , F 2 , F 2 + , F 2 2+ has/have the largest bond energy? (g.) Place the species from the previous question in order of decreasing bond length. (h.) Which of the species from part f is/are paramagnetic? Na , Nat heavy-weight · s os O T No" Hit s Bond order = (8-4) = 2 Fa Fe > Fa > Fas Fat Fat , Fat < Fa
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Daniela Sanchez Giannakis Gen Chem II LG Session 4 ds6547@nyu.edu Question 3: Which of the following molecules depicted below has the higher boiling point? Hint: Think about what factors influence the boiling point of a compound (i.e. size, polarity, strength of intermolecular forces…). Question 4: Calculate the vapor pressure of a solution of 34.0 g of glycerol (C 3 H 8 O 3 ) in 500.0 g of water at 25°C. (Assume ideal behavior.) the linear pentance molecules can pack together more easily this allows for a greater surface area of interactions stwe each molecular compared to the spherical reopentance molecules Therefore , the pentance molecules will experience stronger dispersion forces , resulting in a higher bailing point Prevent = & sevent po - ) 23 . 74 torr solvent 27 . 753 "HaU=N , = 7775 e e 3 Xia HHx+Nglycerol = 0 98687 -- Agagerol- groe = 0 . 369188 mol Psolment 0 9487 (23-76tow) = 23 . 4 four (to 3 s . f . )
Daniela Sanchez Giannakis Gen Chem II LG Session 4 ds6547@nyu.edu Question 5: An aqueous solution of the non-electrolyte D-glucose (C 6 H 12 O 6 ) is prepared to have a molarity of 4.261M at 20.0°C. The solution is observed to have a density of 1.2793g/cm 3 at that temperature. (a.) Calculate the molality of D-glucose (b.) Calculate the mass % of D-glucose (c.) Calculate the mole fraction of water v . B .: Assume you have /L of solution · MH20 1 , 279 . 39-767 444y · "D-gecose = 4 . 261 mol Sin 12) = S11 . 656 g Mp-gencose = 4) may (115gmet) = 644g m= my - solution = " as , i e x 12 = 1 . 2793kg e = 8 328 m (to 4s f ) - volume density (ing(c) 767 644y x 100 % = 20 . 00 % (to 4 s f ) - 1279 . 3 I (r)) et +4 name) = 0 . 8695 ( 4sf ) MH20 Miar+Mp-glucose
Daniela Sanchez Giannakis Gen Chem II LG Session 4 ds6547@nyu.edu (d.) Calculate the normal freezing point (in °C ). Question 6: A biochemical engineer isolates a bacterial gene fragment and dissolves a 10.0-mg sample in enough water to make 30.0 mL of solution. The osmotic pressure of the solution is 0.340 torr at 25°C. (a.) What is the molar mass of the gene fragment? (b.) If the solution density is 0.997 g/mL, how large is the freezing point depression for this solution (K f of water = 1.86°C/m)? RFCH20) = 1 86° by more 1 T = k = z = - 1 84 (8 . 328) =- 15 . 49° (a . ) pU= URT P 0 . 348 too = 4 50x184atm , V = 0 032 , R = 0 082057 , T = 298 . 15K new (0 0300 = s . S15x mol = MMgene fragment - z = 1 . 81x10" gm H = 0 082057 + 298 . IS (b) 1T - kfx M essention = 30 . 0mL (0 . 997g - mi)) = 29 . 91 g : mass of solvent = 29 . 91y-0 01g = 0 . 0299kg -T = -1 84 x otmre=-3 . 43x15 .
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