acidic basic neutral cannot determine CoCl 3 Na 2 Se BaI 2 BaS a Rank the

Acidic basic neutral cannot determine cocl 3 na 2 se

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acidic basic neutral cannot determine CoCl 3 Na 2 Se BaI 2 BaS a) Rank the solutions in order of increasing vapor pressure. Write the formula for one solute in each box. c) Which of the following solutions has the smallest change in entropy due to the salt dissolving? Again assume complete dissolution but consider real solutions. (Circle one) 0.3 m KClO 4 (aq) 0.2 m BaI 2(aq) they experience the same change in entropy G T T f pure d) On the plot to the right: 1) Sketch the Gibbs free energy vs. temperature for an aqueous solution of KClO 4 ( Δ H sol = 51 kJ/mol). 2) Show the y-intercept relative to pure water. 3) Indicate and label the freezing point of the solution. O New freezing point 10 10 4 6 -3 per arrow -2 per mistake 2 pts Δ H 2 pts slope 2 pts freezing point must be below Tm H 2 Te
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CHEM230 _W16 _PREP 3 _key p 27 4 3 6 179) O As HO OH OH As HO OH OH A) B) A is the stronger acid with the lower pK a because the conjugate base can be resonance stabilized as show below. a) Circle the compound with the lower pK a O As HO O OH O As HO O OH c) Explain why isothiocyanic acid (pK a = 1.1) has a lower pK a than cyanic acid (pK a = 3.45) H-S-C N H-O-C N isothiocyanic acid pK a = 1.1 cyanic acid pK a = 3.45 The conjugate base of HSCN has a negative charge on the larger S atom where it is more dispersed and thus more stable whereas the HOCN has the negative chage on the smaller O atom OR HS bond is weaker and easier to break due to the greater disparity in size of H and S than H and O. b) Briefly explain your choice 224) a) Write out the equilibrium that would have a K = 8x10 -11 given that Fe +2 as Fe(H 2 O) 6 +2 has a K a = 8x10 -11 . Included phases for all reactants and products. b) Using one or two sentences, explain why Fe +2 (K a = 8x10 -11 ) has a smaller K a than Fe +3 (K a = 3.5x10 -3 .) 8 Fe(H 2 O) 6 +2 (aq) + H 2 O (l) <--> H 3 O + (aq) + Fe(H 2 O) 5 (OH) +1 (aq) The higher charge on the Fe +3 better stabilizes the negative charge(on the OH-) on the conjugate base making the products more stable and thus made in larger proportion, resulting in a higher K OR gerater the charge, weaken O-H bond in water, making it easier to break the bond and donate a protont - 1 charge -2 incorrect conj. base
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CHEM230 _W16 _PREP 3 _key p 28 214) 50.0 mL of 6.0 M NaOH (aq) (MW = 40) is mixed with 150.0 mL of 4.5 M NaCl (aq) (MW= 58.44) . The density of the resulting solution is 1.027 g/mL. Answer the two questions about the combined solution. a) What is the pH of the combined solution? 200 mL of solution 0.05L 6 molNaOH L ! " # $ % & = 0.3 molNaOH 0.3 molOH 0.3 molOH 0.2 Lsolution = 1.5 M OH pOH = log[ OH ] = log(1.5) = 0.176 pH = 14 pOH = 14 ( 0.176) pH = 14.176 pH = 14.176 b) If the vapor pressure of pure water at 20 o C is 17.5 torr, what is the vapor pressure of the combined solution at 20 o C? P vap,solution = torr 14.25 1 pts mol OH 1 pts volume solution 2 pts [OH - ] (or if used M1V1 = M2V2 then 4 pts here instead of 1+1+2 2 pts pOH 2 pts pH 8 2 pts mass solution 2 pts mass NaOH 2 pts mass NaCl 1 pts mass water; 1 pt mole water 200 mL solution 1.027 g mL !
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