Ionized in solution so there are more particles to

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ionized in solution, so there are more particles to interfere with the freezing process (i= 2). For HF, it is partially ionized and fewer particles are in solution. (50) Part II: Calculations (12) 1. 54.7% Sn (be sure to correct for the vapor pressure of water “collected over water” as you calculate the moles of gas collected. 2. 195 g/mol (8) 3. Kp = 14.6 (15 with 2 sig figs) Find the partial pressures of each component from the total pressure. Total Pressure = (.96-x) + (1.02 –x) + x = 1.22 atm x=0.76 (10) 4. pH = 2.37 Find the molarity of the acid before you calculate the pH (10) 5. Calculate K b = 2.37 x 10-5 (25) Part III: Multiple Choice 5 questions @ 5 points each (5) 1. 11.94 Note that Ca(OH) 2 releases 2 moles of OH–. (5) 2. rate = k [A] 2 [B 2 ] 2 find the equivalent for 2AB from the previous step and substitute (5) 3 0.829 M (5) 4. shift to the left, [NO] decreases, K remains the same (5) 5. is more than 3.0 M because Q is greater than K so the shift is to the left
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(48) Part IV: Multiple Choice. Worth 3 points each. 1. 7.9 x 10 4 2. K p =K c (RT) 2 3. pH 6.27 4. H 2 PO 4 5. H 2 X because the equilbirum is toward the weaker of the two acids and the weaker of the two bases 6. c) Zn(NO 3 ) 2 because it is a transition metal 7. HF 8. d) CO 3 2– 9. B 2 10. bond angles between hybrid orbitals are increasing 11. neon filled 12. b) H 2 <CO<HF <NaCl 13. Cl2 because it has the most unshared electrons to interact with the water dipole 14. FCC 4 atoms 15. 0.0987 Latm 16. Exothermic overall, but activation energy must be positive.
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