1442answerstest1

1442answerstest1 - Answers Test 1 Chapters 12 & 13...

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Answers Test 1 Chapters 12 & 13 Intermolecular Attractions and Properties of Solutions (Silberburg) 1. (a) Hydrogen Bonding (b) London Dispersion (c) ion-ion (d) dipole-dipole (e) London 2. (d) the only one that is non-polar. CH 4 is non-polar. Likes dissolve likes. 3. (c) 4. m = moles solute/ Kg solute = (45g/40g/mol) / .1 kg water = 11.25m M = moles solute / L of solution = (45g/40g/mol) / .0967 L = 11.63 M To get L of solution: total mass of solution is 45g NaOH + 100g water = 145g D = M/V so V = M/D = 145g/1.5 g/ml = 96.7 ml or .0967 L 5. Assume you have 100 g of solution the it is 50g HCl and 50g water. m = moles solute / Kg solvent = ( 50gHCl/ 36.5g/mol) / .05 Kg water = 27.4m 6. i is almost equal to the number of ions a compound will break into in water. (a) NaCl is a soluble ionic compound and breaks into two ions so i = 2 (b) CH 3 CH 2 CH 3 is not ionic and does not ionize in water so i = 1. (c)
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1442answerstest1 - Answers Test 1 Chapters 12 & 13...

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