CHEM MID-TERM (2)

CHEM MID-TERM(2) - Name CHM 152 Midterm Answer Sheet Question:1 Chapter 14 EOC 6 Problems are from your textbook(EOC Points 8 Weight = 0.0512 x

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Y Reaction coordinate P H Δ R Eaaaaaaaa X Energy EnerEnergy Uncatalyzed exothermic reaction Y P Eáaaaaaa H Δ R X Catalyzed exothermic reaction Name_______________________________________ CHM 152 Midterm Answer Sheet Problems are from your textbook (EOC) Question:1 Chapter 14 EOC # 6 Points : 8 Weight = 0.0512 x 85g/mol x 0.5kg = 2.176 Number of moles of NaNO 3 = weight molecular weight = . / 2 176mol85g mol = 0.0256mol (n 1 ) Number of moles of H 2 O = . / = . ( ) 500g18 02g mol 27 78mol n₂ Mole fraction of NaNO 3 = + n₁n₁ n₂ = . . + . = . 0 0256mol0 0256mol 27 78mol 9 2 x 10⁻⁴ Question:2 Chapter 14 EOC # 8 Points : 8 - Mole fraction of ethylene glycol = 0.125 n 1 = # of moles of ethylene glycol n 2 = # of moles of water + = . = = n1n1 n2 0 125 n2 weight of waterMolecular weight of water . / = + = . 955g18 02g mol 53moln₁n₁ 53 0 125 + = . n₁ 53n₁ 10 125 1+ 53 n₁ = 8 = = . n1 537 6 625mol Mass of EG = n x Molecular mass of Ethylene glycol: 6.625 x 62.048 = 411.1g - Molarity = ( ) amount of ethylene glycol mol Amount of ( ) water kg Mass of water = 955g = .955kg . . = . 6 625mol0 955kg 6 937m Question:3 Chapter 14 EOC # 14 Points :5 Acetone can form Hydrogen Bond with water by using O in C=O group of acetone molecule and H in water molecule. H 3 C – C – CH 3 H 2 C = C – CH 3 O O-H
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Name_______________________________________ CHM 152 Midterm Answer Sheet Problems are from your textbook (EOC) Question:4 Chapter 14 EOC # 21 Points : 8 O 2 = K x PO 2 = (1.66 x 10 -6) M/mmHg x 40mmHg = 6.6 x 10 -5 M O 2 6.6 x 10 -5 mol/L x 32.0g O 2 = 2 x 10 -3 g O 2 /L Question:5 Chapter 14 EOC # 29 Points : 8 Benzene normally boils at 80.10°C so, (84.2 – 80.10) = 4.1°C Δt = K bp x m solute . . = . 0 200mol0 125 kg solvent 1 60m K bp for benzene = + 2.53°C Δt = K bp x m solute = +2.53°C x 1.60 = + 4.1°C Question:6 Chapter 14 EOC # 41 Points : 8 (61.82 – 61.70) = 0.12°C Δt = m x K bp 0.12°C = m x +3.63°C . ° + . ° / = . 0 12 C 3 63 C m 0 033 molarity . = . = . 0 033 molarity x mol 0 0250kg CHCl₃ 8 3 x 10⁻⁴mol . . - = / 0 125g compound8 3 x 10 4mol 150 g mol
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Name_______________________________________ CHM 152 Midterm Answer Sheet Problems are from your textbook (EOC) Question:7 Chapter 14 EOC # 98 Points : 5 Methanol and ethanol are quite miscible in water because they have polar O-H bonds and less number of non-polar C-C bonds. In the case of higher alcohols there are longer carbon chains and there are large numbers of C-C bonds and they are non-polar, so they oppose the water molecules and solubility decreases. So higher alcohols are not as miscible in water. Question:8 Chapter 15 EOC # 12 Points : 10 (a) For NO the rate of reaction is second order. NO is halved (from 0.420 to 0.210) then the rate of reaction is decreased by 4 times. -For H
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This note was uploaded on 11/16/2008 for the course CHM 152 taught by Professor Thomas during the Fall '08 term at Pima CC.

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CHEM MID-TERM(2) - Name CHM 152 Midterm Answer Sheet Question:1 Chapter 14 EOC 6 Problems are from your textbook(EOC Points 8 Weight = 0.0512 x

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