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Ms. Richards

Written by Kim Richards

Chem II Review Answers

 

Ch. 5

 

9. 1.73 torr

10. 5.3 atm

 

12. 0.42 mole

 

33. 125 g

 

36. 736 torr; 26.8 mL

 

37. a. 0.79 atm; b. 0.21 atm

 

38. O2 = 0.0831 atm;

H2 = 1.32 atm;

N2 = 0.0952 atm

 

41. 730 torr

 

45. a. no change; b. increase;

c. goes to zero

 

49. Ne diffuses 1.48 x faster.

 

54. a. ideal = 0.243 atm; (don't

do Van der Waals!)

b. ideal = 24.3 atm

 

 

Ch. 8

 

38. a. see-saw b. bent

c. square planar

d. trigonal bypyramidal

e. linear

 

51. A

53. D

54. B

60. A

63. C

65. A

66. B

 

 

 

 

 

 

Ch. 11

 

16. a. second one

b. first one

c. first one

 

18. 440 L atm/mol

 

31. 291 torr

 

35. -1.8oC

 

41. 3.90 atm

 

50. 2.49 atm

 

53. D

55. D

56. D

59. C

61. A

 

 

 

Ch. 6

 

4. a. -1.81 kJ

b. -776 kJ

 

11. 240 kJ

 

18. -890.37 kJ

 

21 e. -125 kJ

 

32. B

35. B

 


Ch. 22 Questions:


1)  Hydrocarbon chains with a double bond are named with what ending?

2)  The -yne ending is given to hydrocarbons with which type of C-C bond?

3) When hydrogen atoms are added to unsaturated hydrocarbons, what kind of reaction takes place?

4) Which kind of hydrocarbon functional group is responsible for sweet fruity odors of foods?

5) What is the functional group structure that carboxylic acids have that alcohols do not?

6) Removing a H on a hydrocarbon chain and replacing it with a halogen is called.......?

7) When a benzene ring is used as a substituent on a hydrocarbon chain, how is it named?

8) Which functional group has two carbon chains attached on either side of a carbonyl group?

9)  Draw structures for

a.  2,3-dimethylhexene               b. ethane                         c. ethyne

d. cyclopentane                         f. 3-ethylcyclooctene

 

ANSWERS CH. 22

1) -ene                      2) triple bond                      3)hydrogenation             4) esters

5) C=O                     6) halogenation                    7) phenyl                       8) aldehydes

CH3

/

9)  a)   CH2=CHCHCH2CH2CH3 b) CH3CH3 c) CH2

/

CH3

d)  draw a pentagon

e) Draw a stop sign (octagon; add extra bonding line inside shape from carbon 1 to 2; add CH2CH3 to third carbon away from where you began double bond.

Last Updated on Friday, 18 May 2012 15:14
 
 
 
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