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Solved Practice Problems on General Chemistry | CHEM 141, Study notes of Chemistry

Material Type: Notes; Professor: Hill; Class: GENERAL CHEMISTRY; Subject: Chemistry; University: Jackson State University; Term: Fall 2008;

Typology: Study notes

Pre 2010

Uploaded on 08/08/2009

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Chapter 7 Practice Problems [Watts, Fall 2008.]
Problem 1.
The false statement is (b). The reason is that cathode rays are electrons. Therefore they
are independent of element. See pages 260-262.
Problem 5
Amplitude is not mathematically related to the other quantities. The other quantities are
related to each other according to: u = λυ. The meanings are: u is the speed; λ is the
wavelength; and υ is the frequency. If we are talking about electromagnetic waves, the
speed is the speed of light, which is given the symbol c. See page 268.
Problem 18
The set of quantum numbers given (n = 2, l = 2, ml = 0) is not allowed.
Reason: since n = 2, the largest allowed value of l is 1 (i.e. n – 1).
Other notes: (1) There is no such thing as a 2d orbital; (2) a 2p orbital has n = 2, l = 1.
Problem 32
λ = 11.5 nm = 11.5 x 10-9 m = 1.15 x 10-8 m
116
8
18
1061.2
1015.1
10998.2
s
m
sm
c
Region of spectrum: ultraviolet. See page 269.
λ = 9.32 cm = 9.32 x 10-2 m
19
2
18
1022.3
1032.9
10998.2
s
m
sm
c
Region of spectrum: radiowaves. See page 269.
λ = 1.93 x 106 m
12
6
18
1055.1
1093.1
10998.2
s
m
sm
c
pf3
pf4

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Chapter 7 Practice Problems [Watts, Fall 2008.] Problem 1. The false statement is (b). The reason is that cathode rays are electrons. Therefore they are independent of element. See pages 260-262. Problem 5 Amplitude is not mathematically related to the other quantities. The other quantities are related to each other according to: u = λυ. The meanings are: u is the speed; λ is the wavelength; and υ is the frequency. If we are talking about electromagnetic waves, the speed is the speed of light, which is given the symbol c. See page 268. Problem 18 The set of quantum numbers given (n = 2, l = 2, m l = 0) is not allowed. Reason: since n = 2, the largest allowed value of l is 1 (i.e. n – 1). Other notes: (1) There is no such thing as a 2d orbital; (2) a 2p orbital has n = 2, l = 1. Problem 32 λ = 11.5 nm = 11.5 x 10-9^ m = 1.15 x 10-8^ m 16 1 8 8 1

  1. 61 10
  2. 15 10

    

  s m c m s

Region of spectrum: ultraviolet. See page 269. λ = 9.32 cm = 9.32 x 10-2^ m 9 1 2 8 1

  1. 22 10
  2. 32 10

    

  s m c m s

Region of spectrum: radiowaves. See page 269. λ = 1.93 x 10^6 m 2 1 6 8 1

  1. 55 10
  2. 93 10

  s m c m s  

Region of spectrum: extra low frequency. See page 269. Problem 38 E (^) photonh ( 6. 626  10 ^34 Js )( 3. 73  1014 s ^1 ) 2. 47  10 ^19 J Compare this energy with that of a photon of light that has λ = 530 nm = 5.30 x 10-7^ m. J m hc Js m s E (^) photon 7 19 34 8 1

  1. 75 10
  2. 30 10

     

 Therefore, the energy of the second photon is greater than that of the first one. Problem 40 J m hc Js m s E (^) photon 2 24 34 8 1

  1. 66 10
  2. 0 10

     

 This is the energy in J of one photon of light with a wavelength of 12.0 cm. To find the energy of one mole of photons, multiply by Avogadro’s number: N E ( 6. 022  1023 mol ^1 )( 1. 66  10 ^24 J ) 1. 00 Jmol ^1  1. 00  10 ^3 kJmol ^1 A photon Problem 46 Calculate the energy levels needed (i.e. energies of the n = 1, n = 2, and n = 7 orbits). J

B J

E^18

18 1 2 1 2.^17910

   

J

B J

E^19

18 2 2 4 5.^447510

   

J

B J

E^20

18 7 2 49 4.^44710

   

Energy for n = 1 → n = 2: ΔE = EE = E 2 – E 1 = 1.63425 x 10-18^ J

(e) Correct. There are two possible values for the spin quantum number, ½ and -½. (f) Incorrect. The electron could be in a 5f or 5g orbital.