Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

16 Problems on Principles of Chemistry - Assignment 6 | CHM 211, Assignments of Chemistry

Material Type: Assignment; Professor: Castellani; Class: Principles of Chemistry I; Subject: Chemistry; University: Marshall ; Term: Fall 2008;

Typology: Assignments

Pre 2010

Uploaded on 07/30/2009

koofers-user-zqf
koofers-user-zqf 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Chapter 6 Homework Solutions
10/18/08
10. a) Wavelength and frequency are inversely proportional.
b) Ultraviolet
13. Wavelength of X-rays < ultraviolet light < green light < red light < infrared light < radio waves
16. a) ν =
s
m 10 x 00.3
m 1
Å 10
Å 0.01
1810 = 3.00 x 1017 s-1
b) λ =
s
m 10 x 00.3
s 10 x .67
18
1-10 = 0.0039 m (3.9 mm)
c) (a) = yes, (b) = no
d) distance = (25.5 fs) 1 300 s
10 fs
x 10 m
s
15
8
. = 7.65 x 10-6 m
22. a) ν =
s
m 10 x 003
m 1
nm 10
nm 895
189 .= 5.09 x 1014 s-1
b) E = (6.6262 x 10-34 J•s)(5.09 x 1014 s-1)
mol
photons 10 x 0226 23
.(0.10 mol) = 2.03 x 104 J
c) E = (6.6262 x 10-34 J•s)
s
m 10 x 003
m 1
nm 10
nm 895
189 . = 3.37 x 10-19 J/photon
d) Probably not. Each element has a unique set of lines. It would be random chance that
strontium had a line at 589 nm.
26. ν =
h
E =
( )
s
J
10
x
626
.
6
molecules 10 x 6.022
mol 1
kJ
J 1000
kJ/mol 941
34-
23
= 2.36 x 1015 s-1
λ =
s
m 10 x .9982
s 10 x 36.2
18
1-15 = 1.27 x 10-7 m (127 nm = ultraviolet light)
34. When an electron moves to a higher energy state, the positively charged nucleus and the
negatively charged electron move away from each other. It requires the input of energy to
separate oppositely charged particles. The reverse is true when an electron falls from a
higher energy state to a lower one.
a) absorbed b) emitted c) absorbed
36. a) ν41 =
2234-
-18
4
1
-
1
1
sJ 10 x 6.626
J 10 x 2.18 = 3.08 x 1015 s-1
pf3

Partial preview of the text

Download 16 Problems on Principles of Chemistry - Assignment 6 | CHM 211 and more Assignments Chemistry in PDF only on Docsity!

Chapter 6 Homework Solutions

  1. a) Wavelength and frequency are inversely proportional. b) Ultraviolet
  2. Wavelength of X-rays < ultraviolet light < green light < red light < infrared light < radio waves
  3. a) ν = (^)  

s

  1. 00 x 10 m 1 m

10 Å

10. 0 Å

= 3.00 x 10^17 s-^1

b) λ = (^)  

s

  1. 00 x 10 m 7 .6x 10 s

10 - 1 = 0.0039 m (3.9 mm)

c) (a) = yes, (b) = no

d) distance = (25.5 fs)

1 s 3 00 10 fs

x 10 m (^15) s

8  

= 7.65 x 10-^6 m

  1. a) ν = (^)  

s

300 x 10 m 1 m

10 nm 589 nm

= 5.09 x 10^14 s-^1

b) E = (6.6262 x 10-^34 J•s)(5.09 x 10^14 s-^1 ) (^)  

mol

  1. 022 x 1023 photons (0.10 mol) = 2.03 x 10^4 J

c) E = (6.6262 x 10-^34 J•s) (^)  

s

300 x 10 m 1 m

10 nm 589 nm

= 3.37 x 10-^19 J/photon

d) Probably not. Each element has a unique set of lines. It would be random chance that strontium had a line at 589 nm.

  1. ν = h

E

  1. 626 x 10 J• s

6.022x 10 molecules

1 mol kJ

1000 J

941 kJ/mol

  • 34

= 2.36 x 10^15 s-^1

λ = (^)  

s

2 .998x 10 m

  1. 36 x 10 s

15 - 1 = 1.27 x 10

  • (^7) m (127 nm = ultraviolet light)
  1. When an electron moves to a higher energy state, the positively charged nucleus and the negatively charged electron move away from each other. It requires the input of energy to separate oppositely charged particles. The reverse is true when an electron falls from a higher energy state to a lower one. a) absorbed b) emitted c) absorbed
  2. a) ν 41 = (^)  

4

6.626x 10 J s

2.18 x 10 J = 3.08 x 10^15 s-^1

λ 41 = (^)  

m

10 nm 3.08x 10 s

3.00 x 10 m/s^9 15 - 1

8 = 97.3 nm (ultraviolet light)

E 41 = (6.626 x 10-^34 J•s)(3.08 x 10^15 s-^1 ) = 2.04 x 10-^18 J (released) In the book, the constant is shown with a negative value. The difference between the book & these problems is that here you first calculate the frequency of the light, which must have a positive value. Thus, the sign convention used by the problems must be ignored initially. Plugging into nearly identical equations. ν 52 = 6.91 x 10^14 s-^1 λ 52 = 434 nm (visible light) E 52 = 4.58 x 10-^19 J (released)

ν 36 = 2.74 x 10^14 s-^1 λ 36 = 1090 nm (infrared light) E 36 = 1.82 x 10-^19 J (absorbed)

  1. a) λ = (^)  

1000 m

1 km 1 hr

3600 s 50 km

1 hr 85 kg

  1. 626 x 10 -34Js = 5.6 x 10-^37 m

b) λ = (^)  

1 kg

1000 g (10.0g)(250m/s)

  1. 626 x 10 -34J s = 2.6 x 10-^34 m

c) λ = (^)  

1 kg

1000 g 1.66x 10 g

1 amu 2.5x 10 m

1 s 6.941amu

  1. 626 x 10 J s 5 - 24
  • = 2.3 x 10-^13 m

d) λ = (^)  

1 kg

1000 g 1.66x 10 g

1 amu 550 m

1 s 48.00amu

  1. 626 x 10 J s
  • 24

= 1.5 x 10-^11 m

  1. a) n = 4: l = 3, 2, 1, 0 b) l = 2: ml = -2, -1, 0, 1, 2 c) l = 2, 3, 4, …
  2. a) 2, 1, 1 2, 1, 0 2, 1, - b) 5, 2, 2 5, 2, 1 5, 2, 0 5, 2, -1 5, 2, -
  3. 2 p , 1 s , not allowed, 3 d , not allowed, not allowed, 4 d , 5 f
  4. a) Cs: [Xe] 6 s^1 c) Se: [Ar] 4 s^2 3 d^10 4 p^4 e) skip

b) Ni: [Ar] 4 s^2 3 d^8 d) Cd: [Kr] 5 s^2 4 d^10 f) Pb: [Xe] 6 s^2 4 f^14 5 d^10 6 p^2

  1. a) Ga: [Ar] 4 s^2 3 d^10 4 p^1 , 1 unpaired e-^ d) I: [Kr] 5 s^2 4d^10 5 p^5 , 1 b) Ca: [Ar] 4 s^2 , 0 e) Y: [Kr] 5 s^2 4 d^1 , 1 c) V: [Ar] 4 s^2 3 d^3 , 3 f) Pt: [Xe] 6 s^2 4 f^14 5 d^8 , 2 g) skip
  2. a) halogens b) Group 4B (4) c) Group 3A (13) d) skip
  3. a) l b) n and l c) ms d) ml