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Chem 110 Exam #4 Key for Spring 2016, Study notes of Chemistry

The answers to Exam #4 for Dr. McCorkle's Chem 110 class, held in Spring 2016. It includes multiple choice questions and short answer problems covering topics such as atomic structure, quantum mechanics, and spectroscopy.

What you will learn

  • What is the process called where electrons fill up the orbitals from low energy to high energy?
  • Which quantum number describes the shape of an orbital?
  • Which quantum number describes the orientation of an orbital?
  • What is the vertical height of a wave called?
  • What happens when waves of equal amplitude from two sources are out of phase when they interact?

Typology: Study notes

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DO NOT OPEN
UNTIL INSTRUCTED TO DO SO
CHEM 110 โ€“ Dr. McCorkle โ€“ Exam #4 KEY
While you wait, please complete the following information:
Name: _______________________________
Student ID: _______________________________
Turn off cellphones and stow them away. No headphones, mp3 players, hats,
sunglasses, food, drinks, restroom breaks, graphing calculators, programmable
calculators, or sharing calculators. Grade corrections for incorrectly marked or
incompletely erased answers will not be made.
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Download Chem 110 Exam #4 Key for Spring 2016 and more Study notes Chemistry in PDF only on Docsity!

DO NOT OPEN

UNTIL INSTRUCTED TO DO SO

CHEM 110 โ€“ Dr. McCorkle โ€“ Exam #4 KEY

While you wait, please complete the following information:

Name: _______________________________

Student ID: _______________________________

Turn off cellphones and stow them away. No headphones, mp3 players, hats,

sunglasses, food, drinks, restroom breaks, graphing calculators, programmable

calculators, or sharing calculators. Grade corrections for incorrectly marked or

incompletely erased answers will not be made.

  1. How many valence electrons do the halogens possess? A) 1 B) 2 C) 5 D) 6 E) 7
  2. Suppose a metal ejects electrons from its surface when struck by red light. What will happen if the surface is struck by blue light? A) No electrons would be ejected. B) Electrons would be ejected, and they would have the same kinetic energy as those ejected by red light. C) Electrons would be ejected, and they would have greater kinetic energy than those ejected by red light. D) Electrons would be ejected, and they would have lower kinetic energy than those ejected by red light.
  3. Identify the isoelectronic elements A) Clโˆ’, Fโˆ’, Brโˆ’, Iโˆ’, Atโˆ’ B) N3โˆ’, S2โˆ’, Brโˆ’, Cs+, Sr2+ C) P3โˆ’, S2โˆ’, Clโˆ’, K+, Ca2+ D) Zn2+, Co2+, Cu2+, Cr2+, Cd2+ E) Ne, Ar, Kr, Xe, He
  4. What period 3 element has the following ionization energies (all in kJ/mol)? IE 1 = 1012 IE 2 = 1900 IE 3 = 2910 IE 4 = 4960 IE 5 = 6270 IE 6 = 22,

A) Mg B) Si C) P D) S E) Cl

  1. Arrange the elements in order of increasing IE 1 : N, F, As A) N < As < F B) As < N < F C) F < N < As D) As < F < N E) F < As < N
  2. Arrange the elements in order of increasing metallic character: P, As, K A) P < As < K B) As < P < K C) K < P < As D) As < K < P E) K < As < P

Calculations โ€“ Write your initials in the upper-right corner of every page that contains work. For full credit show all work and write neatly; give answers with correct significant figures and units. For calculations, place a box around your final answer.

  1. For the following short answer questions, please circle, fill-in, or provide the correct answer. [2 points each]

a. Which transition has a longer wavelength? n = 7 โ†’ n = 5 n = 3 โ†’ n = 1

b. Which has the lowest energy? 6 s 5 d 4 f

c. A gas phase element gaining an electron is a typically endothermic process. True False

d. How many electrons in an atom could have n = 4, l = 2 , ml = +1? ____ 2 _____

e. The two most probable ions of titanium are ___ Ti2+ ____ and ___ Ti4+ ____.

  1. How many photons are contained in a flash of green light (525 nm) that contains 189 kJ of energy? [4 points]

๐Ÿ“๐Ÿ๐Ÿ“ ๐ง๐ฆ ร—

๐Ÿ๐ŸŽโˆ’๐Ÿ—^ ๐ฆ

= ๐Ÿ“. ๐Ÿ๐Ÿ“ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ•^ ๐ฆ

(๐Ÿ”. ๐Ÿ”๐Ÿ๐Ÿ” ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ‘๐Ÿ’^ ๐‰ โˆ™ ๐ฌ)(๐Ÿ. ๐Ÿ—๐Ÿ—๐Ÿ•๐Ÿ— ร— ๐Ÿ๐ŸŽ๐Ÿ–^ ๐ฆ/๐ฌ)

๐Ÿ“. ๐Ÿ๐Ÿ“ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ•^ ๐ฆ

= ๐Ÿ‘. ๐Ÿ•๐Ÿ– ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ๐Ÿ—^ ๐‰/๐ฉ๐ก๐จ๐ญ๐จ๐ง

๐Ÿ๐Ÿ–๐Ÿ— ๐ค๐‰ ร—

๐Ÿ๐ŸŽ๐Ÿ‘^ ๐‰

= ๐Ÿ. ๐Ÿ–๐Ÿ— ร— ๐Ÿ๐ŸŽ๐Ÿ“^ ๐‰

๐Ÿ. ๐Ÿ–๐Ÿ— ร— ๐Ÿ๐ŸŽ๐Ÿ“^ ๐‰

๐Ÿ‘. ๐Ÿ•๐Ÿ– ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ๐Ÿ—^ ๐‰/๐ฉ๐ก๐จ๐ญ๐จ๐ง

= ๐Ÿ“. ๐ŸŽ๐ŸŽ ร— ๐Ÿ๐ŸŽ๐Ÿ๐Ÿ‘^ ๐ฉ๐ก๐จ๐ญ๐จ๐ง๐ฌ

  1. Calculate the de Broglie wavelength in meters of a 46 g golf ball with a velocity of 30. m/s. [3 points]

๐ฆ = ๐Ÿ’๐Ÿ” ๐  ร— ๐Ÿ ๐ค๐  ๐Ÿ๐ŸŽ๐Ÿ‘^ ๐  = ๐ŸŽ. ๐ŸŽ๐Ÿ’๐Ÿ” ๐ค๐ 

๐›Œ = ๐ฆ๐ฏ๐ก = ๐Ÿ”.๐Ÿ”๐Ÿ๐Ÿ”ร—๐Ÿ๐ŸŽ

โˆ’๐Ÿ‘๐Ÿ’ (^) ๐‰โˆ™๐ฌ (๐ŸŽ.๐ŸŽ๐Ÿ’๐Ÿ” ๐ค๐ )(๐Ÿ‘๐ŸŽ. ๐ฆ/๐ฌ) = ๐Ÿ’. ๐Ÿ– ร— ๐Ÿ๐ŸŽ

  1. Explain the trends in atomic radius across a period (from left to right) and down a group (from top to bottom). [4 points]

Atomic radius decreases across a period because electrons are being added

to the same valence shell but the effective nuclear charge increases as

protons are added to the nucleus, causing increased attraction with the

electrons, pulling them closer, decreasing the radius.

Atomic radius increases down a group because electrons are being added

to higher energy shells that are farther away from the nucleus.

  1. Consider the electronic transition within a hydrogen atom from n = 5 to n = 2.

a. Calculate the energy of the photon that is emitted in joules. [2 points]

๐„๐ฉ๐ก๐จ๐ญ๐จ๐ง = ๐‘๐‘ฏ [( ๐ง๐Ÿ

๐Ÿ๐ข๐ง๐š๐ฅ๐Ÿ^

๐ข๐ง๐ข๐ญ๐ข๐š๐ฅ๐Ÿ^

)]

๐„๐ฉ๐ก๐จ๐ญ๐จ๐ง = ๐Ÿ. ๐Ÿ๐Ÿ– ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ๐Ÿ–^ ๐‰ [(

)] = ๐Ÿ’. ๐Ÿ“๐Ÿ– ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ๐Ÿ—^ ๐‰

b. Calculate the frequency in Hertz. [2 points]

E = hฮฝ

๐Ÿ’. ๐Ÿ“๐Ÿ– ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ๐Ÿ—^ ๐‰

๐Ÿ”. ๐Ÿ”๐Ÿ๐Ÿ” ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ‘๐Ÿ’^ ๐‰ โˆ™ ๐ฌ

= ๐Ÿ”. ๐Ÿ—๐Ÿ ร— ๐Ÿ๐ŸŽ๐Ÿ๐Ÿ’^ ๐‡๐ณ

c. Calculate the wavelength in meters. [2 points]

E photon = ๐ก๐œ๐›Œ

(๐Ÿ”. ๐Ÿ”๐Ÿ๐Ÿ” ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ‘๐Ÿ’^ ๐‰ โˆ™ ๐ฌ)(๐Ÿ. ๐Ÿ—๐Ÿ—๐Ÿ•๐Ÿ— ร— ๐Ÿ๐ŸŽ๐Ÿ–^ ๐ฆ/๐ฌ)

๐Ÿ’. ๐Ÿ“๐Ÿ– ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ๐Ÿ—^ ๐‰

= ๐Ÿ’. ๐Ÿ‘๐Ÿ’ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ•^ ๐ฆ

d. Does this emission occur within the visible region of the electromagnetic spectrum? [ points]

๐Ÿ’. ๐Ÿ‘๐Ÿ’ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ•^ ๐ฆ ร—

๐Ÿ ๐ง๐ฆ

๐Ÿ๐ŸŽโˆ’๐Ÿ—^ ๐ฆ = ๐Ÿ’๐Ÿ‘๐Ÿ’ ๐ง๐ฆ^ Yes, this is visible (indigo).

Formulas & Constants (you may or may not need)

1 inch = 2.54 cm (exact) 1 mile = 5280 ft โ‰ˆ 1.609 km 1 kg โ‰ˆ 2.205 lb

1 lb = 16 oz โ‰ˆ 453.6 g 1 gal = 4 qt = 8 pt โ‰ˆ 3.785 L 1 L = 1000 cm^3

K = ๏‚ฐC + 273.15 ๏‚ฐF = 1.8 x ๏‚ฐC + 32 ๏‚ฐC = (๏‚ฐF โ€“ 32)/1.

1 cal = 4.184 J 1 Cal = 1000 cal q = m x C x ฮ”T

Avogadroโ€™s # = 6.022x10^23 Molar volume = 22.4 L R = 0.08206 Lโˆ™atmmolโˆ™K

P 1 V 1 T 1 =^

P 2 V 2 T 2 ๐‘ขrms^ = โˆš

3RT ๐‘€ ๐พ๐ธ =^

1 2 ๐‘š๐‘ฃ

(^2) = 3 2 ๐‘…๐‘‡

1 atm = 760 mmHg 1 mmHg = 1 torr PTotal = P 1 + P 2 + ...

PA = ฮงA ยท PTotal PV = nRT โˆ†E = q + w

w = โˆ’Pโˆ†V q = C ร— โˆ†T q = m ร— s ร— โˆ†T

โˆ† H ยฐrxn = ฮฃ[n โˆ† H fยฐ(products)] โ€“ ฮฃ[n โˆ† H fยฐ(reactants)] R = 8.314 J/mol ยท K

h = 6.626ร—10โˆ’^34 J ยท s c = 2.9979ร—10^8 m/s RH = 2.18ร—10โˆ’^18 J

1 Hz = sโˆ’^1 ฮป = (^) mvh โˆ†x ร— mโˆ†v โ‰ฅ (^) 4ฯ€h

E photon = hฮฝ = hcฮป ๐ธphoton = R๐ป [( (^) n^1 final^2 ) โˆ’ ( (^) n^1 initial^2 )]

Scratch Page

(to be handed in)