CHEM125 – BASIC HOMEWORK REVIEW – CH.1-8

BASIC HOMEWORK REVIEW CH. 1-8

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Question 1

How many protons does copper have?

Periodic Table and protons

Explanation

CHEM125 - BASIC HOMEWORK REVIEW - CH.1-8 1 atomic number = protons = electrons

Question 2

How many electrons does copper have? Remember atoms are neutral.

Explanation

Look Above ^^^^

Question 3

How many neutrons does copper-63 have? Help given in feedback.

Explanation

neutron = atomic mass – atomic number neutron = 63 – 29 = 34

Question 4

What is the mass of 1 mol of CO2 in g? Molar mass example chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

CO2 = 12 + (2 x 16) = 44g

Question 5

What is the mass of 5 molecules of SO3 in g?

Using Avagadro’s Number

chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

1 mol = 6.02 x 1023 (Avogadro’s Number). 5 molecules x ( 1 mol / 6.02 E23) x ( 80g / 1mol ) = 6.64e-22 g

Question 6

Correctly match the element with the property. Periodic Table and Group IA and IIA metals Metal that reacts with cold water to produce OH- and H2(g)Choose…SiCrSNaHe chem125-rev-ch-1-8 - Fluff Butt ReviewsMetal (all conduct electricity)–no reaction with water, but reacts with acid to produce H2 gasChoose…SiCrSNaHe chem125-rev-ch-1-8 - Fluff Butt ReviewsNon-metal–does not conduct electricityChoose…SiCrSNaHe chem125-rev-ch-1-8 - Fluff Butt ReviewsNoble gas–inert, stable electron configurationChoose…SiCrSNaHe chem125-rev-ch-1-8 - Fluff Butt ReviewsSemimetal–poor conductor of electricityChoose…SiCrSNaHe chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

  • Na reacts to produce OH- and H2 ions
  • Transition metal react with acid to produce H2
  • Nonmetal do not conduct electricity
  • He is a nobel gas (look at the periodic table)
  • Silicon is a common semiconductor/ semimetal

Question 7

What is the correct formula for aluminum sulfide? Oxidation states Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Al: +3 S: -2 Now cross the charges to get : Al2S3

Question 8

Which compound is the most ionic? Ionic Compounds Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

The molecule with the greatest difference in electronegativity is the most ionic.

Question 9

Match properties correctly

Quantum Numbers

Indicates energy and size of orbitalChoose…2npsdf chem125-rev-ch-1-8 - Fluff Butt Reviewsl=0Choose…2npsdf chem125-rev-ch-1-8 - Fluff Butt Reviewsl=1Choose…2npsdf chem125-rev-ch-1-8 - Fluff Butt Reviewsl=2Choose…2npsdf chem125-rev-ch-1-8 - Fluff Butt Reviewsl=3Choose…2npsdf chem125-rev-ch-1-8 - Fluff Butt ReviewsNumber of possible electron spins for each orbitalChoose…2npsdf chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

CHEM125 - BASIC HOMEWORK REVIEW - CH.1-8 2 Each single orbital can hold a max of 2 electrons

Question 10

Select the correct set of quantum mechanical numbers for an electron in an s, p, d, and f orbital. (4 correct answers.)

Allowed Combination of Quantum Numbers

Select one or more: chem125-rev-ch-1-8 - Fluff Butt Reviews chem125-rev-ch-1-8 - Fluff Butt Reviews chem125-rev-ch-1-8 - Fluff Butt Reviews chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

CHEM125 - BASIC HOMEWORK REVIEW - CH.1-8 3 magnetic spin can be only +1/2 or -1/2

Question 11

Correctly match the information for the angular momentum quantum number (l) which gives the shape of the orbital. A nodal plane is a region where the electron density is zero, that is, the probability of finding the electron there is zero. The maximum number of electrons in an electron orbital.Choose…2dspf chem125-rev-ch-1-8 - Fluff Butt ReviewsSpherically symmetric orbital–no nodal planesChoose…2dspf chem125-rev-ch-1-8 - Fluff Butt ReviewsOrbital shaped like a peanut or dumbbell–one nodal planeChoose…2dspf chem125-rev-ch-1-8 - Fluff Butt Reviews4 of the 5 orbital shaped like a double peanut–2 nodal planesChoose…2dspf chem125-rev-ch-1-8 - Fluff Butt ReviewsOrbital shaped like flower or fancy–3 nodal planesChoose…2dspf chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

CHEM125 - BASIC HOMEWORK REVIEW - CH.1-8 4

Question 12

What is the ground-state electron configuration for S?

Electron Configurations

Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Count from left to right. CHEM125 - BASIC HOMEWORK REVIEW - CH.1-8 5

Question 13

What is the electron configuration for S2-? Electron Configurations Negative Ions Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Add two addition electron on top of the original configuration 1s22s22p63s23p4 + 2 1s22s22p63s23p6

Question 14

What is the electron configuration for K1+? Electron Configurations Positive Ions Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Subtract an electron from the original from highest nth energy level 1s22s22p63s23p64s1 -1

Question 15

Match atomic radius from largest to smallest.

Atomic Radius

KChoose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius chem125-rev-ch-1-8 - Fluff Butt ReviewsAlChoose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius chem125-rev-ch-1-8 - Fluff Butt ReviewsNaChoose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius chem125-rev-ch-1-8 - Fluff Butt ReviewsArChoose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius chem125-rev-ch-1-8 - Fluff Butt ReviewsSChoose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Follow the trend CHEM125 - BASIC HOMEWORK REVIEW - CH.1-8 6

Question 16

Match ionization energies from smallest to largest.

Ionization Energy

KChoose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy chem125-rev-ch-1-8 - Fluff Butt ReviewsAlChoose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy chem125-rev-ch-1-8 - Fluff Butt ReviewsNaChoose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy chem125-rev-ch-1-8 - Fluff Butt ReviewsArChoose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy chem125-rev-ch-1-8 - Fluff Butt ReviewsSChoose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Follow the trend.

Question 17

What is the correct formula for iron (III) oxide? Multiple oxidation states Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Fe (III) : +3 O: -2 Cross the charges to get Fe2O3

Question 18

Which of the following tin (IV) compounds makes sense using oxidation states and conservation of charge?

Oxidation State and Polyatomic Ions

Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

Tin (IV) : +4 : -1

Question 19

Warmup question. How many moles of Ca(OH)2 are required to neutralize (react completely with) 0.28 moles of HCl? The balanced reaction is: 2HCl(aq) + Ca(OH)2(aq) → 2H2O(l) + CaCl2(aq)

Using Balanced Equations

chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

0.28 mol HCl x ( 1 mol Ca(OH)2 / 2 mol HCl) = 0.14

Question 20

What volume in L of 0.42 M Ca(OH)2 is required neutralize 0.29 moles of HCl? The balanced reaction is: 2HCl(aq) + Ca(OH)2(aq) → 2H2O(l) + CaCl2(aq)

Using Balanced Equations II

chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

0.29 mol HCl x ( 1 mol Ca(OH)2 / 2 mol HCl ) x ( 1 L / 0.42mol Ca(OH)2) = 0.34

Question 21

What volume in L of chlorine at 30oC and 750 torr is required to react completely with 8.73 g aluminum: 2Al(s) + 3Cl2(g) → 2AlCl3(s) Calculate the moles chlorine needed (as in the previous problem), then use the ideal gas law–PV = nRT. Additional help is given in the feedback.chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

moles Al = 8.73 g / 26.9815 g/mol=0.3236

the ratio between Al and Cl2 is 2 : 3

moles Cl2 required = 0.3236 x (3 / 2) = 0.485
p = 750/760 = 0.987 atm
T = 30 + 273 = 303 K

V = nRT/P = 0.485 x 0.08206 x 303 / .987 = 12.31

Question 22

The following reaction takes place at a certain elevated temperature: Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(s) What is the percent yield of iron if 53.6 g Fe2O3 in excess CO produces 18.3 g Fe? The M.W. of Fe2O3 is 159.7 g/mol and the M.W. of CO is 28.01 g/mol. Recall that the percent yield is the actual yield divided by the theoretical yield times 100%. Additional help is given in the feedback.chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

53.6g / 159.7 = .3356 mol Fe2O3 .3356 mol Fe2O3 x ( 2 mol Fe / 1 mol Fe2O3 ) x ( 56g / 1 mol Fe) = 37.59 g 37.59 g (theoretical) 18.3 g (actual) 18.3 / 37.59 = 0.4882 x 100 = 48.82 % CHEM125 - BASIC HOMEWORK REVIEW - CH.1-8 7

Question 23

What is the mass percent of carbon in sugar, C12H22O11? M.W. sugar = 342 g/mol. Help is given in the feedback.chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

C = 12 x 12 = 144 144/ 342 = 0.42 x 100 = 42%

Question 24

Analysis of 3.500 g sample of an unknown compound shows that it contains 1.400 g C, 0.235 g H, and the rest O. What is the empirical formula of this compound? Hint, given in feedback.Select one: chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

C : 1.400 g / 12 g/mol = 0.1167 H: 0.235 g / 1.008 g/mol = 0.2331 O : 1.865 / 16.00 g/mol = 0.1166 C : 0.1167 / 0.1166 = 1 H : 0.2331 / 0.1166 = 2 O : 0.1166/ 0.1166 = 1

Question 25

Hemoglobin has a mass of about 67,000 g/mol. It is 0.3335% iron by mass. How many iron atoms are in a molecule of hemoglobin? Help is given in the feedback.chem125-rev-ch-1-8 - Fluff Butt Reviews

Explanation

67,000 g/mol x 0.003335 = 223.445 g/mol of iron 223.445g/mol x (1/55.85 mol/g ) = 4 atoms