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

Answer: 

Explanation

atomic number = protons = electrons

Question 2

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

Explanation

Look Above ^^^^

Question 3

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

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 Answer:

Explanation

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

Question 5

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

Using Avagadro’s Number

Answer:

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)Answer 1Choose…SiCrSNaHe Metal (all conduct electricity)–no reaction with water, but reacts with acid to produce H2 gasAnswer 2Choose…SiCrSNaHe Non-metal–does not conduct electricityAnswer 3Choose…SiCrSNaHe Noble gas–inert, stable electron configurationAnswer 4Choose…SiCrSNaHe Semimetal–poor conductor of electricityAnswer 5Choose…SiCrSNaHe 

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:a. AlSb. Al2Sc. AlS2d. Al3S2e. Al2S3 

Explanation

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

Question 8

Which compound is the most ionic? Ionic Compounds Select one:a. NaF b. H2Oc. FeOd. NaN3e. IF

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 orbitalAnswer 1Choose…2npsdf l=0Answer 2Choose…2npsdf l=1Answer 3Choose…2npsdf l=2Answer 4Choose…2npsdf l=3Answer 5Choose…2npsdf Number of possible electron spins for each orbitalAnswer 6Choose…2npsdf 

Explanation

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:a. 3s: n=”3,” l=”0,” ml = 0, and ms = 1b. 3s: n=”3,” l=”0,” ml = 0, and ms = -1/2 c. 5p: n=”5,” l=”0,” ml = 1, and ms = +1/2d. 5p: n=”5,” l=”1,” ml = -1, and ms = +1/2 e. a d orbital: n=”2,” l=”2,” ml = 0, and ms = +1/2f. a d orbital: n=”4,” l=”2,” ml = 0, and ms = +1/2 g. 5f: n=”5,” l=”3,” ml = -2, and ms = +1/2 h. 5f: n=”5,” l=”4,” ml = 3, and ms = +1/2

Explanation

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.Answer 1Choose…2dspf Spherically symmetric orbital–no nodal planesAnswer 2Choose…2dspf Orbital shaped like a peanut or dumbbell–one nodal planeAnswer 3Choose…2dspf 4 of the 5 orbital shaped like a double peanut–2 nodal planesAnswer 4Choose…2dspf Orbital shaped like flower or fancy–3 nodal planesAnswer 5Choose…2dspf 

Explanation

Question 12

What is the ground-state electron configuration for S?

Electron Configurations

Select one:a. 1s22s22p63s23p4 b. 1s22s22p63s23p6c. 1s22s22p63p6d. 1s42p63s23p4

Explanation

Count from left to right.

Question 13

What is the electron configuration for S2-? Electron Configurations Negative Ions Select one:a. 1s22s22p63s23p4b. 1s22s22p63s23p6 c. 1s22s22p63s23p44s2d. 1s22s22p63p8

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:a. 1s22s22p63s23p64s2b. 1s42s42p63s4c. 1s22s22p63s23p6 d. 1s22s22p63s23p64s1

Explanation

Subtract an electron from the original from highest nth energy level 1s22s22p63s23p64s1 -1  1s22s22p63s23p6

Question 15

Match atomic radius from largest to smallest.

Atomic Radius

KAnswer 1Choose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius AlAnswer 2Choose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius NaAnswer 3Choose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius ArAnswer 4Choose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius SAnswer 5Choose…Fourth largest atomic radiusLargest atomic radiusSecond largest atomic radiusSmallest atomic radiusThird largest atomic radius 

Explanation

Follow the trend

Question 16

Match ionization energies from smallest to largest.

Ionization Energy

KAnswer 1Choose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy AlAnswer 2Choose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy NaAnswer 3Choose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy ArAnswer 4Choose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy SAnswer 5Choose…Largest ionization energySmallest ionization energyThird smallest ionization energyFourth smallest ionization energySecond smallest ionization energy 

Explanation

Follow the trend.

Question 17

What is the correct formula for iron (III) oxide? Multiple oxidation states Select one:a. FeOb. Fe3Oc. FeO3d. Fe2O3 e. Fe3O2

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:a. Sn(BaOH)4b. Sn(BaSO4)4c. Sn(BaPO4)4 d. Sn(C2H3O2)41-

Explanation

Tin (IV) : +4 (BaPO4): -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

Answer:

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

Answer:

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.Answer:

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.Answer:

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 %

Question 23

What is the mass percent of carbon in sugar, C12H22O11? M.W. sugar = 342 g/mol. Help is given in the feedback.Answer:

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:a. HCOOHb. CH2O c. C2H5OHd. CH4O4e. C3H7OH

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.Answer:

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