3. 4 - pbworkschemnotes.pbworks.com/f/solution_pdf.pdf · 3.weight 4.mass 5.density correct...

Devillez (ld2653) – Test 1 Review – Devillez – (99998) 1

This print-out should have 133 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

A chemist investigates theI) melting point

II) boiling pointIII) flammabilityof acetone, a component of fingernail polishremover. Which is/are physical?

1. III only

2. I, II and III

3. II and III only

4. I and II only correct

5. I only

6. I and III only

7. II only

8. None of these

Explanation:

Flammability is a chemical change; melt-ing and boiling (change of state) are physicalchanges.

002 10.0 pointsConsider the statement:

“The temperature of the land is an impor-tant factor for the ripening of oranges, be-cause it affects the evaporation of water andthe humidity of the surrounding air.”

How many of these factors,temperature ripeningevaporation humidity

are physical properties or changes?

1. 2

2. None

3. 4

4. 3 correct

5. 1

Explanation:

The temperature and humidity are physicalproperties and the evaporation of water is aphysical change.

003 10.0 pointsAll of the following are characteristic proper-ties of phosphorus. Which one is a chemicalproperty?

1. The white form is soluble in liquid carbondisulfide, but is insoluble in water.

2. The red form melts at about 600◦C andthe white form melts at 44◦C.

3. Both red phosphorus and white phospho-rus exist in solid allotropic forms.

4. When exposed to air, white phosphoruswill burn spontaneously, but red phosphoruswill not. correct

5. The red form of phosphorus is insolublein both water and carbon disulfide.

Explanation:

Chemical properties are exhibited as mat-ter undergoes chemical changes. The burningof white or red phosphorus explains its inter-action with chemical reactions and changes incomposition which justifies its being a chemi-cal property.

004 10.0 pointsWhich of the following is an extensive physicalproperty?

1. reactivity

2. color

3. density


4. mass correct

5. refractive index

Explanation:

Extensive properties are those which de-pend on the amount of substance present.Mass depends on and is directly proportionalto the amount of matter present.

005 10.0 pointsWhich of the following is an intensive prop-erty?1. number of moles of molecules

2. volume

3. weight

4. mass

5. density correct

Explanation:

Intensive properties are independent of theamount of matter present. The density of asubstance will be the same regardless of thesize (large or small) of the sample.

006 10.0 pointsWhich is NOT a pure substance?

1. water

2. chlorine

3. copper

4. brass correct

Explanation:

Brass is an alloy (a physical mixture of themetals copper and zinc).

007 10.0 pointsWhich diagram(s)

I II

III IV

best characterize(s) a mixture?

1. I, II and III only

2. I and III only

3. III and IV only correct

4. IV only

Explanation:

The dark and light spheres represent atomsof different elements. A mixture consists ofdifferent substances combined together phys-ically. Although some atoms are chemicallybonded in the figures, the samples do not con-sist of all of the same type of bonded atoms.


best characterize(s) an element?

I II


III IV

best characterize(s) an element?

1. III and IV only

2. II only correct

3. I only

4. I and II only

Explanation:

The dark and light spheres represent atomsof different elements. An element consists ofonly one kind of atom. In II the atoms happento form diatomic molecules, but only one typeof atom is present in II.


best characterize(s) a compound?

I II

III IV

1. II and III only

2. II only

3. I and IV only

4. I only correct

Explanation:

The dark and light spheres represent atomsof different elements. A compound is a com-bination of two or more elements, chemicallybound together, with the same compositionthroughout.


I II

III IV

best represent(s) molecules?

1. II and IV only

2. I only


4. I and III only

5. III only

6. II and III only

7. Another combination

8. II only

9. IV only

Explanation:

The dark and light spheres represent atoms


of different elements. Molecules consist of twoor more atoms chemically bonded together.


I II

III IV

best represent(s) atoms?

1. I and III only

2. I and II only

3. I only

4. II only

5. IV only

6. III only correct

7. Another combination

8. II and IV only

9. II and III only

Explanation:

The dark and light spheres represent atomsof different elements. Atoms are single andunattached to each other.

012 10.0 pointsWhich of the following statements concerningsubstances is NOT true?1. A pure substance can only be a homoge-

neous solution. correct

2. A pure substance cannot be heteroge-neous.

3. Heterogeneous solutions cannot be puresubstances.

4. Mixtures can be homogeneous.

Explanation:

A solution, although homogeneous, cannotbe a pure substance since it consists of twocomponents: the solute and the solvent.


I II

III

best characterize(s) a pure substance that isan element?1. I and III only

2. I and II only

3. II and III only

4. None of these

5. II only correct

6. All of these

7. III only


8. I only

Explanation:

014 10.0 pointsWhich of the following is a pure substance?

1. a puddle of milk

2. a rod of steel

3. a block of wood

4. a cube of sugar correct

5. a strap of leather

Explanation:

A cube of sugar is an example of a purecompound.

015 10.0 pointsWhich of these substances

I) carbon dioxide gasII) distilled water

III) jellois a pure substance?


2. I only

3. II only

4. II and III only

5. All of these

6. III only

7. None of these

8. I and III only

Explanation:

016 10.0 pointsWhat is the proper solution to the followingequation?

(150 + 2.5 + 36.75) ÷ (6.60 + 0.173) =

1. 30

2. 27.942

3. 27.94183

4. 28 correct

5. 27.9

6. 27.94

7. 27.9418

Explanation:

The last position retained for the first addi-tion problem will be the tens position givingit 2 sigfigs. The last position retained forthe second addition problem will be the hun-dredths position giving it 3 sigfigs. Thereforethe final answer 27.94 needs to be rounded to2 sigfigs: 28.

017 10.0 pointsWhich of the following numbers has the great-est number of significant figures?

1. 130

2. 0.690 correct

3. 1.2

4. 33,000

5. 0.0032

6. 4.3 × 102

Explanation:

Zeros to the left of the decimal are notsignificant. Zeros used to place the decimalare not significant. 0.690 has 3 significantfigures. All the other choices have 2 significantfigures.

018 10.0 pointsWhat is the length of the line segment in-dicated on the scale provided using proper


significant figures?

0 1 2 3

1. 3.00 units correct

2. 3.000 units

3. 3.0 units

4. 3 units

Explanation:

The scale is in units of tenths, so hundrethsis the uncertain digit.

019 10.0 pointsWhat is the proper measurement of the vol-ume represented?

1

2

3

4

5mL

1. 2.5 mL correct

2. 2.0 mL

3. 2.50 mL

4. 2.500 mL

Explanation:

The scale is in units of 1, so tenths is theuncertain digit.

020 10.0 pointsKeeping in mind the rules for significant dig-its, calculate the sum of 2.636 g and 0.7948g.

1. 3.4 g

2. 3.43080 g

3. 3.4308 g

4. 3.431 g correct

5. 3.43 g

6. 3 g

Explanation:

sum of given values = ?

2.636 g + 0.7948 g = 3.431 g

021 10.0 pointsKeeping in mind the rules for significant fig-ures, divide 92.80 g by 3.91111 mL.

1. 23.73 g/mL correct

2. 23.7 g/mL

3. 23.727 g/mL

4. 23.7273 g/mL

5. 24 g/mL

Explanation:

92.80 g (4 sf)

3.91111 mL (6 sf)= 23.73 g/mL

(round to 4 sig. figs)

022 10.0 pointsFor the conversion of 0.0003140 kilograms toounces, the conversion factors are 103 gramsper kilogram, 1 pound per 453.6 grams and16 ounces per pound.

The correctly expressed answer is1. 1.108 × 10−5 oz

2. 4.329 × 10−5 oz

3. 1.108 × 10−2 oz correct

4. 1.108 × 102 oz

5. 8.902 × 10−1 oz


Explanation:

? ounces = 0.0003140 kg

×1000 g

kg×

1 pound

453.58 g

×16 ounces

1 pound

= 1.108 × 10−2 ounces

Zeroes at the end of numbers containing a dec-imal point are considered to be significant, so0.0003140 has 4 significant figures. The 1 kgto 1000 g and 1 pound to 16 ounces conver-sions are exact definitions, so these numbersdo not affect how many significant figures theanswer should have.

023 10.0 pointsHow many milliliters are in a 67.8 ft3 box?

1. 1.92×106 mL correct

2. 13300 mL

3. 0.00239 mL

4. 2070 mL

5. 7150 mL

Explanation:

? mL = 67.8 ft3 ×(12 in)3

1 ft3

×(2.54 cm)3

1 in3×

1 mL

1 cm3

= 1.92 × 106 mL

024 10.0 pointsAn Olympic hurdler accelerates at a rate of15.5 m/s2. What is the rate in miles/min2?

1. 31.8 miles/min2

2. 6.89 miles/min2

3. 0.581 miles/min2

4. 2.69 miles/min2

5. 34.7 miles/min2 correct

6. 1.15 miles/min2

7. 1.49 miles/min2

Explanation:

Acceleration rate = 15.5 m/s2.

15.5m

s2×

1 km

1000 m×

1 mile

1.609 km

×(60 s)2

(1 min)2= 34.7

miles

min2

025 10.0 pointsGiven:

1 pound (lb) = 453.6 grams1 pound = 16 ounces (oz)How many ounces are equal to 0.0003140

kilograms? Taking significant figures into ac-count, the correctly expressed answer is1. 1.1076 × 10−2 oz

2. 1 × 10−2 oz

3. 1.108 × 10−2 oz correct

4. 1.1 × 10−2 oz

5. 1.11 × 10−2 oz

Explanation:

? ounces = 0.0003140 kg

×1000 g

kg×

1 pound

453.58 g

×16 ounces

1 pound

= 1.108 × 10−2 ounces

Zeroes at the end of numbers containing a dec-imal point are considered to be significant, so0.0003140 has 4 significant figures. The 1 kgto 1000 g and 1 pound to 16 ounces conver-sions are exact definitions, so these numbersare not included when deciding how manysignificant figures the answer should have.


026 10.0 pointsAn oxygen molecule at 25◦C has a speed ofabout 927 meters per second. Express thisspeed in miles/hour. (1 mi = 5280 ft)1. 13400 miles/hour

2. 0.000482 miles/hour

3. 20700 miles/hour

4. 2070 miles/hour correct

5. 0.000160 miles/hour

6. 0.576 miles/hour

7. 3340000 miles/hour

Explanation:

T = 25◦C v = 927m

s

927m

s×

100 cm

1 m×

1 in

2.54 cm×

1 ft

12 in

1 mi

5280 ft×

3600 s

1 hr= 2073.64 mi/h

027 10.0 pointsCalculate the density of methanol (CH3OH) ifyou know that there are 4.2 moles of methanolin 5.67 L.

1. 0.741 g/mL

2. 0.0237 g/mL correct

3. 1334 g/mL

4. 0.0431 g/mL

5. 0.131 g/mL

6. 11.34 g/mL

Explanation:

n = 4.2 mol V = 5.67 L

d =4.2 mol CH3OH

5.67 L

×32 g CH3OH

1 mol CH3OH×

1 L

1000 mL

= 0.0237 g/mL

028 10.0 pointsPolycarbonate plastic has a density of 1.3g/cm3. A photo frame is constructed fromtwo 6.2 mm sheets of polycarbonate. Eachsheet measures 28 cm by 24 cm.

What is the mass of the photo frame?

1. 730 g

2. 1100 g correct

3. 1000 g

4. 850 g

5. 1200 g

Explanation:

d = 1.3 g/cm3 ℓ = 28 cmw = 24 cm h = 2(6.2 mm) = 1.24 cm

V = length × width × height

d =m

Vm = d V = d (ℓ w h)

= (1.3 g/cm3) (28 cm) (24 cm) (1.24 cm)

= 1083.26 g

029 10.0 pointsThe density of gold is 19.3 g/mL. What is themass of a gold nugget which has a volume of3.28 mL?

1. 5.88 g

2. 0.170 g

3. 63.3 g correct


4. 30.4 g

Explanation:

030 10.0 pointsA solid has a mass of 95 grams. A graduatedcylinder contains 35 mL of water. When thesolid is completely submerged in the water,the new water level reads 84 mL.

What is the density of the solid?1. 2.7 g/mL

2. 0.80 g/mL

3. 1.9 g/mL correct

4. -1.9 g/mL

5. 1.1 g/mL

6. 190 g/mL

7. 0.52 g/mL

Explanation:

031 10.0 pointsOn the Kelvin scale, a temperature of 43◦C is1. 43 K.

2. 316 K. correct

3. 273 K.

4. 230 K.

Explanation:

032 10.0 pointsCopper melts at 1083 ◦C. What is its meltingtemperature in ◦F?

1. 583 ◦F

2. 1981 ◦F correct

3. 619 ◦F

4. 1324 ◦F

5. 797 ◦F

Explanation:

? ◦F = 1083◦C ×1.8◦F

1.0◦C+ 32◦F = 1981◦F

033 10.0 pointsIdentify the incorrect statement regardingDalton’s Atomic Theory.

1. A given compound always has the samerelative numbers and types of atoms.

2. All atoms of a given element are identi-cal.

3. All matter is composed of atoms.

4. All matter is composed of protons, elec-trons, and neutrons. correct

Explanation:

In the early 1800s when Dalton formulatedhis atomic theory, it was believed that allatoms of the same element were the same.The fundamental particles of the atom –protons, neutrons, and electrons – were notknown at this time.

034 10.0 pointsThe following are all proposals of Dalton’satomic theory. Which part of Dalton’s atomictheory was shown to be wrong by the discov-ery of natural radioactivity?1. Atoms combine in simple numerical ratios

to form compounds.

2. Atoms are permanent, unchanging, indi-visible bodies. correct

3. Two or more kinds of atoms may combinein different ways to form more than one kindof chemical compound.

4. Atoms of a given element all weigh thesame.

5. Compounds consist of collections of


molecules made up of atoms bonded to-gether.

Explanation:

The discovery of natural radioactivity pro-vided evidence that atoms can change. Theycan undergo radioactive decay, emitting elec-trons, protons, or neutrons. They can alsoundergo nuclear fission, in which a heavy nu-cleus splits into two lighter nuclei.

035 10.0 pointsThe discovery and characterization of cathoderays was important in the development of theatomic theory because

1. it led to the suggestion of the existence ofthe neutron.

2. it indicated that all matter contained elec-trons. correct

3. it indicated that all matter contains pro-tons.

4. None of the these is correct.

5. it indicated that all matter contains alphaparticles.

Explanation:

The cathode ray experiment results werethe same, even when different types of gas,composition of electrodes, and power sourceswere used.

036 10.0 pointsRutherford’s α-scattering experiment showedthat most α particles directed toward a thinmetallic foil passed through with only slightdeviations. From this evidence we can con-clude that

1. the mass of the atom is concentrated in avery small area. correct

2. an α particle is a type of light whichshould not be deflected.

3. the foil was too thin.

4. α particles are uncharged.

5. α particles are too small to hit anything.

Explanation:

In Rutherford’s gold foil experiment in1910, α (alpha) particles were fired at goldfoil, and the resulting deflection of the parti-cles were observed. Most of the α particleswent through the sample undeflected, sug-gesting that much of the atom was emptyspace. But of the few α particles that weredeflected, they were deflected at all angles,including some very wide angles! The widedeflections suggested a very hard (dense) pos-itively charged core in the atom. However,this core, or nucleus, must be small in relationto the overall size of the atom, since so fewof the α particles were deflected in the firstplace. It also suggests that the nucleus is sur-rounded by a cloud of electrons at relativelygreat distances from the nuclei.

037 10.0 pointsWhat experimental evidence led Thomson toconclude that cathode rays were negativelycharged?

1. Cathode rays were deflected towards anegatively charged plate.

2. Cathode rays pushed a paddle wheelcounterclockwise.

3. Cathode rays pushed a paddle wheelclockwise.

4. False; cathode rays are positively charged,not negatively charged.

5. Cathode rays were deflected towards apositively charged plate. correct

Explanation:

Cathode rays are repelled by negativeplates and attracted to positive plates. Sinceopposites attract, cathode rays must be nega-tively charged.


038 10.0 pointsIdentify the experimental evidence that indi-cates an atom has a positively charged nu-cleus.

1. Electrons can be ejected from a metallicsurface with high energy light.

2. α particles are deflected at large angleswhen projected toward a thin sheet of metal.correct

3. Cathode rays produce diffraction pat-terns.

4. Line spectra are produced from elementalgases in a gaseous discharge tube.

5. Cathode rays are attracted to the positiveplate of an applied electrical field.

Explanation:

In Rutherford’s gold foil experiment in1910, α (alpha) particles were fired at goldfoil, and the resulting deflection of the parti-cles were observed. Most of the α particleswent through the sample undeflected, sug-gesting that much of the atom was emptyspace. But of the few α particles that weredeflected, they were deflected at all angles,including some very wide angles! The widedeflections suggested a very hard (dense) pos-itively charged core in the atom. However,this core, or nucleus, must be small in relationto the overall size of the atom, since so fewof the α particles were deflected in the firstplace. It also suggests that the nucleus is sur-rounded by a cloud of electrons at relativelygreat distances from the nuclei.

039 10.0 pointsWhich of the following is FALSE?

1. The mass of a proton is much greater thanthe mass of an electron.

2. The mass of protons and neutrons areabout the same.

3. The mass of neutrons and electrons are

about the same. correct

4. The mass of a hydrogen atom is about thesame as a proton.

Explanation:

A proton weighs 1 amu, as does a neutron.

An electron weighs1

2000amu. Hydrogen

has one proton, one electron, and usually noneutrons, so H weighs about 1 amu.

040 10.0 pointsHow many electrons are present in one neonatom?1. 10 correct

2. 9

3. 11

4. 8

5. 12

Explanation:

The atomic number of an element is equalto the number of protons in an atom. In aneutral atoms the number of protons is equalto the number of electrons. Neon has anatomic number of 10.

041 10.0 pointsHow many protons are present in one iron(III)ion?1. 25

2. 28

3. 26 correct

4. 27

5. 29

Explanation:

The atomic number of an element is equalto the number of protons in that element’snucleus. The number of protons defines an el-ement and cannot be changed without chang-ing the element.


Iron has an atomic number of 26.

042 10.0 pointsHow many protons are present in one Ca2+

ion?1. 19

2. 22

3. 20 correct

4. 21

5. 18

Explanation:

The atomic number of an element is equalto the number of protons in that element’snucleus. The number of protons defines an el-ement and cannot be changed without chang-ing the element.

Calcium has an atomic number of 20.

043 10.0 pointsWhat is the symbol for the element whichcontains 5 electrons in the neutral state?1. C

2. B correct

3. Be

4. Li

Explanation:

Boron has 5 protons and in its neutral state,has 5 electrons as well.

044 10.0 pointsA magnesium (Mg) atom with a mass num-ber of 25 contains1. 13 protons, 12 neutrons and 12 elec-

trons.

2. 12 protons, 13 neutrons and 12 electrons.correct

3. 25 protons, 25 neutrons and 25 elec-trons.

4. None of these

5. 25 protons, 13 neutrons and 25 elec-trons.

Explanation:

045 10.0 pointsIdentify the isotope that has atoms with 12neutrons, 10 protons, and 10 electrons.

1. 22Na

2. 10S

3. None of these

4. 22Ne correct

5. 10P

6. 12Mg

7. 12Al

Explanation:

The element Ne is identified by the numberof protons (atomic number), and the syntax

is n+ppNep−e.

046 10.0 pointsWrite the hyphen notation for the elementthat contains 15 electrons and 15 neutrons.

1. copper-60

2. phosphorus-30 correct

3. silicon-30

4. fluorine-15

5. sulfur-31

6. phosphorus-31

7. oxygen-15

8. zinc-60

Explanation:


#e− = 15 #n = 15hyphen notation for given element = ?

atomic number = number of protons

= number of electrons

mass number= number of protons + number of neutrons

atomic number = 15 (element is phosphorus)mass number

= 15 protons + 15 neutrons = 30nuclide is phosphorus-30

047 10.0 pointsChoose the correct nuclear symbol and hy-phen notation for the isotope which has amass number of 28 and atomic number of 14.

1. 1428S, sulfur-14

2. 1414P, phosphorus-14


4. 1414Si, silicon-14


6. 2814Si, silicon-28 correct

7. 1428Si, silicon-14

8. 1414S, sulfur-14

9. 2814S, sulfur-28

Explanation:

mass number = 28 atomic number = 14nuclear symbol and hyphen notation

for given nuclide = ?atomic number = number of protonselement with 14 protons: silicon (Si)nuclear symbol −→ 28

14Sihyphen notation −→ silicon-28

048 10.0 pointsA newly discovered element has two isotopes.One has an atomic weight of 120.9038 amu

with 57.25% abundance. The other has anatomic weight of 122.8831 amu. What is theatomic weight of the element?

1. 122.38 amu

2. 123.45 amu

3. 121.17 amu

4. 122.15 amu

5. 121.54 amu

6. 121.75 amu correct

Explanation:

Average Atomic Mass

= 0.5725(120.9038) + 0.4275(122.8831)

= 121.75 amu

049 10.0 pointsCalculate the molar mass of the noble gaskrypton in a natural sample, which is 0.81%78Kr (molar mass 77.92 g/mol), 1.82% 80Kr(molar mass 79.91 g/mol), 10.92% 82Kr (mo-lar mass 81.91 g/mol), 12.42% 83Kr (molarmass 82.92 g/mol), 56.65% 84Kr (molar mass83.91 g/mol), and 17.38% 86Kr (molar mass85.91 g/mol).

Correct answer: 83.7949 g/mol.

Explanation:

MMkrypton = 0.0081 (77.92 g/mol)

+ 0.0182 (79.91 g/mol)

+ 0.1092 (81.91 g/mol)

+ 0.1242 (82.92 g/mol)

+ 0.5665 (83.91 g/mol)

+ 0.1738 (85.91 g/mol)

= 83.7949 g/mol

050 10.0 points


Which of the following statements about anonmetal is true?

1. A nonmetal forms covalent bonds withmetals.

2. A nonmetal has metallic bonding.

3. A nonmetal tends to form cations.

4. A nonmetal has more than 3 electrons inits outer shell. correct

Explanation:

Nonmetals start in group VI of the periodictable, so they have more than 3 electrons intheir outer shells.

051 10.0 pointsWhich characteristic describes a metal?

1. brittle

2. malleable correct

3. good insulator

4. low conductivity

5. tend to be reduced

Explanation:

Metals are lustrous, malleable, ductile,good conductors, and tend to be oxidized.

052 10.0 points

Identify the transition metals.A. hafniumB. radiumC. radon

1. B and C only

2. A only correct

3. A, B, and C

4. A and B only

5. C only

6. B only

7. None of these

8. A and C only

Explanation:

The transition metals lie in the shaded area

053 10.0 pointsClassify the elements Si, Sr, Sn, Sb.

1. metal; nonmetal; metal; nonmetal

2. metalloid; metal; metal; nonmetal

3. nonmetal; nonmetal; metal; metalloid

4. None of these

5. metalloid; metal; metal; metalloid cor-

rect

Explanation:

Silicon (Si) is a group 14 metalloid, stron-tium (Sr) is a group 2 metal, tin (Sn) is agroup 14 metal and antimony (Sb) is a group15 metalloid.

054 10.0 pointsList the elements that are members of thealkali metals.

1. C, Si, Ge, Sn, Pb

2. Li, Na, K, Rb, Cs, Fr correct

3. B, Al, Ga, In, Tl

4. None of these


5. Be, Mg, Ca, Sr, Ba, Ra

6. N, P, As, Sb, Bi

7. F, Cl, Br, I, At

8. He, Ne, Ar, Kr, Xe, Rn

9. O, S, Se, Te, Po

Explanation:

The name for the group I metal is alkali metals

055 10.0 points

Consider the periodic table

X YZ

Identify the nonmetal.

1. Z correct

2. X

3. Y

Explanation:

The nonmetals lie in the shaded area:

X YZ

056 10.0 pointsIn the periodic table, which elements typicallyhave similar properties?

1. those in the same columns correct

2. those related diagonally

3. those on opposite sides

4. those in the same rows

Explanation:

Elements in the same group have the samenumber of valance (outer) electrons and oftenreact in similar ways.

057 10.0 pointsThe letters in the periodic table are used toidentify four different elements.

AB

CD

Based on the organization within the peri-odic table, which element is expected to have2+ charge when it forms an ion?

1. C

2. D

3. B

4. A correct

Explanation:

Group II elements form ions with a 2+charge.

058 10.0 pointsOf the following elements, the one which isNOT a metal is1. potassium.

2. iron.

3. cobalt.

4. arsenic. correct

5. silver.

Explanation:

Iron, cobalt, silver, and potassium are all


metals since they are to the left of the linedividing metals and non-metals. Arsenic isthe only element to the right of that line.

059 10.0 pointsOf the following elements, the one which isNOT a nonmetal is

1. phosphorous.

2. bismuth. correct

3. sulfur.

4. silicon.

5. iodine.

Explanation:

Only bismuth is a metal in the list of el-ements because it is to the left of the lineseparating the metals and non-metals.

060 10.0 pointsWhich of the following elements would beexpected to resemble strontium (Sr) mostclosely in chemical properties?1. Si

2. Ba correct

3. Li

4. Cs

5. Sc

Explanation:

Elements with similar chemical propertiesare arranged in the same group or vertical col-umn. Thus strontium (Sr) would be expectedto have similar properties to barium (Ba).

061 10.0 pointsIn the periodic table, a horizontal row of ele-ments is called1. a group.

2. a period. correct

3. a transition.

4. a lanthanide.

5. an orbital.

Explanation:

Horizontal rows on the periodic table arecalled periods. The vertical columns aregroups or families.

062 10.0 pointsIn the periodic table, a vertical column ofelements is called1. a lanthanide.

2. a group. correct

3. a period.

4. an orbital.

5. a transition.

Explanation:

Horizontal rows on the periodic table arecalled periods. The vertical columns aregroups or families.

063 10.0 pointsArsenic (As) is classified as

1. a transition element.

2. an actinide.

3. a lanthanide.

4. a noble gas.

5. a representative element. correct

Explanation:

Representative elements are found in the AGroups of the periodic table. As is a GroupVA element.

064 10.0 pointsWhat is the formula for the ionic compoundformed from sodium and oxide ions?


1. NaO2

2. None of these

3. Na3O2

4. NaO

5. Na2O correct

6. Na2O3

Explanation:

The sodium ion is Na+ and the oxide ion isO2−, so the formula is Na2O.

065 10.0 pointsWhat is the formula for the ionic compoundformed from lithium and sulfide ions?

1. LiS

2. None of these

3. Li4S

4. Li2S correct

5. Li2SO4

6. Li2S3

Explanation:

The lithium ion is Li+ and the sulfide ion isS2−, so the formula is Li2S.

066 10.0 pointsWrite the formula of calcium nitride.

1. Ca2N

2. Ca3N2 correct

3. Ca2N3

4. None of these

5. CaN2

6. CaN3

7. Ca3N

Explanation:

Ca is in Group 2 so it forms a +2 ion; N isin Group 5 so it forms a −3 ion. To balancecharges you need three Ca2+ for each two N3−.

067 10.0 pointsWhat is the formula of the compound madeby combining calcium (Ca) with sulfur (S)?

1. SCa2

2. CaS correct

3. SCa

4. CaS2

Explanation:

Ca has two valence electrons and forms a2+ ion. S has six valence electrons and formsa 2− ion. The ratio will be 1 : 1, so theformula is CaS.

068 10.0 pointsWhat is the formula for the salt which con-tains both Ca and NO3 ions? (Note that thecharges of the ions are not indicated.)

1. Ca(NO3)2 correct

2. Ca3(NO3)2

3. Ca2(NO3)2

4. Ca2NO3

5. CaNO3

Explanation:

The calcium ion is Ca2+ and the nitrateion is NO−

3 . Two NO−

3 are needed to balance

the charge on each Ca2+. The formula isCa(NO3)2.

069 10.0 pointsWhich of the following ionic compounds hasthe wrong chemical formula?


1. NaO2 correct

2. KBr

3. K2S

4. CaO

5. MgCl2

Explanation:

The sodium ion is Na1+ and the oxide ionis O2−. The compound should thus be Na2O.

070 10.0 pointsWhich of the following species has the greatestnumber of electrons?1. Cl−

2. F−

3. O2−

4. S2−

5. I− correct

Explanation:

The atomic number is the number of pro-tons in an atom. In a neutral atom this alsoequals the number of electrons. For ions youmust account for the lost or gained electronsto find the total number. O2− has 10 elec-trons, F− has 10 electrons, S2− has 18 elec-trons, Cl− has 18 electrons, and I− has 54electrons.

071 10.0 pointsWhich of the following species has the greatestnumber of electrons?1. Al

2. Na+

3. F−

4. Fe2+ correct

5. Ne

Explanation:

072 10.0 pointsAn element (call it X) in Group IA of the peri-odic table forms a compound with an element(call it Y) of Group VIA. What would be themost likely formula and nature of the com-pound?1. X2Y; covalent

2. XY2; ionic

3. XY6; ionic

4. X6Y; ionic

5. X2Y; ionic correct

6. XY2; covalent

7. XY6; covalent

Explanation:

Metals reacts with non-metals to form ioniccompounds, which involve the transfer of elec-trons.

A metal X in group IA would have onevalence electron. It would lose one electronto become isoelectronic with the precedingnoble gas. This metal would now have onefewer electron than protons and would have a+1 charge.

An element Y in group VIA would have sixvalence electrons. It would gain two electronsto become isoelectronic with the nearest noblegas. This element would now have two moreelectrons than protons and would have a −2charge.

We need twice as many X+ ions as Y2− ionsto equalize the overall charge, so the ratio ofX to Y ions is 2:1, and the formula would beX2Y.

073 10.0 pointsChoose that pair of names and formulas thatdo not match.

1. Fe2(SO4)3 : iron(III) sulfate

2. SnCl4 : tin(IV) chloride


3. MnCO3 : manganese(II) carbonate

4. Cu2O : copper(II) oxide correct

5. SnO : tin(II) oxide

Explanation:

The copper(II) ion is Cu2+; the oxide ionis O2−. One O2− is needed to balance thecharge of each Cu2+, so the correct formulafor copper(II) oxide is CuO.

In Cu2O, the total charge from copper ionsmust be +2 to balance the −2 charge fromthe oxide ion. Each copper ion would havea charge of +1. This compound is copper(I)oxide.

The manganese(II) ion is Mn2+; the car-bonate ion is CO3

2−. One CO32− is needed

to balance the charge of each Mn2+, so theformula is MnCO3.

The tin(IV) ion is Sn4+; the chloride ionis Cl−. Four Cl− are needed to balance thecharge of each Sn4+, so the formula is SnCl4.

The tin(II) ion is Sn2+; the oxide ion is O2−.One O2− is needed to balance the charge ofeach Sn2+, so the formula is SnO.

The iron(III) ion is Fe3+; the sulfate ion isSO4

2−. Three SO42− are needed to balance

the charge of every two Fe3+. (This gives atotal anion charge of −6 and a total cationcharge of +6.) The formula is Fe2(SO4)3.

074 10.0 pointsWhich of the following pairs of names andformulas are correct?Z1) dinitrogen tetroxide : N2O4

Z2) calcium hydroxide : CaOHZ3) tin : Sn

1. Z1) and Z3) correct

2. Z1) only

3. Z1) and Z2)

4. Z1), Z2), and Z3)

5. Z3) only

Explanation:

The calcium ion is Ca2+; the hydroxide ionis OH−. Two OH− are needed to balance thecharge on each Ca2+, so the formula shouldbe Ca(OH)2.

N2O4 is a covalent compound and is cor-rectly named using prefixes to indicate thenumber of atoms of each element in the com-pound.

Sn is the symbol for tin.

075 10.0 pointsThe binary compound PCl3 is called

1. phosphorus chloride.

2. None of these.

3. monophosphorus trichloride.

4. triphosphorus chloride.

5. phosphorus trichloride. correct

Explanation:

PCl3 is a covalent compound and is there-fore named using prefixes to indicate the num-ber of each type of atom. The prefix for threeis “tri” so this molecule is named phosphoroustrichloride. The “mono” prefix is commonlyleft off of the first element name in a binarycompound.

076 10.0 pointsWhat is the name of the compound with theformula CCl4?

1. carbon tetrachloride correct

2. carbon chloride

3. chlorine carbonide

4. carbon(IV) chloride

Explanation:

This is a covalent compound and should benamed using prefixes to indicate the numberof atoms of each element in the compound.Binary covalent compounds are named withthe name of the first element followed by the


name of the second element with the suffix“-ide” added. CCl4 is carbon tetrachloride.

077 10.0 pointsWrite the formula of perchloric acid.

1. HClO2

2. HClO

3. HClO5

4. HClO4 correct

5. HClO3

Explanation:

Change ‘-ic’ to ‘-ate’. The perchlorate an-ion is ClO−

4 . Now add enough H+ to makethe overall species neutral.

078 10.0 pointsWhat is the chemical formula for hypopophos-phorous acid?

1. H2PO

2. H2PO4

3. H2PO5

4. HPO5

5. HPO8

6. HPO

7. H2PO2

8. H2PO3

9. H2PO6

10. H3PO2 correct

Explanation:

079 10.0 pointsPick the name and formula that do NOTmatch.

1. H2SO4 : sulfurous acid correct

2. HIO (or HOI) : hypoiodous acid

3. HNO3 : nitric acid

4. HMnO4 : permanganic acid

Explanation:

Sulfuric acid is H2SO4. Sulfurous acidhas one fewer oxygen atom, so its formulais H2SO3.

Iodic acid is HIO3. Hypoiodous acid hastwo fewer oxygen atoms, so its formula is HIO.

MnO−

4 is the permanganate ion. HMnO4 ispermanganic acid.

NO−

3 is the nitrate ion. HNO3 is nitric acid.

080 10.0 pointsPick the name and formula that do NOTmatch.

1. HCl : hydrochloric acid

2. H2CO3 : carbonic acid

3. H2SO4 : sulfuric acid

4. HClO4 : chloric acid correct

Explanation:

Cloric acid is HClO3. Perchloric acidhas one more oxygen atom, so its formulais HClO4.

HCl is hydrogen chloride. In aqueous solu-tion, HCl is known as hydochloric acid.

SO2−4 is the sulfate ion so H2SO4 is sulfuric

acid.CO2−

3 is the carbonate ion so H2CO3 iscarbonic acid.

081 10.0 pointsWhich one has the greatest number of atoms?

1. 3.05 moles of CH4 correct

2. 3.05 moles of water

3. 3.05 moles of helium


4. All have the same number of atoms

5. 3.05 moles of argon

Explanation:

For 3.05 moles of water:

? atoms = 3.05 mol H2O ×6.02 × 1023 molec

1 mol

×3 atoms

1 molecule= 5.51 × 1024 atoms

For 3.05 moles of CH4:

? atoms = 3.05 mol CH4 ×6.02 × 1023 molec

1 mol

×5 atoms

1 molecule= 9.18 × 1024 atoms

For 3.05 moles of helium:

? atoms = 3.05 mol He ×6.02 × 1023 atoms

1 mol= 1.84 × 1024 atoms

For 3.5 moles of argon:

? atoms = 3.05 mol Ar ×6.02 × 1023 atoms

1 mol= 1.84 × 1024 atoms

082 (part 1 of 2) 10.0 points

How many moles of Ag+ are in 3 g ofAgCl?Correct answer: 0.0209322 mol.

Explanation:

mAgCl = 3 g

MWAgCl = 107.87 g/mol + 35.45 g/mol

= 143.32 g/mol

nAgCl =3 g

143.32 g/mol= 0.0209322 mol

The number of moles of Ag+ ions equals thenumber of moles of AgCl.

083 (part 2 of 2) 10.0 points

How many moles of H2O are in 1.8 g ofAuCl3 · 2 H2O?Correct answer: 0.0106085 mol.

Explanation:

mAuCl3·2 H2O = 1.8 g

MWAuCl3·2 H2O = 196.97 g/mol

+ 3 (35.45 g/mol)

+ 4 (1.00079 g/mol)

+ 2 (16.0 g/mol)

= 339.352 g/mol

nH2O =

(

1.8 g AuCl3 · 2 H2O

339.352 g/mol

)

×

(

2 mol H2O

1 mol AuCl3 · 2 H2O

)

= 0.0106085 mol H2O

084 10.0 pointsHow many fluorine atoms are in 4.0 moles offluorine molecules?

1. 1.5 × 1023 atoms

2. 6.6 × 10−24 atoms

3. 2.4 × 1024 atoms

4. 4.8 × 1024 atoms correct

Explanation:

nF = 4.0 molFluorine is diatomic. Each F2 molecule

contains two fluorine atoms. We can useAvogadro’s number and the ratio of F atomsto F2 molecules to find the number of fluorineatoms:

? atoms F = 4.0 mol F2

×6.022 × 1023 F2

1 mol F2

×2 atoms F

1 molec F2

= 4.8 × 1024 atoms F


085 10.0 pointsEach mole of Al(NO3)3 contains how manymoles of oxygen atoms?

1. 12 mol

2. 6 mol

3. 1 mol

4. 3 mol

5. 9 mol correct

Explanation:

There are 9 O atoms in each molecule ofAl(NO3)3. This same ratio holds for the num-ber of moles of O in one mole of Al(NO3)3.

? mol O

mol Al(NO3)3=

9 O

1 mol Al(NO3)3

Therefore there are 9 mol O in one mole ofAl(NO3)3.

086 10.0 pointsHow many atoms of hydrogen are containedin 1 mole of methane (CH4)?

1. 6.02 × 1023 atoms

2. The correct answer is not given.

3. 2.41 × 1024 atoms correct

4. 4 atoms

5. 3.01 × 1024 atoms

Explanation:

n = 1 molEach methane molecule contains 4 hydro-

gen atoms. There are Avogadro’s number ofmethane molecules in one mole of methanemolecules:

nH = 1 mol CH4

×6.02 × 1023 molec CH4

1 mol CH4

×4 H atoms

1 molec CH4

= 2.41 × 1024 H atoms

087 10.0 pointsHow many moles of carbon are in 8.00 molesof C12H22O11?

1. 8.00 mol

2. 88.0 mol

3. 96.0 mol correct

4. 10.0 mol

Explanation:

n = 8.00 molIn 1 mole of C12H12O11, there are 12 moles

of C, 22 moles of H, and 11 moles of O. If youhave 8 moles of C12H12O11, then

12 mol C

1 mol C12H22O11=

x

8 mol C12H22O11

x =12 mol C

1 mol C12H22O11

× 8 mol C12H22O11

= 96 mol C

088 10.0 pointsHow many sulfur atoms are in 78.4 g of sul-fur?Correct answer: 1.47235 × 1024 atoms.

Explanation:

mSu = 78.4 gTo solve this problem, we will need both the

atomic mass of sulfur (S, 32.066 g/mol) andAvogadro’s number (6.022×1023 atoms/mol).First convert grams to moles of sulfur usingthe atomic mass:

? mol S = 78.4 g S ×1 mol S

32.066 g S= 2.44496 mol S

Now convert moles to the number of atomsusing Avogadro’s number:

? atoms S = 2.44496 mol S

×6.022 × 1023 S atoms

1 mol S= 1.47235× 1024 atoms S


089 10.0 pointsHow many moles are in 418.8 g of Ba(OH)2?Correct answer: 2.44412 mol.

Explanation:

mBa(OH)2 = 418.8 g n = ?

1 mol Ba ×137.33 g Ba

mol Ba= 137.33 g

2 mol O ×16.00 g O

mol O= 32.00 g

2 mol H ×1.01 g H

mol H= 2.02 g

Molar mass = 171.35 g /mol

nBa(OH)2 =418.8 g

171.35 g/mol Ba(OH)2

= 2.44412 mol Ba(OH)2

090 10.0 pointsHow many moles are in 158.55 g ofFe3(PO4)2?Correct answer: 0.443447 mol.

Explanation:

mFe3(PO4)2 = 158.55 g n = ?

3 mol Fe ×55.85 g Fe

mol Fe= 167.6 g

2 mol P ×30.97 g P

mol P= 61.94 g

8 mol O ×16.00 g O

mol O= 128.00 g


nFe3(PO4)2 =158.55 g

357.54 g/mol Fe3(PO4)2

= 0.443447 mol

091 10.0 pointsAspirin has a formula C9H8O4. What is themolar mass of aspirin?1. 95 g

2. 220 g

3. 325 g

4. 180 g correct

Explanation:

092 10.0 pointsAlthough coal is not a pure chemical com-pound, its elemental composition may beapproximated by the formula C135H96NO9S.What is the approximate percentage by massof sulfur in coal?

1. 1.68% correct

2. 0.413%

3. 42.8%

4. 3.11%

Explanation:FWcoal = 135(12.0107 g/mol)

+ 96(1.00794 g/mol)

+ 1(14.0067 g/mol)

+ 9(15.9994 g/mol)

+ 1(32.065 g/mol)

= 1908.27 g/mol

% S in coal =1(32.065 g/mol)

1908.27 g/mol× 100%

= 1.68032 %

093 (part 1 of 2) 10.0 pointsZinc chloride (ZnCl2) is 52.02% chlorine bymass. What mass of chlorine is contained in61.7 g of ZnCl2?

Correct answer: 32.0963 g.

Explanation:

% mass of chlorine in ZnCl2 = 52.02%mCl in 61.7 g of ZnCl2 = ?

mCl = mass % × mass ZnCl2

= 0.5202 × 61.7 g

= 32.0963 g Cl

094 (part 2 of 2) 10.0 pointsHow many moles of Cl is this?Correct answer: 0.905397 mol.

Explanation:


nCl =mass Cl

molar mass of Cl

=32.0963 g Cl

35.45 g Cl

= 0.905397 mol Cl

095 10.0 pointsWhat is the percentage of hydrogen inC7H8O? The molecular weight of C is12.011 g/mol, of H 1.00794 g/mol, and ofO 15.9994 g/mol.Correct answer: 7.45656%.

Explanation:

MWC = 12.011 g/molMWH = 1.00794 g/mol

MWO = 15.9994 g/mol

MWtotal = #C · MWC + #H · MWH

+ #O · MWO

= 7(12.011 g/mol)

+ 8(1.00794 g/mol)

+ 1(15.9994 g/mol)

= 108.14 g/mol ,

so the percentage of hydrogen in 0-cresol is

% =#H · MWH

MWtotal100.0%

=8 (1.00794 g/mol)

108.14 g/mol100.0%

= 7.45656%

096 10.0 pointsWhat mass of rhodium contains as manyatoms as there are iron atoms in 28 g of iron?

Correct answer: 51.596 g.

Explanation:

mFe = 28 g

mRh =

(

28 g Fe

55.847 g/mol N

)

× (102.91 g/mol Rh)

= 51.596 g Rh

097 (part 1 of 3) 10.0 pointsa) How many CaH2 formula units are presentin 6.341 g of CaH2?Correct answer: 9.08837×1022 formula units.

Explanation:

mCaH2= 6.341 g

MWCaH2= 40.0 g/mol + 2 (1.0079 g/mol)

= 42.0158 g/mol

n =6.341 g

42.0158 g/mol

×

(

6.022 × 1023 formula units/mol)

= 9.08837 × 1022 formula units

098 (part 2 of 3) 10.0 pointsb) Find the mass of 6.22 × 1024 formula unitsof NaBF4 (sodium tetrafluroborate).Correct answer: 1134.1 g.

Explanation:

nNaBF4= 6.22 × 1024 formula units

MWNaBF4= 22.99 g/mol + 10.81 g/mol

+ 4 (19.0 g/mol)

= 109.8 g/mol

mNaBF4=

6.22 × 1024 formula units

6.022 × 1023 formula unitsmol

× (109.8 g/mol)

= 1134.1 g

099 (part 3 of 3) 10.0 pointsc) How many moles of CeI3 are in 9.66 ×

1021 formula units of CeI3 (cerium(III) iodide;a bright, water, soluble solid).Correct answer: 0.0160412 mol.

Explanation:

nCeI3 = 9.66 × 1021 formula units


nCeI3 =9.66 × 1021 formula units

6.022 × 1023 formula units/mol

= 0.0160412 mol

100 (part 1 of 3) 10.0 pointsFind the mass of 2.72 mol Cl2.Correct answer: 193.12 g.

Explanation:

nCl2 = 2.72 mol m = ?

m = (2.72 mol Cl2)

(

71 g Cl2mol Cl2

)

= 193.12 g Cl2

101 (part 2 of 3) 10.0 pointsFind the mass of 2.65 × 1023 molecules H2S.Correct answer: 14.9618 g.

Explanation:

nH2S = 2.65 × 1023 molec m = ?

m =(

2.65 × 1023 molec)

·

(

mol

6.022 × 1023 molec

) (

34 g H2S

mol

)

= 14.9618 g H2S

102 (part 3 of 3) 10.0 pointsFind the mass of 17 molecules SO2.Correct answer: 1.80671× 10−21 g.

Explanation:

nSO2= 17 molec m = ?

m = (17 molec)

(

mol

6.022 × 1023 molec

)

·

(

64 g SO2

mol

)

= 1.80671× 10−21 g SO2

103 10.0 pointsHow many molecules are in 1.20 g of SO2?Correct answer: 1.12789× 1022 molec.

Explanation:

mSO2= 1.20 g n = ?

1 mol S ×32.07 g S

mol S= 32.07 g

2 mol O ×16.00 g O

mol O= 32.00 g


nSO2=

1.20 g

64.07 g/mol SO2

×6.022 ×1023 molec

1 mole= 1.12789 × 1022 molec SO2

104 10.0 pointsA can of mineral water contains 345 mL ofwater. How many molecules of water does thecan contain? (water = 1.00 g/mL)

1. 57.6 molec

2. 3.74 × 1027 molec

3. 3.46 × 1025 molec

4. 1.15 × 1025 molec correct

5. 19.2 molec

Explanation:

VHCl = 345 mLWe can use the density of water (H2O) to

convert from mL of water to grams of water:

? g H2O = 345 mL H2O ×1 g H2O

1 mL H2O= 345 g H2O

Before we can use Avogadro’s number toconvert to the number of water molecules,we must first convert from grams water tomoles of water. We do this using the formulaweight, which can be determined from theatomic weights on the periodic table:

? FWH2O = 2 mol H ×1.01 g H

mol H

+ 1 mol O ×16.00 g O

mol O


=18.02 g H2O

mol H2O

? mol H2O = 345 g H2O ×1 mol H2O

18.02 g H2O= 19.1 mol H2O

Now we can use Avogadro’s number to findthe number of water molecules:

? molec H2O = 19.1 mol H2O

×6.02 × 1023 H2O

1 mol H2O

= 1.15 × 1025 molec H2O

105 10.0 points

How many aluminum atoms are in198.273 g of Al2O3?Correct answer: 2.34209× 1024 atoms.

Explanation:

m = 198.273 gEach Al2O3 molecule contains two alu-

minum atoms. There are Avogadro’s num-ber of Al2O3 molecules in one mole of Al2O3.We need the formula mass of Al2O3 so we canconvert grams of Al2O3 to moles Al2O3.

Molecular mass of Al2O3:

? g/mol = 2(26.98 g/mol)

+ 3(16.00 g/mol)

= 101.96 g/mol

We can use this formula mass to convert gAl2O3 to mol Al2O3:

? mol Al2O3 = 198.273 g Al2O3

×1 mol Al2O3

101.96 g Al2O3

= 1.94462 mol Al2O3

We can now use Avogadro’s number and theratio of Al atoms to Al2O3 molecules to findthe number of aluminum atoms:

? atoms Al = 1.94462 mol Al2O3

×6.022 × 1023 Al2O3

1 mol Al2O3

×2 atoms Al

1 molec Al2O3

= 2.34209× 1024 atoms Al

106 10.0 pointsA 46 gram sample of NH3 contains how manygrams of H?Correct answer: 8.17958 g.

Explanation:

m = 46 gTo solve this problem we need the formula

weight of NH3:

FW = 3 mol H ×1.01 g H

mol H

+ 1 mol N ×14.01 g N

mol N

=17.04 g NH3

mol NH3

This formula weight is used to convertgrams NH3 to moles NH3:

? mol NH3 = 46 g NH3 ×1 mol NH3

17.04 g NH3

= 2.69953 mol NH3

Each mole of NH3 contains 3 mol of H. Wecan use this ratio to convert from moles NH3

to moles H:

? mol H = 2.69953 mol NH3 ×3 mol H

1 mol NH3

= 8.09859 mol H

Finally, the atomic weight of H can be usedto convert from moles H to grams H:

? g H = 8.09859 mol H ×1.01 g H

1 mol H= 8.17958 g H

107 10.0 pointsIron forms a compound called ferrocene. It


has the composition 64.56% C, 5.42% H, and30.02% Fe. What is the empirical formula offerrocene?1. FeC6H5

2. Fe2C10H11

3. FeC10H10 correct

4. FeC11H9

5. Fe3C11H12

6. FeC7H8

Explanation:

For 100 g of compound,

moles of C =64.56 g

12.01 g/mol= 5.376 mol

moles of H =5.42 g

1.0079 g/mol= 5.38 mol

moles of Fe =30.02 g

55.85 g/mol= 0.5375 mol

Dividing by 0.5375 mol gives 10.0 C : 10.0H : 1.0 Fe.

The empirical formula is FeC10H10.

108 (part 1 of 2) 10.0 pointsOsmium forms a number of molecular com-pounds with carbon monoxide. One light-yellow compound was analyzed to give thefollowing elemental composition: 15.89% C,21.18% O, and 62.93% Os. What is the em-pirical formula of this compound?

1. OsCO

2. Os2C4O5

3. OsC3O5

4. OsCO4

5. OsC4O4 correct

6. Os2CO4

Explanation:

For 100 g of osmium carbonyl compound,

moles of C =15.89 g

12.01 g/mol= 1.323 mol

moles of O =21.18 g

16.0 g/mol= 1.324 mol

moles of Os =62.93 g

190.2 g/mol= 0.3309 mol

Dividing each number by 0.3309 mol givesa ratio of 4.0 C : 4.0 O : 1.0 Os .

The empirical formula is OsC4O4.

109 (part 2 of 2) 10.0 pointsFrom the mass spectrum of the compound,the molecule was determined to have a molarmass of 907 g/mol. What is its molecularformula?

1. Os3C12O12 correct

2. Os3C3O12

3. Os3C9O15

4. Os4C8O11

5. Os4C2O8

6. Os4C8O10

Explanation:

The formula mass of OsC4O4 is 302.24g/mol. The molar mass is 907 g/mol whichis 3 times the formula mass, so the molecularformula is Os3C12O12.

110 10.0 pointsPaclitaxel, which is extracted from the Pacificyew tree Taxus brevifolia, has antitumor ac-tivity for ovarian and breast cancer. It is soldunder the trade name Taxol. On analysis,its mass percentage composition is 66.11% C,6.02% H, and 1.64% N, with the balance be-ing oxygen. What is the empirical formula ofpaclitaxel?

1. C47H51NO14 correct

2. C45H50NO13

3. C3H5NO4

4. C4H5NO9


5. C35H41NO8

6. C49H53NO16

Explanation:

For 100 g of Paclitaxel,

moles of C =66.11 g

12.01 g/mol= 5.504 mol

moles of H =6.02 g

1.0079 g/mol= 5.97 mol

moles of N =1.64 g

14.01 g/mol= 0.117 mol

mass of O = 100 − 66.11 − 6.02 − 1.64

= 26.23 g

moles of O =26.32 g

16.0 g/mol= 1.639 mol

Dividing by 0.117 mol gives 47.0 C : 51.0 H: 1.0 N : 14.0 O.

The empirical formula is C47H51NO14.

111 10.0 pointsWhat is the molecular formula of the moleculethat has an empirical formula of CH2O and amolar mass of 120.12 g/mol?

1. CH2O

2. C3H6O3

3. C4H8O4 correct

4. C5H10O5

5. C2H4O2

6. C6H12O6

Explanation:

empirical formula: CH2Omolar mass = 120.12 g/mol

x =molecular formula mass

empirical formula massEmpirical formula mass of CH2O

= 12.01 amu + 2.02 + 16.00 amu

= 30.03 amu

x =120.12 g/mol

30.03 g/mol= 4.000 → 4

molecular formula: CxH2xOx = C4H8O4.

112 10.0 pointsAn analysis of nicotine (MW = 162 g/mol)gives 74.0% carbon, 8.65% hydrogen, and17.3% nitrogen. What is the TRUE(MOLECULAR) formula for nicotine?

1. C12H4N

2. C4H5N2

3. C10H14N2 correct

4. C9H12N3

5. C6H8N5

6. C5H7N

7. C8H10N4

8. C4H5N4

9. C11H16N

Explanation:

% C = 74.0% % H = 8.65%% N = 17.3% MW = 162 g/mol

Assume we have 100 g of nicotine. (In fact,we can assume any mass, but choosing 100 gmakes the math easier.) If we have 100 g ofnicotine, 74.0 g (74.0%) of it is carbon, 8.65 g(8.65%) of it is hydrogen, and 17.3 (17.3%) ofit is nitrogen.

This means that, in this 100 g sample:

? mol of C = 74.0 g C ×1 mol C

12.011 g C= 6.16 mol C

? mol of H = 8.65 g H ×1 mol H

1.0079 g H= 8.58 mol H

? mol of N = 17.3 g N ×1 mol N

14.0067 g N= 1.24 mol N

The smallest number of moles here is then1.24, and we can use this number to detem-ine the empirical (simplest) formula for thiscompound. For each element, we should now


divide the number of moles we just calculatedby this smallest number of moles to force themole ratio to integers.So for C, we have

6.16 mol C

1.24= 5 mol C

For H, we have

8.58 mol H

1.24= 7 mol H

And for N, we have

1.24 mol N

1.24= 1 mol N

This means this compound’s empirical for-mula is C5H7N1, better written as C5H7N.

However, the question asked what thecompound’s TRUE molecular formula was.For this, we need to compare the empiricalformula’s apparent molecular weight to theknown molecular weight of the compound.The apparent molecular weight of the empir-ical formula is 81 g/mol, exactly half of theknown molecular weight. This means thatthe TRUE formula is double the empiricalformula, i.e., C10H14N2.

113 10.0 pointsCalculate the nuclear binding energy of a 32

16Satom. The measured atomic mass of 32

16S is31.972 070 amu.

1. 2.62 × 10−11 J

2. 4.36 × 10−11 J correct

3. None of these

4. 4.36 × 1011 J

5. 2.73 × 10−12 J

6. 1.36 × 10−12 J

7. 2.62 × 1013 J

Explanation:

measured atomic mass of 3216S= 31.972 070 amu

nuclear binding energy of 3216S = ?

16 protons: (16×1.007 276 amu)= 16.116 416 amu

16 neutrons: (16×1.008 665 amu)= 16.138 640 amu

16 electrons: (16×0.000 548 6 amu)= 0.008 777 6 amu

total combined mass: 32.263 834 amumass defect

= 32.263 834 amu − 31.972 070 amu

= 0.291 764 amu

= (0.291 764 amu)

(

1.6605 × 10−27 kg

1 amu

)

= 4.8447 × 10−28 kg

E = mc2

= (4.8447× 10−28 kg)(3.00 × 108 m/s)2

= 4.36 × 10−11 kg·m2/s2

= 4.36 × 10−11 J

114 10.0 pointsThe mass of a 7

3Li atom is 7.016 00 amu.Calculate its mass defect.

1. None of these

2. 7 amu

3. 0.294 91 amu

4. 0.042 13 amu correct

5. 3 amu

6. 1 amu

7. 0.126 39amu

Explanation:

measured atomic mass of 73Li

= 7.016 00 amumass defect = ?

3 protons: (3×1.007 276 amu)= 3.021 828 amu

4 neutrons: (4×1.008 665 amu)= 4.034 66 amu

3 electrons: (3×0.000 5486 amu)= 0.001 646 amu


total combined mass: 7.058 134 amu

mass defect = 7.058 134 amu− 7.016 00 amu

= 0.042 13 amu

115 10.0 pointsWhat is the mass defect of 28

14Si which has anactual mass of 27.97693 g/mol?

1. 0.0102 g

2. 0.52416 g

3. 0.1004 g

4. 0.25406 g correct

Explanation:

116 10.0 pointsWhen 131I emits a β particle, what nuclide isproduced?

1. 131Xe correct

2. 131Te

3. 130I

4. 130Te

5. 127Sb

Explanation:

117 10.0 pointsA nuclide undergoes proton emission to form52Fe. What is the nuclide?

1. 56Ni

2. 52Mn

3. 53Co correct

4. 52Co

5. 53Fe

Explanation:

118 10.0 pointsWhat type of particle is emitted in the trans-formation

201Pt → 201Au?

1. positron

2. β particle correct

3. γ particle

4. No particle is emitted because electroncapture occurs.

5. α particle

Explanation:

119 10.0 pointsWhat nuclide is formed when 24Na undergoesβ decay?

1. 24Ne

2. 28P

3. 28S

4. 24Mg correct

Explanation:

The nuclear reaction is

2411Na →

AZ? + 0

−1e

A = 24 − 0 = 24

Z = 11 − (−1) = 12

2412? is magnesium:

2411Na →

2412Mg + 0

−1e

120 10.0 pointsWhat is the least massive particle?1. proton


2. α particle

3. electron correct

4. neutron

Explanation:

121 10.0 pointsUranium-238 decays by emission of an alphaparticle. The product of this decay is

1. 23488Ra .

2. 23491Pa .

3. 23490Th . correct

4. 23492U .

Explanation:

122 10.0 pointsBalance the nuclear equation

3215P + ? −→

3315P

1. None of these

2. ? = 0−1β

3. ? = 3415P

4. ? = 42He

5. ? = 10n correct

6. ? = 115P

7. ? = 11p

Explanation:

mass number: 33 − 32 = 1atomic number: 15 − 15 = 0

3215P + 1

0n −→3315P

123 10.0 pointsFor the nuclear reaction

23994Pu →

23592U + ?,

the missing particle is

1. 1n.

2. e−.

3. 42He. correct

4. 1p.

Explanation:23994Pu →

23592U +4

2He

124 10.0 pointsThe half-life of carbon-14 is 5,730 years. Ifyou started with 100.0 g of carbon-14, howmuch would remain after 4 half-lives?

1. 12.5 g

2. 57.3 g

3. 25.0 g

4. 6.25 g correct

Explanation:

After 4 half lives the amount of carbon hasbeen halved 4 times, so

A =A0

2n=

100 g

24= 6.25 g

125 10.0 pointsIf you have 200.0 g of radioisotope with a half-life of 5 days, how much isotope would remainafter 15 days?

1. 40.0 g

2. 13.3 g

3. 12.5 g

4. 25.0 g correct

Explanation:


The half life is 5 days, so after 15 days,three half lives have elapsed. The amount ofisotope remaining is

A =A0

2n=

200.0 g

23= 25 g

126 10.0 pointsThe half-life of P-32 is 14 days. How long aftera sample is delivered can a laboratory wait touse a sample in an experiment if they need atleast 10 percent of the original radioactivity?

1. 42 days correct

2. 56 days

3. 14 days

4. 28 days

5. 70 days

Explanation:

After one half life 50% remains; aftertwo half lives 25% remains; after threehalf lives 12.5% remains. So, after threehalf-lives, over 10% of the 32P remains.3(14 days) = 42 days, so the laboratory canwait about 42 days but shouldn’t wait longerthan that if at least 10% of the original ra-dioactive substance 32P is needed.

127 10.0 pointsThe isotope 90Sr has a half-life of 28 years. Ifyou have 16 g of 90Sr today, how much will beleft in 112 years?

1. 1 g correct

2. 8 g

3. 2 g

4. 4 g

Explanation:112

18= 4, so 112 years is 4 half-lives.

16 g to 8 g is the first half-life;8 g to 4 g is the second half-life;

4 g to 2 g is the third half-life;2 g to 1 g is the fourth half-life.

1 g of 90Sr will be left after four half-lives(112 years).

128 10.0 pointsCalculate the energy change when one 235Unucleus undergoes the fission reaction

235U + n →142Ba + 92Kr + 2 n .

The masses needed are 235U : 235.04 u;142Ba : 141.92 u; 92Kr : 91.92 u; n : 1.0087 u.

1. −2.8 × 10−11 J correct

2. −1.7 × 10−10 J

3. +1.8 × 10−10 J

4. +2.9 × 10−11 J

5. −3.2 × 10−28 J

Explanation:

129 10.0 pointsWhat nuclear reaction probably does not oc-cur in a conventional fission nuclear reactordesigned to produce energy?

1. 235U + n →239Pu correct

2. 235U + n →103Mo + 131Sn + 2 n

3. 235U + n →90Sr + 143Xe + 3 n

4. 235U + n →141Ba + 92Kr + 3 n

Explanation:

If 235U captures a neutron, it forms an un-stable species that breaks into barium andkrypton. To make plutonium, 238U is neededas the starting material.

130 10.0 pointsIn the fission process

23992U →

14356Ba + 86

36Kr + X n,what is X?1. 5


2. 20

3. 15

4. 10 correct

5. 25

Explanation:23992U →

14356Ba + 86

36Kr + 10 10n

131 10.0 pointsIn nuclear fission, which is true?

1. One large atom splits into two or moresmaller atoms. correct

2. Two or more smaller atoms combine toform one larger atom.

3. Individual protons split into smallerpieces.

4. Electrons combine to form larger, nega-tively charged particles.

Explanation:

132 10.0 pointsThe nuclear equation

21H + 3

1H →42He + 1

0nis an example of

1. a branching chain.

2. nuclear fission.

3. nuclear fusion. correct

4. a chain reaction.

Explanation:

Two individual atoms combined to formone atom. This is nuclear fusion.

133 10.0 pointsWhich reaction best describes nuclear fusion?

1. 2 C2H6 + 3 O2 → 4 C + 6 H2O + energy

2. 10n + 235

92U →9038Sr + 143

54Xe + energy

3. 21H + 3

1H →42He + 1

0n + energy correct

4. 42He + 235

92 U →23894 Pu + energy

Explanation:

Fusion starts with multiple atoms whichform a single atom.

3. 4 - pbworkschemnotes.pbworks.com/f/solution_pdf.pdf · 3.weight 4.mass 5.density correct...

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