Chem 1151Exam 1100 pointsSummer 2005

Name______KEY______

1 in = 2.54 cmN = 6.022E+23

I.Multiple Choice (16 questions @ 3 points = 48 points)

1.The answer to (3.478 x 1.164) + 0.286 = 4.04839/2.5 + 0.286

2.5

= 1.6193 + 0.286 = 1.9053 = 1.9

A.1.9054

B.1.905

C.1.91

**D.1.9

E.2.

2.Which of the following statements describes a chemical property of phosphorus?

A.Red phosphorus and white phosphorus are solids at room temperature.

**B. When exposed to air, white phosphorus burns spontaneously but the red form does not.

C.The white form is soluble in liquid CS2 but is insoluble in water.

D.Red phosphorus melts at 873 K and the white form melts at 317 K.

E.Red phosphorus is insoluble in water and CS2.

3.The volume of a milk carton is about 200. in3. What is the volume in m3?

A.5.08E + 5 m3 200. in3 x (2.54 cm/in)3 x (1 m/100 cm)3 = 3.28E - 3 m3

B.5.08 m3

C.0.787 m3

**D.3.28E - 3 m3

E.2.00E - 4 m3

4.What is the total concentration of ions in a 0.420 M solution of K2CO3?

**A.1.26 MK2CO3 2K+ + CO32- ; one mol cmp  3 mol ions

B.0.840 M3 x 0.420 M = 1.26 M

C.0.420 M

D.0.210 M

E.0.140 M

5.How many protons, neutrons and electrons are present in 119Sn2+?

Sn has 50 protons. The 2+ ion has 48 electrons. 119-50 = 69 neutrons.

# protons / # neutrons / #electrons
A / 119 / 50 / 119
B / 50 / 69 / 52
**C / 50 / 69 / 48
D / 69 / 50 / 69
E / 50 / 119 / 48

6.All of the following are bases except

A.NaOH

B.NH3

C.Mg(OH)2

D.KOH

**E.CH3CHOsee defn of base

7.Element X (Z = 114) is expected to be a

**A.metal and behave like PbElement 114 lies below Pb

B.metal and behave like Pt

C.nonmetal and behave like Hg

D.non metal and behave like Pb

E.be a metal and behave like Hg

8.The formulas for the hydroxide ion, nitrite ion and phosphate ion are

A.OH-, NO3-, PO43-

B.H-, NO3-, PO43-

**C.OH-, NO2-, PO43-See Table 2.3

D.O-, NO2-, PO33-

E.OH-, NO2-, P3-

9.0.5 mole of NH4Cl has

A.3.011E+23 H atoms

B.6.022E+23 H atoms

C.9.033E+23 H atoms

**D.1.204E+24 H atoms= 0.5 mol cmp (4 mol H/1 mol cmp) N

E.1.084E+25 H atoms

10.All of these compounds have ionic bonds except

A.MnO2

B.KIO4

C.KI

D.Li3N

**E.PCl3P and Cl are nonmetals

11.When 5.0 mol copper reacts with 13. mol nitric acid, the number of moles of water produced is

3Cu(s) + 8HNO3(aq)  3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

5.0 mol Cu x (8 mol acid/3 mol Cu) = 13.33 mol acid needed. But only 13.0 mol acid available, so nitric acid = LR.

13.0 mol acid x (4 mol water/8 mol acid) = 6.5 mol water

A.13. mol

B.6.7 mol

**C.6.5 mol

D.5.0 mol

E.4.0 mol

12.2.25 mol of NaBr is dissolved in water to make a 350. mL solution. The molarity of the solution is

A.6.25E-2 M

B.2.25 M

**C.6.43 M= # mol solute/L soln = (2.25 g/0.350 L)

D.6.43E-3 M

E.0.789 M

13.The empirical formula of a compound is BH3. If its molar mass is 55.2 amu, then the molecular formula is

BH3 molar mass = 13.8 g/mol. 55.2/13.8 = 4 so there are 4 units of BH3 in the molecular formula

A.BH3

B.B2H6

C.B3H3

**D.B4H12

E.B5H15

14.Molarity is defined as

A.#mol solute/liter solvent

**B.# mol solute/liter solutionSee defn of molarity

C.# mol solvent/liter solution

D.Total # mol/liter solvent

E.# mol solute/# mol solvent

15.Which statements are true when a solution is diluted?

1.solvent is addedTRUE

2.the number of moles of solute stays the sameTRUE

3.the number of moles of solvent decreases

4.the volume of the solution increasesTRUE

5.the molarity increases

A.1 and 3

B.2 and 4

C.1, 2 and 5

**D.1, 2 and 4

E.3, 4 and 5

16.The molar mass of the molecule C2H8N2 and its empirical formula are

A.30.0 amu and CH4N

**B.60.0 amu and CH4NC2H8N2 molar mass is 60.0 g/mol

C.30.0 amu and C2H8N2Reduce subscripts by dividing by 2

D.60.0 amu and C2H8N2

E.none of these

II.Problems

1.(3)Balance this chemical equation

Al4C3+12H2O 4Al(OH)3 +3CH4

2.(5)There are only two naturally occurring isotopes of chlorine.

They are Cl-35 with a mass of 34.968 amu and Cl-37 with a mass of 36.956 amu. Calculate the abundances of these isotopes.

Let x = fractional abundance of Cl-35 and y = (1-x) = frac. abun. of Cl-37

Atomic mass of Cl (from Periodic Table) = 35.453 g/mol.

34.968x + 36.956 (1-x) = 35.453

This rearranges to give 1.998x = 1.503.

Then solve for x = 0.756 for Cl-35 and y = 0.244 for Cl-37

3.(5)Samples A and B contain carbon and hydrogen

Sample / Mass carbon (g) / Mass hydrogen (g) / Mass C/mass H
A / 5.70 g / 1.90 g / 3
B / 4.47 g / 0.993 g / 4.5

Are Samples A and B the same compound? What Law justifies your answer?

Now use these data to illustrate the Law of Multiple Proportions.

Since the mass ratios of Mass C/Mass H are not equal, the Law of Def. Proportions says that Samples A and B are different.

The Law of Multiple Proportions says that 3/ 4.5 should be a ratio of small whole numbers. 3/ 4.5 = 6/9 or 2/3.

4.(5)When a bright orange crystal was analyzed by mass, it was found to contain 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula for this compound?

Na17.5g/23g/mol = 0.76086; divide by 0.76086 to get 1 mol Na

Cr39.7g/52 g/mol = 0.76346; divide by 0.76086 to get 1 mol Cr

O42.8g/16 g/mol = 2.675; divide by 0.76086 to get 3.5 mol O

The mol ratios are 2: 2: 7 or Na2Cr2O7

5.(14)2Ca3(PO4)2(s) + SiO2(s) + 10C(s)  P4(g) + 6CaSiO3(l) +10CO(g)

a.(7)Name these reactants and products and tell what kind of bonding each one contains.

Name / Ionic of Covalent bonding?
Ca3(PO4)2 / Calcium phosphate / I
C / carbon / NA
P4 / Phosphorus / C
CO / Carbon monoxide / C

b.(5) How many grams of SiO2 (molar mass = 60.1 g/mol) are needed to produce 350. g of CaSiO3 (molar mass = 116.2 g/mol)?

(350. g/116.2g/mol) x (1 mol SiO2 /6 mol of CaSiO3) (60.1 g/mol) = 30.2 g SiO2

c.(2)If only 25.4 g of CaSiO3 are produced, what is the reaction yield?

% yield = 100 x (actual mass/theoretical mass) = 100 x 25.4g/350 g = 7.26%

Name______

I.______(48)

II.______(32)

Total______(80) or ______(100)

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