Keys for Tests Given Fall 2003

Test 1 1. a.) No. Theories attempt to explain facts in the context of some problem. Facts are physical observables and theories are attempts at understanding the phenomena of the observable. For example, the freezing point of pure water is 0oC, and the freezing point of a definite concentration of salt in water is whatever it is (a meansurable quantity). A theory would be needed to try to understand why the freezing point of the salt water is less than that of pure water. We would want to do more measurements of different concentrations of salt in the water to give more freezing point data and then develop a theory based upon our understanding of the molecular stucture of water and the salt solutions.
b.) Facts are physical observables such as your temperature.
c.)^oF = ^oC + 32 = 72.7^oF
d.) ^oC = 5/9*( ^oF-32) = 123 ^oC
e.)T(K) = t ^oC + 273 = 396 K

2. a.) 223 lbs*454g/lb*1kgm/1000g = 101 kg
b.) V = m/d = 71.6g/7.86g/cm³ = 9.11 cm³. Now take the cube root to get length = 2.09 cm
c.) the ratio of the mass of solid or liquid to an equal volume of distilled water at 4^oC.
d.) q = 1.00cal/g^oC*352g*(95-25)^oC = 25.3 kcal
e.) 235 cal = .11cal/g^oC*121g*(T - 25)^oC = 13.31T -333
or 901 = 13.31T --> T = 67.7^oC

3.a.) 1:1:4 b.) H,N,O,F,Cl,Br,I c.) 58 d.) ¹⁷₃₇Ar e.) atoms with the same A but different Z

4. a.) 0.7899*23.99 + 0.10*24.00 + 0.1101*25.98 = 24.21
b.) F, Cl, Br, I, At
c.) K and V are metals and N and I are nonmetals
d.) Elements 1A, 2A, .... 8A except for transition elements

5. a.) N: 1s²2s²2p³
Mg: 1s²2s²2p⁶3s²
Ti: 1s²2s²2p⁶3s²3p⁶4s²3d² most of you missed this by not putting in the 4s² electrons.
P:1s²2s²2p⁶3s²3p³
b.) O with 6 electron "dots", N with 5 electron "dots", B with 3 electron "dots"
c.) IE is the energy required to remove the outermost electron from an atom.
d.) An ion is an atom that has had electrons either added or removed. The example for Mg would be Mg²⁺ because the IIA atoms always loose 2 electrons to form the ion.
e.) C O F

Test 2 1. a.) cation b.) ionic c.) 3s d.) nonpolar covalent e.) ionic (10 points)

2. a.) Al³⁺ b.) true c.) KCl d.) FeO e.) sodium nitrate (15 points)

3. a.) copper(II)oxide b.) iron(III)carbonate c.) aluminum slufate
d.) polar, nonpolar, polar e.) S, O (16 points)

4. a.) tetrahedral b.) linear c.) bent d.) pyramidal (12 points)

5. a.) H: 1, N: 3, O: 2, Br: 1, C: 4
b.) 8, 32, 18, 20
c.) hydrogen carbonate, nitrite, sulfate
d.) not stable, does not achieve Nobel gas structure (24 points)

6. a.) iron(II)hydroxide b.) aluminum oxide c.) magnesium sulfide d.) potassium iodide e.) lead(II)acetate (12 points)

7. a.) KBr b.) Li₂SO₄ c.) Fe₂(SO₄)₃ d.) CaO (12 points)

Test 3 1. a.) 16.2 b.) 9.76x10²⁴ c.) 4.67x10^-23

2. a.) 194.193 b.) 2.4x10^-4 c.) 1.4x10²⁰

3.
a.) 2C₄H₁₀(g) + 13O₂(g) --> 8CO₂(g) + 10H₂O(g)
b.)2B₄H₁₀(s) + 11O₂(g)--> 4B₂O₃(s) + 10H₂O(g)
c.) 2Fe + 3CO₂ --> Fe₂O₃ + 3CO
d.) KClO₃ ---> 2KCl + 3O₂

4. 0.240 g MgO

5. % yield = 33.1

6. a.) limiting reagent = SbF₃ b.) 101 g Freon

7. Ag⁺(aq) + Cl^-(aq) --> AgCl(s)

8. a.) Al is oxidized, Fe is reduced, Fe is oxidizing agent Al is reducing agent
b.) Cr is oxidized, Ni is reduced, Ni is oxidizing agent Cr is reducing agent
c.) exothermic d.) combustion

Test 4

1. a.) PV = nRT --> n = PV/(RT) = 3.97 atm*(6.92l)/(.0821latm/molK)*(1174.1K)
so n = 0.285 Below I don't write units, but you should!!!
b.)P = nRT/V = .921*.0821*684K/5.610 = 4.26 atm * 760 torr/atm = 3240 torr.
c.) V = nRT/P + .435*.0821*684/(6924 torr*(atm/760torr)) = 2.68 L = 2680 ml

2. a.) V = const/P; n,T constant b.) V = constant*T; n,p constant
c.) V = constant*n; P,T constant d.) PV = nRT

3. a.) P = n_totalRT/V
n_total = 1 g(1molH₂/2.0 g H₂) + 5 g He (1molHe/4.0gHe) = .5 mols H₂ + 1.25 mol He = 1.75 moles
P = 1.75moles*(.0821 latm/molK)*293K/5.0 l = 8.4 atm

b.) X_H₂ = .5/1.75 = 0.29; X_He = 0.71
so P_H₂ = .29 * 8.4 atm = 2.4 atm and P_He = 0.71*8.4 atm = 6.0 atm

4. 4.44 liters

5. a.) increase b.) should show oxgens dotted to hydrogens on different molecules of water.
c.) resultant force or surface of a liquid that stabilizes it. d.) increases

6. a.) greater. IF increases as size increases sinde no dipole moments
b.) vapor pressure = external pressure c.) vapor pressure = standard atmospheric pressure
d.) crystallization

7. a. Figure 5.18 in your book b.) homogeneous c.) decreases
d.) 26.3g*1mg/84.0g = 0.313 mol NaHCO₃
M = 0.313mol/.200L = 1.57 M

8. a.) M_NH₄⁺ = 1.0 M M_SO₄^2- = 0.50M
b.) 2.0 L sol*1.5molNa₂CO₃/1.0L = 3.0 mol Na₂CO₃
3.0 mol*106g/mol = 320 g or 3.2 x10² g d.) 48g/mol 61.790g/48g/mol = 1.29 mol .670 = 1.29/V --> V = 1930 ml

9.a.) strong weak b.) hydrogen bonds c.) size d.) lower

10. a.) 264g/126.05g/mol = 2.095 moles Na₂SO₃
Have 500 g water so: 2.095*2 = 4.19 moles/kgm
deltaT = 3particles*4.19moles/kgm* 1.860kgmK/mol = 23.38K = 23.38⁰C so new T = -23.38 ⁰C
b.) semipermeable c.) reverse d.) anions

Test 5

1. a.) 10 b.) true c.) effective d.) activation

2. a.) transition b.) lowering c.) rate d.) reversible

3. a.) temperature b.) right c.) left d.) increase

4. a.) 10⁴ b.) K_c = [NO]²[O₂]/[NO₂]² c.) K_c= [H₂][Cl₂]/[HCl}² = .21*.43/(8.7)²= 1.2x10^-3
d.)+50 kj sketch: see book

5. a.) k = rate/[H₂O₂] = -0.01/min b.) increase concentration, increase temperature, add catalyst
c.) No, speed depends upon the energy of activation, not the exo character.
d.) yes, endo reactions always have high energies of activation and so always must be slower

THIS WAS JUST LIKE AN OLD TEST. MAKE SURE THAT YOU REALLY UNDERSTAND THE ANSWERS AND DON'T DEPEND UPON OTHER TESTS BEING REPEATS OF OLD TESTS.