### Answers to Study Guide for CHEM 101 Test 4, Fall 2007

1. If 29.3 L of a gas at 25o C and 1.00 atm is heated to 273oC at 1.00 atm what is the new volume?

P is constant, use Charles' Law, V1/T1 = V2/T2
Rearrange to V2 = V1T2/T1
Calculate temperatures in Kelvin!!!!
T1 = 25+273 = 298 K; T2 = 273+273 = 546 K
V2 = V1T2/T1 = (29.3 L)(546 K)/(298 K) = 53.7 L

2. A gas occupies 10.92 L at 35o C and 766 torr. What will be its new volume at 2,000 torr and 50.0oC?

More than one thing is changing! Use the combined gas law. You must change temperatures to Kelvin T1 = 273+35 = 308 K; T2 = 273+50 = 323 K P1V1/T1 = P2V2/T2; rearrange to V2 = P1V1T2/T1P2
V2 = [(766 torr)(10.92 L)(323 K]/[(308 K)(2000 torr)] = 4.39 L

3. How many moles of SO2 are contained in a volume of 39.0 L at 27oC and 965 torr?

Use the ideal gas law, PV = nRT, rearrange to n = PV/RT
Convert P to atmospheres P = 965 torr/(760 torr/atm)=1.27atm
Convert T to K; 273+27=300 K
n = [(1.27 atm)(39.0 L)]/[(0.0821 L atm/{mol K})(300 K) =
n = 2.01 mol

4. Using Graham's Law predict how much faster dinitrogen, N2, should effuse through a pin hole in a tank of gas than sulfur dichloride, SCl2?

Graham's Law is v1/v2 = [m2/m1]1/2
m2 = 32.06 + 2(35.45) = 102.96 g/mol
m1 = 2(14.01) = 28.02 g/mol
v1/v2 = [102.96/28.02]1/2 = [3.675]1/2 = 1.917

5. Wet oxygen, 123 mL, is collected over water at a pressure of 751 torr at 27o C. What is the pressure of the oxygen?

 Temp., oC PH20, torr 20 17.5 21 18.7 22 19.8 23 21.1 24 22.4 25 23.8 26 25.2 27 26.7 28 28.3 29 30.0 30 31.8

Dalton's Law of partial pressure: PT = p1 + p2 + ....
PT = pwater + poxygen
poxygen = PT - pwater = 751 torr - 26.7 torr = 724 torr

6. The density of an inert gas at 800 torr and 30oC is 0.1694 g/L. What is the inert gas?

Assume 1 mole of gas and solve for a molar volume at these conditions using the Ideal Gas Law, PV =nRT, rearranged to V = nRT/P. Remember pressure must be in atmospheres, and temperature must be in Kelvin.
P = 800 torr/(760 torr/atm) = 1.05 atm, T = 30 + 273 = 303 K
V = [(1.00mol)(0.0821Latm/molK)(303K)]/1.05 atm = 23.68 L
(23.68 L/mol)(0.1694 g/L) = 4.01 g/mol, the gas is helium.

7. If 36.0 L of a gas at 25oC and 1.90 atm is compressed to 11.2 L at 25oC, what is the new pressure?

Use Boyle's Law, rearrange P1V1 = P2V2 to P2 = P1V1/V2

P2 = (1.90 atm)(36.0 L)/(11.2 L) = 6.11 atm

8. If 29.3 L of a gas at 25oC and 1.00 atm is heated to 193oC at 1.00 atm what is the new volume?

45.8 L, see No. 1 for method.

9. A gas occupies 2.92 L at 15oC and 706 torr. What will be its new volume at 9,000 torr and 30.0oC?

0.241 L, see No. 2 for method.

10. How many moles of CO2 are contained in a volume of 30.0 L at 72oC and 695 torr?

0.969 mol, see No.3 for method.

11. Wet oxygen, 1023 mL, is collected over water at a pressure of 731 torr at 24oC. What is the pressure of the oxygen?

709 torr, see No. 5 for method.

12. Using Graham's Law predict how much faster dioxygen, O2, should effuse through a pin hole in a tank of gas than dichlorine gas, Cl2.

1.489 times as fast, see No. 4 for method.

13. The density of a certain gaseous fluoride of phosphorus is 5.63 g/L at STP. Calculate the molecular weight of this fluoride. Knowing that there is only one phosphorus per molecule calculate the molecular formula.

At STP a molar volume is 22.4 L/mol.
(22.4 L/mol)(5.63 g/L) = 126 g/mol = FW
Since there is only one phosphorus that accounts for 31 g of the FW: 126 - 31 = 95 g
Dividing the remaining 95 g by the atomic weight of fluorine will tell us how many fluorines are in the compound:
95 g/19 = 5, thus the formula must be PF5.

14. Skip this problem

15. Define:

1. vapor pressure: pressure of the vapor phase of a matrerial above the liquid phase of the material. Vapor pressure increases withn increasing temperature.

2. standard temperature & pressure: 273.15 K (0.0000oC and 1.00 atm

3. normal or standard boiling point: the temperature at which the vapor pressure above a liquid is equal to 1.00 atm

4. viscosity: a measure of teh resistance to flow of a liquid. Can bethought of as relative thickness of liquid

5. surface tension: the attraction between molecules of a liquid at the surface of the liquid

6. hydrogen bonding: attraction between hydrogens bonded to a more electronrgative atomm (F, O, N, Cl) to the more electronegative element in another molecule of the substance

7. physical state change or phase change: change from solid to liquid, liqid to gas or solid to gas, or vice versa

8. unit cell: smallest repeat unit in a crystalline substance

9. van der Waals forces: small attractive/repulsive forces between atoms or molecules

10. dipole-dipole forces: attractive forces between polar molecules

11. vapor pressure: see a.

12. crystalline lattice: geometric arrangement of ions or moecules in a crystalline substance

16. State and explain:

1. Boyle's Law: P1V1 = P2V2 at constant T

2. Gay-Lussac's Law: P1/T1 = P2/T2 at constant V

3. Dalton's Law of Partial Pressures: PT = p1 + p2 + ..., the total pressure is equal to the sum of the partial pressures of the gases present

4. the Ideal Gas Law: PV = nRT

5. Raoult's Law: skip

6. Charles' Law: V1/T1 = V2/T2 at constant T

7. Combined Gas Law: P1V1/T1 = P2V2/T2

8. Avagadro's Law: equal volumes of gases the same temperature and pressure contain equal numbers of gas molecules

9. Graham's Law of Effusion: v1/v2 = (m2/m1)1/2

17. State the properties of each of the three common physical states of matter, solid; liquid; and gas.

gases: completely fill the container, are very highly compressible

liquids: take on shape of container, only slightly compressible

solids: have definite shape. only vedry, very, slightly compressible

18. What happens to the temeprature during a physical state (phase) change?

During a physical state change the temperature is constant.

19. What is the smallest repeat unit in a crystalline solid known as?

The smallest repeat unit in a crystalline solid is known as the "unit cell."

20. Find the nuclear binding energy per mole for lithium-7 (at. mass = 7.01600 g/mol); chlorine-35 (at. mass = 34.95952 g/mol); and bismuth-209 (at. mass = 208.9804 g/mol). Then find the nuclear binding energy per atom, and finally, the nuclear binding energy per nucleon.

total particle mass= 3(mass of a proton)+3(mass of electron)+4(mass of neutron)
total particle mass = 3(1.67252x10-24g)+3(9.1095x10-28g)+4(1.67495x10-24g)
total particle mass = 1.172 x 10-23 g/atom
total particle mass per mole =(1.172 x 10-23g/atom)(6.022 x 1023 atoms/mol) total particle mass per mole = 7.05784 g/mol
mass defect = 7.05784 g/mol - 7.01600 g/mol = 0.04184 g/mol
mass defect in kg/mol = 0.04184 g/mol x 1 kg/1000 g = 4.184 x 10-5 kg/mol

E = mc2, c = speed of light = 3.00 x 108 m/s
E/mol = (4.184 x 10-5 kg/mol)(3.0 x 108m/s)2 = 3.766 x 1012 J/mol
binding energy per atom = 6.254 x 10-12 J/atom (above divided by Avagadro's number, 6.022 x 1023)
binding energy per nucleon = 8.934 x 10-13 J/nucleon (nuclear binding energy per atom divided by 7, the total number of protons and neutrons)

All other problems are set up similarly

For Cl-35 binding energy per mole = 2.951 x 1013 J/mol
nuclear binding energy per atom = 4.900 x 10-11 J/atom
nuclear binding energy per nucleon = 1.400 x 10-12 J/nucleon

For Bi-209 binding energy per mole = 1.577 x 1014 J/mol
nuclear binding energy per atom = 2.619 x 10-10 J/atom
nuclear binding energy per nucleon = 1.25 x 10-12 J/nucleon

21. The half-life of 239Pu is 24,000 yr. What fraction of the plutonium waste present today will have decayed in 1000 yrs?

Number of half-lives = 1000/24,000 = 0.0416666
Amount left = [(0.5)0.0416666 ]100% = 97.15%
Amount decayed = 100.00% - 97.15 = 2.85%

22. A 0.500 g sample of strontium-90 diminishes to 0.393 g in 10.0 yrs. What is the half-life of strontium-90?

ln(C0/Ct) = kt
k = [ln(C0/Ct)]/t = [ln(0.500g/0.393g)]/10.0y = 0.024079848 1/y
t1/2 = ln 2/0.024079848 = 28.8 y

23. Balance the nuclear equations below by supplying the missing reactant or product.

1. 94Be + 42He ------------------> 10n + 126C

2. 24395Am + 42He --------------->24697Bk + 10n

3. 42He + 136C ---------------> 168O + 10n

4. 136C --------------------> 42He + 84Be + 10n

24. How many neutrons are in the nucleus of the isotopes below?

1. 105B, 5 n

2. 146C, 8 n

3. 7935Br, 44 n

4. 6529Cu, 36 n

5. 20983Bi, 126 n

6. 23592U, 143 n

7. 23390Th, 143 n

8. 4019K, 21 n

9. 10847Ag, 61 n

10. 9038Sr, 52 n

25. Provide the names from the formula below:
?
1. HfO2 - hafnium (IV) oxide

2. KIO4 - potassium periodate

3. SnBr2 - tin (II) bromide

4. Sb2O5 - antimony (V) oxide

5. PbCl2 - lead (II) chloride

6. ReCl2 - rhenium (II) chloride

7. RaI - radium iodide

8. HgCl2 - merculry (II) chloride

9. Hg2Cl2 - mercury (I) chloride

10. TlCl - thallium (I) chloride

26. Provide the fomula from the name for the compounds below:

1. platinum (IV) chloride - PtCl4

2. osmium (VIII) oxide - OsO4

3. tungsten (VI) chloride - WCl6

4. tin (II) fluoride - SnF2
5. lead (II) oxide - PbO

6. tellurium diiodide - TeI2

7. xenon tetrafluoride - XeF4

8. iodine pentachloride - ICl5

9. gold (III) fluoride - AuF3

10. bismuth (III) chloride - BiCl3

27. Define:
1. halflife - The time it takes for one half of an istope present to decay. Halflives are invariant for each radioactive isotope. Halflives may be calcylated using the equation t1/2 = ln2/k, where kis the rate decay constant from ln(C0/Ct) = kt .

2. fusion - The nuclear process in which two lighter nuclei are fused together to make a heavier nucleus. This is the process which fuels the sun, with the major reaction being

31H + 21H ---------> 42He + 10n

3. fission - The nuclear process in which a heavier isotope is split into two or more lighter nuclei. In chain reaction fission, a neutron splits a heavier nucleus which produce more neutrons. For chain reaction fission the number of neutrons averaged per disintegration must be approximately 2.5.

4. β (beta) particle - A particle ejected from the nucleus, which is essentially an electron.

5. α (alpha) particle - A particle ejected from the nucleus which is essentially the nucleus of a helium atom.

6. positron - A particle ejected from the nucleus that has the same mass as that of an electron, but a positive charge, 0+1e .

7. β (beta) decay - A common type of decay in a decay series, in which a β particle is ejected from the nucleus. The mass of the isotope decaying stays the same and vthe atomic number goes up by one. An example is shown below:

23490Th ---------> 23491Pa + 0-1e

8. α (alpha) decay - A common type of decay in which an α particle is ejected from the nucleus. The mass of the isotope decaying goes down by four and the atomic number goes down by two. An example is shown below:

23892U ----------> 23490Th + 442He

9. γ (gamma) ray - High energy ray, a form of light, often ejected from raidoative nuclei.

10. mass defect - the difference in mass between the sum of all the masses of all the particles in an atom and the actual mass of the atom.

11. nuclear binding energy - The energy necessary to hold the nucleus together, which is derived from the mass defect, and may be calculated using Einstein's equation, E = mc2 in which the mass, m, is in kg and the c, the speed of light is approximately 3.00 x 108 m/s. the energy is then in joules, J, the SI (metric) unit for energy.