- If 29.3 L of a gas at 25
^{o}C and 1.00 atm is heated to 273^{o}C at 1.00 atm what is the new volume?

P is constant, use Charles' Law, V_{1}/T_{1}= V_{2}/T_{2}

Rearrange to V_{2}= V_{1}T_{2}/T_{1}

Calculate temperatures in Kelvin!!!!

T1 = 25+273 = 298 K; T2 = 273+273 = 546 K

V_{2}= V_{1}T_{2}/T_{1}= (29.3 L)(546 K)/(298 K) = 53.7 L

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

More than one thing is changing! Use the combined gas law. You must change temperatures to Kelvin T_{1}= 273+35 = 308 K; T_{2}= 273+50 = 323 K P_{1}V_{1}/T_{1}= P_{2}V_{2}/T_{2}; rearrange to V_{2}= P_{1}V_{1}T_{2}/T_{1}P_{2}

V_{2}= [(766 torr)(10.92 L)(323 K]/[(308 K)(2000 torr)] = 4.39 L

- How many moles of SO
_{2}are contained in a volume of 39.0 L at 27^{o}C 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

- Using Graham's Law predict how much faster dinitrogen, N
_{2}, should effuse through a pin hole in a tank of gas than sulfur dichloride, SCl_{2}?

Graham's Law is v_{1}/v_{2}= [m_{2}/m_{1}]^{1/2}

m_{2}= 32.06 + 2(35.45) = 102.96 g/mol

m_{1}= 2(14.01) = 28.02 g/mol

v_{1}/v_{2}= [102.96/28.02]^{1/2}= [3.675]^{1/2}= 1.917

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

Temp., ^{o}CP _{H20}, torr20 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: P_{T}= p_{1}+ p_{2}+ ....

P_{T}= p_{water}+ p_{oxygen}

p_{oxygen}= P_{T}- p_{water}= 751 torr - 26.7 torr = 724 torr

- The density of an inert gas at 800 torr and 30
^{o}C 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.

- If 36.0 L of a gas at 25
^{o}C and 1.90 atm is compressed to 11.2 L at 25^{o}C, what is the new pressure?

Use Boyle's Law, rearrange P_{1}V_{1}= P_{2}V_{2}to P_{2}= P_{1}V_{1}/V_{2}

P_{2}= (1.90 atm)(36.0 L)/(11.2 L) = 6.11 atm

- If 29.3 L of a gas at 25
^{o}C and 1.00 atm is heated to 193^{o}C at 1.00 atm what is the new volume?

45.8 L, see No. 1 for method.

- A gas occupies 2.92 L at 15
^{o}C and 706 torr. What will be its new volume at 9,000 torr and 30.0^{o}C?

0.241 L, see No. 2 for method.

- How many moles of CO
_{2}are contained in a volume of 30.0 L at 72^{o}C and 695 torr?

0.969 mol, see No.3 for method.

- Wet oxygen, 1023 mL, is collected over water at a pressure of 731 torr at 24
^{o}C. What is the pressure of the oxygen?

709 torr, see No. 5 for method.

- Using Graham's Law predict how much faster dioxygen, O
_{2}, should effuse through a pin hole in a tank of gas than dichlorine gas, Cl_{2}.

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

- 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 PF_{5}.

- Skip this problem

- Define:

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

- standard temperature & pressure: 273.15 K (0.0000
^{o}C and 1.00 atm

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

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

- surface tension: the attraction between molecules of a liquid at the surface of the liquid

- 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

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

- unit cell: smallest repeat unit in a crystalline substance

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

- dipole-dipole forces: attractive forces between polar molecules

- vapor pressure: see a.

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

- vapor pressure: pressure of the vapor phase of a matrerial above the liquid phase of the material. Vapor pressure increases withn increasing temperature.
- State and explain:

- Boyle's Law: P
_{1}V_{1}= P_{2}V_{2}at constant T

- Gay-Lussac's Law: P
_{1}/T_{1}= P_{2}/T_{2}at constant V

- Dalton's Law of Partial Pressures: P
_{T}= p_{1}+ p_{2}+ ..., the total pressure is equal to the sum of the partial pressures of the gases present

- the Ideal Gas Law: PV = nRT

- Raoult's Law: skip

- Charles' Law: V
_{1}/T_{1}= V_{2}/T_{2}at constant T

- Combined Gas Law: P
_{1}V_{1}/T_{1}= P_{2}V_{2}/T_{2}

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

- Graham's Law of Effusion: v
_{1}/v_{2}= (m_{2}/m_{1})^{1/2}

- Boyle's Law: P
- 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

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

During a physical state change the temperature is constant.

- 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."

- 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^{-24}g)+3(9.1095x10^{-28}g)+4(1.67495x10^{-24}g)

total particle mass = 1.172 x 10^{-23}g/atom

total particle mass per mole =(1.172 x 10^{-23}g/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 = mc^{2}, c = speed of light = 3.00 x 10^{8}m/s

E/mol = (4.184 x 10^{-5}kg/mol)(3.0 x 10^{8}m/s)^{2}= 3.766 x 10^{12}J/mol

binding energy per atom = 6.254 x 10^{-12}J/atom (above divided by Avagadro's number, 6.022 x 10^{23})

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 10^{13}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 10^{14}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

- The half-life of
^{239}Pu 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%

- 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(C_{0}/C_{t}) = kt

k = [ln(C_{0}/C_{t})]/t = [ln(0.500g/0.393g)]/10.0y = 0.024079848 1/y

t_{1/2}= ln 2/0.024079848 = 28.8 y

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

^{9}_{4}Be +^{4}_{2}He ------------------>^{1}_{0}n +^{12}_{6}C

^{243}_{95}Am +^{4}_{2}He --------------->^{246}_{97}Bk +^{1}_{0}n

^{4}_{2}He +^{13}_{6}C --------------->^{16}_{8}O +^{1}_{0}n

^{13}_{6}C -------------------->^{4}_{2}He +^{8}_{4}Be +^{1}_{0}n

- How many neutrons are in the nucleus of the isotopes below?

^{10}_{5}B, 5 n

^{14}_{6}C, 8 n

^{79}_{35}Br, 44 n

^{65}_{29}Cu, 36 n

^{209}_{83}Bi, 126 n

^{235}_{92}U, 143 n

^{233}_{90}Th, 143 n

^{40}_{19}K, 21 n

^{108}_{47}Ag, 61 n

^{90}_{38}Sr, 52 n

- Provide the names from the formula below:

?

- HfO
_{2}- hafnium (IV) oxide

- KIO
_{4}- potassium periodate

- SnBr
_{2}- tin (II) bromide

- Sb
_{2}O_{5}- antimony (V) oxide

- PbCl
_{2}- lead (II) chloride

- ReCl
_{2}- rhenium (II) chloride

- RaI - radium iodide

- HgCl
_{2}- merculry (II) chloride

- Hg
_{2}Cl_{2}- mercury (I) chloride

- TlCl - thallium (I) chloride

- HfO
- Provide the fomula from the name for the compounds below:

- platinum (IV) chloride - PtCl
_{4}

- osmium (VIII) oxide - OsO
_{4}

- tungsten (VI) chloride - WCl
_{6}

- tin (II) fluoride - SnF
_{2}

- lead (II) oxide - PbO

- tellurium diiodide - TeI
_{2}

- xenon tetrafluoride - XeF
_{4}

- iodine pentachloride - ICl
_{5}

- gold (III) fluoride - AuF
_{3}

- bismuth (III) chloride - BiCl
_{3}

- platinum (IV) chloride - PtCl
- Define:
- 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 t
_{1/2}= ln2/k, where kis the rate decay constant from ln(C_{0}/C_{t}) = kt .

- 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

^{3}_{1}H +^{2}_{1}H --------->^{4}_{2}He +^{1}_{0}n

- 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.

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

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

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

- 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:

^{234}_{90}Th --------->^{234}_{91}Pa +^{0}_{-1}e

- 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:

^{238}_{92}U ---------->^{234}_{90}Th +^{4}4_{2}He

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

- 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.

- 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 = mc
^{2}in which the mass, m, is in kg and the c, the speed of light is approximately 3.00 x 10^{8}m/s. the energy is then in joules, J, the SI (metric) unit for energy.

- 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 t