CONCEPTS and FORMULA 2 The total number of electrons at the n energy level is = 2π Ex: 2 at first energy level: 2(1) = 2 πππππ‘ππππ at fourth energy level: = 32 πππππ‘ππππ 2 at seventh energy level: 2(7) = 98 πππππ‘ππππ Colligative properties Beer’s Law A = Ο΅bC A - absorbance Ο΅ - molar absorptivity B - length of light path C- concentration Partial pressure - concentration relationship P = hX P - partial pressure h - Henry’s constant (shift 7 > 06) X - mole fraction Energy of photon -wavelength - wavenumber - frequency relationship E = hf c = λf w = 1/λ E - energy of photon in J h - Henry’s constant (shift 7 > 06) f - frequency in hz λ - wavelength in m 8 c - speed of light = 3π₯10 ππππππππ w = wavenumber Number of photon - wavelength - power relationship Nhc = tPλ N - number of photon h - Henry’s constant (shift 7 > 06) λ - wavelength in m 8 c - speed of light = 3π₯10 ππππππππ P - power in w t = time in s Velocity - mass - power relationship vmc = tP v - speed/velocity in m/s m - mass in kg 8 c - speed of light = 3π₯10 ππππππππ P - power in w t = time in s Charge - current - mass relationship Q = It = znF π π = π§ ππ πΉ Q - charge in C I - current in A t - time in s z = valency number/charge of ions n = mole F = Faraday’s constant (shift 7 > 22) Speed of approach formula π = π (1 − λπππ λ ) s - speed of approach λobs - wavelength of observed light (light that appears) λ- wavelength c = 3x108 MEAN VELOCITY MOST PROBABLE VELOCITY ROOT MEAN SQUARE V Entropy of Mixing βπππ₯π = −βπππ₯πΊ π Gibbs Energy of Mixing βπππ₯πΊ = ππ π(π₯π΄ππ π₯π΄ + π₯π΅ππ π₯π΅) or βπππ₯πΊ = ππ(π₯π΄ππ π₯π΄ + π₯π΅ππ π₯π΅) Where xA and xB are the compositions (fraction/percentage) P - pressure in Pa V - volume in m3 T - temperature in K R = 8.314 Difference in adjacent energy levels βπΈ = πΈπ£+1 − πΈπ£ = β( π 1/2 ) π Translational partition functions (ratio of translational partition functions of two elements) ππ ππ πππππππ‘ 1 3/2 ππ‘ππππ = ( ππ ππ πππππππ‘ 2 ) Atomic mass - valency relationship (FARADAY’S LAW) π= πππ π ππ‘ππππ πππ π Forward reaction - backward reaction CONSTANTS relationship ππ = ππ ππ ππ = equilibrium constant ππ = rate constant for FORWARD RXN ππ = rate constant for BACKWARD RXN PHYSICS FORMULAS F = ma ππ − ππ = 2 2 2 2 1 2 (π£π + π£π) π‘ π£π − π£π = 2π (ππ − ππ) π£π − π£π = 2πβ ππ − ππ - distance between 2 points π£π - initial velocity π£π - final velocity π - acceleration π£π = π£π + ππ‘ β = πππ‘ + 1 2 2 ππ‘ β - height π£π - initial velocity π - acc due to gravity = 9.81 m/s π‘ - time π‘ππ θ = ππ¦ ππ₯ π₯ = π£π₯π π‘ = π£π₯π πππ θ π‘ Time to reach maximum height π‘ = π£ππ ππθ π π¦ = π£π¦π π‘ = π£π¦π π ππθ π‘ Horizontal Distance Maximum height 2 π = π£π π ππ2θ π 2 βπππ₯ = 2 π£π π ππ θ 2π 2β π π = π£π × πΉπππ‘ = πΉ1 − πΉ2 πΊ π1 π2 πΉ = 2 π πΊ - gravity constant SHIFT 7 >39 r - distance between 2 points π1 π2 - mass 1 and mass 2 πΎπΈ = 1 2 2 ππ£ Escape Velocity π£ππ π = 2πΊπ π M - mass of earth = 5.972 x1024 R - radius of earth = 6.378 x106 G - gravity constant = shift 7>39 Weight in the moon vs Weight in the earth W = mg πππππ = πππππ‘β π₯ 1.622 9.81 Molar Volume - Volume & Molar Mass - Mass relationship V = n (xaVa + xbVb) V = naVa + nbVb M = n (xaMa + xbMb) M = naMa + nbMb Note: na + nb =1 ππ = ππ ππ ππ = ππ ππ VALENCY OF SULFUR ( 2 or 6) β In Hydrogen Sulfide (H2S): V =2 S is more electronegative than H β In Sulfuric Acid (H2SO4): V =6 O is more electronegative than S SAMPLE PROBLEMS Only two isotopes of copper are present in naturally occurring copper: 62Cu (62.9298 u) and 65Cu (64.9278 u). Calculate the percent composition of naturally occurring 62Cu using atomic mass of 63.546 u. Let x = composition of 62Cu (1-x) = composition of 65Cu 62. 9298π₯ + 64. 9278(1 − π₯) = 63. 546 Solving x: π₯ = 0. 6910 ππ 69. 10% 62Cu 1 − π₯ = 0. 3090 ππ 30. 90% 65Cu A special-purpose piece of laboratory glassware, called a pycnometer, is used to measure liquid densities. It has a volume of 25 cc, and a mass of 17.24 g when it is full of air. When filled with a liquid of unknown density, its mass = 45.0 g. What is the density of this liquid? π π ππππ ππ‘π¦ = = 45 − 17.24 π 25 ππΏ ππππ ππ‘π¦ = 1. 1104 π/ππΏ How many milliliters of water must be added to 5 mL of 12 M HCl solution to make a 3 M HCl solution? 12 πππ = 0. 06 πππ π»πΆπ 1000 ππΏ 0.06 πππ π»πΆπ 3 π = πΏ π πππ’π‘πππ 0.06 πππ π»πΆπ πΏ π πππ’π‘πππ = = 0. 02 3π 5 ππΏ π₯ πΏ = 20 ππΏ π πππ’π‘πππ 20 ππΏ π πππ’π‘πππ = ππΏ π πππ’π‘π (π»πΆπ) + ππΏ π πππ£πππ‘ (π»2π) 20 ππΏ π πππ’π‘πππ = 5 ππΏ π»πΆπ + ππΏ π πππ£πππ‘ (π»2π) ππΏ π πππ£πππ‘ (π»2π) = 15 ππΏ A current of 0.1000 ampere is passed through a copper sulfate solution for 10 minutes using platinum electrodes. Calculate the number of grams of copper deposited at the cathode from CuSO4 solution I = 0.1 A t = 10 min = 600s z = 2 (from CuSO4 2-) m =? π = πΌπ‘ = π§ππΉ = π§ πΌπ‘ = π§ π ππ π ππ πΉ πΉ (0. 1)(600π ) = 2 π 159.61 (96485) π = 0. 0496 π A photon-powered spacecraft of mass 10.0 kg emits radiation of wavelength 225 nm with a power of 1.50 kW entirely in the backward direction. To what speed will it have accelerated after 10.0 years if released into free space? π£ππ = π‘π π‘π π£ = ππ (10 π₯ 365 π₯ 24 π₯ 60 π₯ 60) π (1500 π) π£ = 8 (10 ππ)( 3 π₯ 10 ) π£ = 157. 68 π/π The emf of the cell Bi | Bi2S3(s) | Bi2S3(aq) | Bi is −0.96 V at 25°C. Calculate the solubility product of Bi2S3 (Bi3+)2(S2+)3 = (2s)2(3s)3 = 108s5 5 108π = − 0. 96 5 π = − 0.96 108 At what speed of approach would a red (660 nm) traffic light appear green (520 nm)? π = π (1 − λπππ λ ) 8 π = (3π₯10 ) (1 − 520 ππ 660 ππ ) 7 π = 6. 36 π₯10 π/π Consider a container of volume 250 cm3 that is divided into two compartments of equal size. In the left compartment there is argon at 100 kPa and 0°C; in the right compartment there is neon at the same temperature and pressure. Calculate the entropy when the partition is removed. Assume that the gases are perfect. −βπππ₯πΊ βπππ₯π = π βπππ₯πΊ = ππ(π₯π΄ππ π₯π΄ + π₯π΅ππ π₯π΅) ( since there are two compartments of equal size, xA = 1/2 and xB = ½ ) 3 3 βπππ₯πΊ = (100, 000 ππ) (250 ππ π₯ βπππ₯πΊ = − 17. 32867951 π½ βπππ₯π = −(−17.32867951) π½ 273.15 πΎ 1π 1 3 (100 ππ) )( 2 ππ 1 2 + 1 2 ππ 1 2 ) −2 βπππ₯π = 0. 0634 π½/πΎ = 6. 34 π₯ 10 π½/πΎ For a harmonic oscillator of effective mass 2.88 × 10−25 kg, the difference in adjacent energy levels is 3.17 zJ. Calculate the force constant of the oscillator. π 1/2 βπΈ = πΈπ£+1 − πΈπ£ = Δ§( π ) Δ§ >> π βπππ‘ 7 > 09 −21 3. 17π₯ 10 −34 = 1. 05 π₯10 ( 1/2 π −25 2.88 π₯ 10 ) π = 260. 23 π/π Calculate the ratio of the translational partition functions of xenon and helium at the same temperature and volume ππ ππ πππππππ‘ 1 3/2 ππ‘ππππ = ( ππ ππ πππππππ‘ 2 ) ππ‘ππππ = ( 131.29 3/2 ) 4 ππ‘ππππ = 188 Salt of A (atomic mass 15), B (atomic mass 27) and C (atomic mass 48) were electrolyzed using the same amount of charge. It was found that when 4.5 g of A was deposited, the mass of B and C deposited were 2.7 g and 9.6 g. The valencies of A, B and C were respectively: πππ π ππ‘ππππ πππ π π= ππ = 4.5 15 = 0. 3 ππ = 2.7 27 = 0. 1 ππ = 9.6 48 = 0. 2 Valencies: 3, 2, 1 In a chemical reaction, the rate constant for the backward reaction is 7.5×10–4 and the equilibrium constant is 1.5. The rate constant for the forward reaction is ππ ππ = ππ ππ 1. 5 = −4 7.5×10 −3 ππ = 1. 125 π₯ 10 The number of ions produced from one molecule of [Pt(NH3)5Br]Br3 in the aqueous solution will be: When dissociated, ions that will be formed are 3 Br 11 [Pt(NH3)5Br] 3+ and No. of ions = 1 +3 No of ions = 4 An electron in a cathode-ray tube accelerates uniformly from 2.00 × 104 m/s to 6.00× 106 m/s over 1.50 cm. In what time interval does the electron travel this 1.50 cm? 1 2 ππ − ππ = 0. 015 = 1 2 (ππ + ππ) π‘ 4 6 (2π₯10 + 6π₯10 ) π‘ −9 π‘ = 4. 9834 π₯10 π An electron of mass 9.11 × 10-31 kg has an initial speed of 3.00 × 105 m/s. It travels in a straight line, and its speed increases to 7.00 × 105 m/s in a distance of 5.00 cm. Assuming its acceleration is constant, determine the magnitude of the force exerted on the electron F = ma 2 2 π£π − π£π = 2π (ππ − ππ) 5 5 7π₯10 − 3π₯10 = 2π (0. 05) 12 2 π = 4π₯10 π/π 12 πΉ = (9. 11π₯10 − 31) (4π₯10 ) −18 πΉ = 3. 64 π₯ 10 π A 200-kg object and a 500-kg object are separated by 4.00 m. Find the net gravitational force exerted by these objects on a 50.0-kg object placed midway between them. Illustration: A (200 kg) —------------ 50 kg object —------------B (500 kg) 2m 2m πΉπππ‘ = πΉπ΄ − πΉπ΅ πΊ π1 π2 πΉ = πΉπππ‘ = πΊ(50)(500−200) πΉπππ‘ = = 2. 5028 π₯10 2 π 2 2 −7 π Suppose you want to launch a 1 000-kg spacecraft from the surface of the Earth at the escape speed, determine how much energy that would require. 1 2 πΎπΈ = 2 ππ£ π£ππ π = 2πΊπ π 24 2πΊ(5.972π₯10 ) π£ππ π = πΎπΈ = 6 6.378π₯10 1 2 = 11179. 82102 2 (1000)(11179. 82012) 10 πΎπΈ = 6. 2494 π₯ 10 π½ A man drops a rock into a well. The man hears the sound of the splash 2.40 s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water? 2.40 s = t1 + t2 Where t1 is the time the rocks reaches the water t2 is the time the sound bounced back and heard Speed = d/t = 336 m/s For t1: For t2: β = πππ‘ + 1 2 2 ππ‘ π ππππ = π π‘ = 336 1 2 β=0+ π‘2 = β 336 2β 9.81 π‘1 = 2β 9.81 2. 4 = 2 (9. 81)π‘ + β 336 β = 26. 4311 π A long jumper leaves the ground at an angle of 20.0° above the horizontal and at a speed of 11.0 m/s. How far does he jump? 2 π£π π ππ2θ π = π 2 π = (11) π ππ(2π₯20) (9.81) π = 7. 9284 π A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. a. How far away from the end of the bar does the mug hit the floor? π = π£π × π = 1. 5 2β π 2(1.2) 9.81 π = 0. 7419 π b. What is the speed and direction of the mug at impact? 2 2 π£π − π£π = 2πβ 2 2 π£π − (1. 5) = 2(9. 81)(1. 2) 2 π£π = 5. 0788 π/π ππ¦ Direction: π‘ππ θ = ππ₯ π‘ππ θ = ππ¦ ππ₯ = 2πβ ππ = 2(9.81)(1.2) 1.5 θ = 72. 8216 πππ The partial molar volumes of two liquids A and B in a mixture in which the mole fraction of A is 0.3713 are 188.2 cm3 mol−1 and 176.14 cm3 mol−1, respectively. The molar masses of A and B are 241.1 g mol−1 and 198.2 g mol−1. What is the volume of a solution of mass 1.000 kg? V = n (xaVa + xbVb) V = n [(0.3713(188.2) +(1-0.3713)(176.14)] M = n (xaMa + xbMb) 1000 = n [(0.3713(241.1) + (1-0.3713)(198.2)] n = 4.6701 sto A V = 4.6701 [(0.3713(188.2) +(1-0.3713)(176.14)] V = 843. 5 cm3 At 20°C, the density of a 20 per cent by mass ethanol–water solution is 968.7 kg m−3. Given that the partial molar volume of ethanol in the solution is 52.2 cm3 mol−1, calculate the partial molar volume of the water. 20% by mass ethanol-water solution = 20 g ethanol + 80 g water Let a = ethanol b = water ππ = 20 π πππ = ππ ππ‘βππππ = 46. 06 π ππ = 80 π πππ = ππ π€ππ‘ππ = 18. 02 π V = n (xaVa + xbVb) V = naVa + nbVb π = ρ= π ππ π π −3 968. 7 ππ π −3 = 0. 9687 π ππ −3 π = 103. 2311 ππ 103. 2311 = 20 46.06 −3 = 100 π (ππ π π’ππ) π π π‘π π΄ (52. 2) + 80 18.02 (ππ) ππ = 18. 15 ππ A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal with an initial speed of 20.0 m/s. The height from which the stone is thrown is 45.0 m above the ground. How long does it take the stone to reach the ground? β = πππ‘ + 1 2 2 ππ‘ ALWAYS REMEMBER TO USE NEGATIVE IN G β = πππ ππθ π‘ + 1 2 ππ‘ 2 − 45 = 20π ππ(30) π‘ + 1 2 2 (− 9. 81)π‘ π‘ = 4. 22 π A playground is on the ο¬at roof of a city school, 6.00 m above the street. The vertical wall of the building is h=7.00 m high, forming a 1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of 53.0° above the horizontal at a point d=24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. a) Find the speed at which the ball was launched. π₯ = π£π₯π π‘ 24 = π£π₯π πππ (53)(2. 2) π£π₯π = 18. 13 π/π b) Find the vertical distance by which the ball clears the wall. β = πππ‘ + β = πππ‘ + 1 2 1 2 2 ππ‘ 2 ππ‘ β = (18. 13 π ππ 53)(2. 2) + 1 2 (− 9. 81)(2. 2) 2 β = 8. 1142 π β = 8. 1142 π − 7 π β = 1. 1142 π c) Find the horizontal distance from the wall to the point on the roof where the ball lands. —-----------------------bnkgkmvkmfkfngng KEY CONCEPTS TO READ Avogadro's principle states that: At the same temperature and pressure, the volumes occupied by the same number of molecules are the same At what temperature does an ideal gas have zero kinetic energy? Ans. at 0 K or -273 deg C A balloon filled with methane gas is pricked with a sharp point and quickly plunged into a tank of hydrogen at the same pressure. After sometime, the balloon will be ENLARGED Reason: the rate of effusion of H2 is greater than the rate of effusion of CH4 Which will take less fuel for a spacecraft to travel, from Earth to Moon or from Moon to Earth? The trip from Moon to Earth due to lower escape velocity If there are no forces acting on an object then it can either be moving in constant velocity or at rest as stated in the first law of motion. Switching off gravity would let the atmosphere evaporate away, but switching off the atmosphere has no effect on the planet’s gravitational field. Suppose you are talking by interplanetary telephone to a friend who lives on the Moon. He tells you that he has just won a newton of gold in a contest. Excitedly, you tell him that you entered the Earth version of the same contest and also won a newton of gold! Who is richer? Ans. Your friend, because if you bring 1 N from moon, its actual weight in the earth is heavier than 1 N Coprecipitates (inclusions, occlusions, and surface adsorbates) are a common problem in gravimetric analysis, but can be controlled by: β β β β Carefully controlling the solution conditions Reprecipitation of the solid Digestion of the precipitate Thoroughly washing and drying the filtrate Volatilization gravimetry would be most useful in determining The amount of water in epsom salts The equivalence point of an acid-base titration is the point where there is an equivalent amount of titrant and titrand. The stoichiometry of EDTA with metal ions is always 1:1, because EDTA forms a cage-like structure around the metal ion. The advantage of using a silver chloride sample cell rather than a sodium chloride sample cell for IR spectroscopy is Aqueous samples can be measured as AgCl is not water soluble. Fluorescence occurs as a result of - relaxation from a singlet excited state to a singlet ground state Phosphorescence occurs as a result of - from a triplet excited state to the singlet ground state Sunflowers were used at Chernobyl to help reduce radioactive substances from groundwater. What method of reducing pollutants is this? Phytoremediation What accounts for the small size of the cells The rate of diffusion The nitrogen cycle includes the following three compounds Nitrates, ammonia, and urea Plants are useful in managing pollution because they Absorb, concentrate, and store pollutants Heterogeneity can also arise if DNA is damaged before amplification. Which of the following does not cause DNA damage? Amination of bases The reason that fats contain more energy than simple sugars, is fats have many more HYDROGEN ATOMS FROM PREVIOUS YEARS ORGCHEM 1. Upon hydrogenation, which of the following alkenes releases the least heat per mole? (E)-3,4-dimethyl-3-hexene 2. This best describes the most stable conformation of trans-1, 3-dimethylcyclohexane One methyl group is axial, the other equatorial. 3. Carbon–carbon double bonds do not undergo rotation as do carbon–carbon single bonds. The reason is that Overlap of the p orbitals of the carbon–carbon pi bond would be lost 4. Identical properties for constitutional isomers molecular formula and molecular weight 5. Fraction of petroleum that distils at the highest temperature fuel oil 6. molecule that has the most negative heat of combustion in kcal/mol Butane 7. A statement that correctly describes the general reactivity of alkynes An alkyne is an electron-rich molecule and therefore reacts as a nucleophile 8. A transition state β Possesses a definite geometry β Maximum on the potential energy diagram β Cannot be isolated 9. Statements that apply to an SN1 reaction β The rate limiting step of the reaction involves only the alkyl halide β There is an intermediate carbocation 10. correct ranking in decreasing order of relative Boiling Point of carbonyl containing compounds primary amide > carboxylic acid >> ester ~acyl chloride ~ aldehyde ~ ketone 11. relative stabilities of the eclipsed and staggered forms of ethane The ECLIPSED form has the most TORSIONAL strain. 12. reactions of aldehydes and ketones nucleophilic addition; free radical addition 13. factors that affect the rate of SN1 reactions β the nature of the alkyl halide β the nature of the leaving group β the concentration of the alkyl halide β the value of the rate constant 14. When petroleum is distilled, the fraction that contains compounds with 5 to 11 carbons is known as Gasoline 15. This specie can best serve as a radical initiator for radical polymerization ROOR 16. major product from an elimination reaction starting with 2-bromopentane trans-2-pentene 17. account for the remarkable catalytic ability of enzymes proper positioning and specificity of active site and substrate 18. Names for vitamin B β thiamine (B1) β riboflavin (B2) β niacin (B3) β pantothenic acid (B5) β pyridoxine (B6) β biotin (B7) β folate or ‘folic acid’ when included in supplements (B9) β cyanocobalamin (B12). 19. The eclipsed and staggered forms of ethane are said to differ in: Conformation 20. What is the major 2-methyl-2-pentene? 2-methyl-2-pentanol product from the acid-catalyzed hydration of 21. A mixture of equal amounts of two enantiomers __________ β is called a racemic mixture β is optically inactive 22. In solution, glucose exists as - an equilibrium mixture of the open-chain form and cyclic hemiacetal forms. 23. An aqueous solution of glucose behaves as an aldehyde because: - its cyclic hemiacetal, the predominant form, is in equilibrium with the free aldehyde form. 24. Its cyclic hemiacetal, the predominant form, is in equilibrium with the free aldehyde form. Bent 25. The following are used as primary standard for sodium hydroxide β KHP β Benzoic acid β Potassium bromide 26. a(alpha)-bromination of carboxylic acid by a mixture of Br2 and PBr3 is called Hell-Volhard Zelinskii (HVZ) reaction 27. Conjugated alkenes are characterized by Alternating pi and sigma bonds 28. physical property that differ for each of a pair of enantiomers direction of rotation of plane-polarized light 29. What is the basis for choosing the right indicator for a given acid-base titration pH at the equivalence point 30. Which of the following could best lessen cost but still get accurate results during standardization process: get aliquot portion of the solution containing the sample 31. It is preferable to have a solid rather than a as the final product in a gravimetric analysis. crystalline, colloidal solid precipitate, 32. The pH value at equivalence point when titrating a weak acid against a strong base is Greater than 7 33. Ethylenediaminetetraacetic acid (EDTA) is a _________ ligand. Hexadentate 34.
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