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87 Cards in this Set
- Front
- Back
- 3rd side (hint)
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Which of the following is not a strong acid? |
A: Strong acids are acids that are completely ionized in water to produce good electrical conductors. Memorize these common strong acids...HCl, HBr, HI, HNO3, H2SO4, HClO4
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The principle quantum number of the atom is symbolized
A. ms B. ml C. n D. l E. s |
C: The principle quantum number is the value "n" and can only be a positive whole number integer. |
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Kinetic Energy is proportional to which of the following?
A. Pressure B. Friction C. Molecular mass D. Density E. Volume |
C: Moving objects that have the capacity to do work possess kinetic energy. Therfore, Ek=1/2 mv2. m = molecular mass and v = velocity. Therefore, the answer to this question is molecular mass. |
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Arrange the following in decreasing atomic radius: O, S, K, Ca, Mg
A. K, Ca, Mg, S, O B. K, Mg, Ca, O, S C. Ca, K, Mg, S, O D. O, S, Ca, Mg, K E. S, O, Mg, Ca, K |
A: When determining Atomic Radius, a simple rule may be followed. Atomic Radius will increase as you move from right to left on the periodic table. It will also increase as you move from top to bottom. So the periodic table will give us the answer. |
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Assume that 2 cubic feet of air is heated from 0 degrees C to 300,000 degrees C. What volume does the air expand?
A. (300,000 X 2)/(273) B. (300,000 X 2)/(0) C. (273 X 2)/(300,273) D. (273 X 2)/(300,000) E. (300,273 X 2)/(273) |
E: Charles' Law is used therefore (V1/T1)=V2/T2. You need to solve for V2
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The density of alcohol is 0.8 g/ml. What volume of alcohol would weigh 65.0g?
A. (65.0)(0.8) B. (65.0)/(800) C. (0.8)/(65.0) D. (0.8)(65.0) E. (65.0)/(0.8) |
E: Density = (weight/Volume) Make sure your units are in g/ml.
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How do you determine the Lewis structure of nitrogen trifluoride?
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Determine the number of valence electrons in one nitrogen atom (group 5A on the periodic table) and three atoms of Flouride (group 7A). 1 Nitrogen has a total of 5 and 3 Flourides has a total of 21. We have a grand total of 26 valence elcetrons. Write a sketletal structure and connect the bonded atoms by dashes, representing covalent bonds. Place pairs of electrons as lone pairs around the terminal atoms until they have an octect (eight electrons). Assign any extra electrons as lone pairs around the central atom.
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Which element is the least electronegative?
A. F B. P C. Ca D. Rb E. Ni |
D: When determining electronegativity, a simple rule may be followed. Electronegativity will increase as you move from left to right on the periodic table. It will also increase as you move from bottom to the top. As F is in the upper right of the table, it is the most electronegative.
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Convert 93.4 degrees F to degrees C.
A. (9/5)(93.4 + 32) B. (5/9)(93.4 - 32) C. (9/5)(93.4 - 32) D. (5/9)(93.4 + 32) E. (9/5)(93.4 + 273.15) |
B: Degrees C = (5/9) (degrees F - 32)
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If a solution boils at 3 degrees K, what is the boiling point on the Fahrenheit scale?
A. (9/5)(-270) + 32 B. (5/9)(-270) + 32 C. (5/9)(3 X 32) D. (9/5)(270) + 32 E. (9/5)(270) - 32 |
A: The temperature in Kelvin = Celsius + 273. Solve for Celsius by subtracting 273 from Kelvin. This gives you -270 degrees C. Now convert Celsius to Fahrenheit by using this equation: F = (9/5)(degrees C) + 32
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When 7.0 g of Si is exploded with 32.0 g of O2 forming SiO2, how many grams of O2 remain uncombined? The reaction equation is Si + O2 -> SiO2
A. 0.75 X 32 B. 0.25 X 32 C. 1.5 X 32 D. 0.5 X 32 E. 1.25 X 32 |
A: You can see from the equation that Si and O2 react in a 1:1 ratio. Therefore, 1 mole of Si will react with 1 mole of O2. To find out the amount of unreacted O2, calculate the number of moles of Si and O2 present and then subtract the number of moles of Si from the number of moles of O2. Finally, take this mole amount of O2 and multiply it by the molecular weight of O2 to give you the number of grams of unreacted O2.
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The value of Kc = 8.0 in the equation 2SO 2(g) + O2(g) ---> 2SO3(g). What is the [SO3] at equilibrium if the [O2] is 2.0 M and the [SO2] is 3.0 M?
A. 8.0 M B. 12.0 M C. 4.0 M D. 14.0 M E. 2.0 M |
B: The equation used is Kc = [SO3]^2/ ([SO2]^2 x [O2]. Plug in the numbers given in the question and you get: 8.0 = x^2/ [(3.0 M)^2 x 2.0 M.) Solving for this equation the answer is 12.0 M
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Which of the following is NOT true of a catalyst?
A. Catalysts will decrease the activation energy of the reaction B. Catalysts will be influenced by the pH of the substance C. Catalysts do not change the thermodynamics of the reaction D. Catalysts cause a change in the equilibrium constant of the reaction E. Catalysts are never consumed during the reaction |
D: A catalyst will increase the rate of a reaction without being consumed by that reaction. It will lower the activation energy of the reaction. However, the equilibrium constant will always remain the same during the reaction. |
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What is the pressure in a 60 liter container that contains 15 moles of helium gas at a temperature of 25 degrees C?
A. (15 X 0.082 X 298)/(60) B. (15 X 0.082 X 25)/(60) C. (60)/(15 X 0.82 X 298) D. (298)/(15 X 0.082 X 60) E. (25 X 60)/(15 X 0.082) |
A: The ideal gas law states that PV=nRT. You must convert T to Kelvin by adding 273 to the temp which give you 298 K. Now solve for P.
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Which of the following scenarios will always produce a spontaneous reaction?
A. Positive ΔH and a positive ΔS B. Positive ΔH and a negative ΔS C. Negative ΔH and a positive ΔS D. Negative ΔH and a negative ΔS E. None of the above |
C: Two questions you need to ask yourself when determining if a reaction is spontaneous or not. A reaction will be spontaneous if: 1) it is an exothermic reaction (Δ H is negative) and 2) the system becomes more disordered. This is an increase in entropy (delta S is positive).
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Which of the following elements are isoelectronic?
A. N and P B. Mg+ and Na C. K- and Ca+ D. H+ and He E. F- and O+ |
B: Isoelectronic is when two species have the same number of electrons. When Mg has a positive charge it is lacking an electron, which makes Na and Mg+ isoelectronic.
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What is the correct formal charge for SO2?
A. S, +1 O, +1 B. S, +1 O, -1 C. S, +2 O, +1 D. S, -1 O, -1 E. S, -1 O, 0 |
I think the answer to this is wrong- S should be = 0 and O should be = 0. May depend on how the lewis structure is drawn.
B: When figuring out formal charges, you must first look at each atom individually. Find out which group each atom is in and that will be the base number you compare to. Count the total number of bonds and elctrons surrounding the atom. subtract this number from the group number. The difference will be the charge on the atom. It may be positive or negative. Example. Sulfur is in group 6. In the molecule SO2 it has 2 free electrons and 3 bonds, for a total of 5. 6-5 = +1. |
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What is the molality of a solution prepared by dissolving 0.2 mol of ethanol into 500 g of water?
A. 0.2 mol/kg B. 0.2 mol/g C. 0.4 mol/kg D. 0.4 mol/g E. 0.02 mol/kg |
C: Molality is moles of solute divided by kg of solvent. Decide which is the solute and which is the solvent. Do the necessary calculations to get the correct units for each and then solve for the answer.
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After heating NaNO3, it decomposes into NaNO2 and O2. How much NaNO3 would you have to decompose to produce 1.5 g of O2?
A. [(32/2) x 1.5](85) B. [(1.5/32) x 2](85) C. [(85/1.5) x 2](32) D. [(85/32) x 1.5](2) E. [(2/1.5) x 32](85) |
B: The first thing to do in these types of problems is to write out the equation and balance it. Therefore, the equation is 2NaNO3 -> 2NaNO2 + O2. This shows that for every mole of O2 produced, 2 moles of NaNO3 must be decomposed. You are given that 1.5 g of O2 is formed, so now you determine the number of moles by dividing by the molecular weight of 32 g/mol. Now the number of moles of NaNO3 can be found by multiplying the number of moles of O2 by 2. Finally, to find the weight of NaNO3 needed, you multiply the number of moles of NaNO3 by the molecular weight of NaNO3 (which is 85).
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What is the mole fraction of NaCl in a solution of made by dissolving 58.5 g of NaCl in 9.0 g of H2O? Molecular weight of NaCl = 58.5 g. Molecular weight of H2O = 18.0 g.
A. 1.5 B. 0.97 C. 0.33 D. 0.67 E. 0.49 |
D: Mole fraction problems are similar to % Compostion problems. A mole fraction of a compound tells what fraction of 1 mole of solution is due to that particular compound. So, mole fraction of solute=(moles of solute)/(moles of solute + moles of solution). You must first find the moles of the compound and solution...moles of NaCl = 58.5/58.5. This gives you 1 mole. You must now do the same for water, which gives you 9g/18g = .5 mole. So the mole fraction of NaCl = (1)/(1 + .5) => 0.67.
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What are the oxidation numbers of CaH2?
A. Ca +2, H2 -1 B. Ca +1, H2 -1 C. Ca +2, H2 -2 D. Ca -2, H2 +2 E. Ca -1, H2 +1 |
A: Rules of oxidation. 1 For atoms, molecules and neutral formulas, the total oxidation number is 0. Oxidation numbers are equal to the charge of the ion. 2 Group IA metals have an oxidation number of +1 and Group IIA metals have +2. 3 In compounds, the oxidation number of flourine is -1. 4 In compounds, the oxidation number of hydrogen is +1. 5 In most of its compounds, oxygen has an oxidation number of -2. 6 In two-element compounds with metals, Group 7A elements have an oxidation number of -2, and Group 5A elements have an oxidation number of -3.
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An atom with an atomic number of 12 and a mass number of 24 consists of
A. 12 protons, 12 electrons, and 12 neutrons B. 12 protons, 12 electrons, and 24 neutrons C. 24 protons, 24 electrons, and 12 neutrons D. 24 protons, 12 electrons, and 12 neutrons E. 12 protons, 24 electrons, and 12 neutrons |
A: Atomic number is the number of protons in the nucleus. Mass number is the sum of protons and neutrons.
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Consider the following equation which is NOT balanced: WO3(s) + H2(g) -> W(s) + H2O(l) How many moles of H2O can be produced from 1.00 mole of WO3?
A. 1.00 B. 2.00 C. 3.00 D. 4.00 E. 5.00 |
C: The first thing that needs to be done is to balance the equation. Do this by counting up the number of each element on each side. Find a common denominator to make sure there are the same number of elements on each side.
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Mg3 N2 + 6H2 O --> 3Mg(OH) 2 + 2NH3 How many moles of NH3 are liberated when 2 moles of Mg3 N2 are reacted with 9 moles of H2 O?
A. 1 B. 2 C. 3 D. 4 E. 5 |
C: The ratio is 1:6 = Mg2 N2:H2 O. So 2 moles of Mg3 N2 would require 12 moles of H2 O to fully react. Water is therefore the limiting reactant. Since two moles of NH3 are produced for every 6 moles of H2 O, 9 Moles of H2 O should yield 3 moles of NH3.
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Consider the following equation which is NOT balanced: KO2(s) + H2O(l) -> KOH(s) + O2(g) How many grams of KO2 must be provided to react completely with 180.0 g of H2O?
A. 1422 B. 374.9 C. 2755 D. 3289 E. 2899 |
A: The first thing that needs to be done is to balance the equation. Do this by counting up the number of each element on each side. Find a common denominator to make sure there are the same number of elements on each side. From here, you can solve many problems just by conversion.
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180 * 2 = 360
360 * 4 = 1440 |
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A solution is prepared by dissolving 10.0 g of NaOH in sufficient H2O to produce 250 mL of solution. Calculate the molarity of this solution.
A. 0.250 M B. 0.500 M C. 0.750 M D. 1.00 M E. 1.25 M |
D: Molarity is the number of moles of a compound in 1 liter of solution. Figure out which is the solution and which is the compound and do the necessary calculations to get mol/L.
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Diethyl ether has the formula (C2H5)2O. Calculate the molar mass of this ether
A. 53.4 g/mol B. 69.7 g/mol C. 45.1 g/mol D. 74.1 g/mol E. 83.2 g/mol |
D: Molar mass is the mass of 1 mole of that substance. So for diethyl ether, there is a total of 4 carbons (12.01 g/mol), 10 hydrogens (1.00 g/mol), and 1 oxygen (15.99 g/mol). Add the atomic mass of each molecule of these three elements to get 74.1.
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A solution is prepared by dissolving 25.0 g of ZnSO4 in 100.0g. of H2O. This solution has a density of 1.124 g/mL. Calculate the molarity of this solution.
A. 1.39 M B. 1.82 M C. 1.04 M D. 1.72 M E. 1.61 M |
A: Molarity is the number of moles of a compound in 1 liter of solution. Figure out which is the solution and which is the compound and do the necessary calculations to get mol/L. |
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V=m/D 111L=125g / 1.124g/ml grams cancel 2nd 25g / 161g/mol = .155moles ZnSO4 3rd .155moles / .1112L = 1.394M |
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Which of the following is a strong base?
A. NH3 B. CH3NH2 C. NH2OH D. KOH E. None of the above |
D: Memorize these strong bases...LiOH, NaOH, KOH, RbOH, CsOH, Mg(OH)2, Ca(OH)2, Sr(OH)2, Ba(OH)2
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As one moves down a column of the periodic table:
A. Electron affinity increases B. Ionization energy increases C. Oxidation state will always increase D. Metalic character increases E. Atomic radius decreases |
D: When determining Metalic character, a simple rule may be followed. Metalic Character will increase as you move from right to left on the periodic table. It will also increase as you move from top to bottom. |
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How many moles of hydrogen gas are in a 100 L container if the pressure is 20 atmospheres and the temperature is 27 degrees C? R= .082 liter-atm/mole degrees K.
A. (0.082 X 300)/(20 X 100) B. (20 X 273)/(0.082 X 100) C. (20 X 100)/(0.082 X 27) D. (20 X 100)/(0.082 X 300) E. (273 X 100)/(0.082 X 20) |
D: The ideal gas law states that PV=nRT. You must convert T to Kelvin by adding 273 to the temp which give you 300 K. Now solve for n.
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Find the Molecular formula for the compound with 12.06% of Hydrogen, 71.95% of Carbon, and 15.97% Oxygen
A. C6 H12 O B. C6 H12O6 C. C6 H6 O H2 D. C4 H10 O H3 E. C3 H6 O3 |
A: Molecular formula is the exact number of atoms for the given compound. The easiest way to find the Molecular formula take is to multiply each percentage given by 100 grams. Then divide the new quantity (in grams) by the atomic mass (grams per mole) of each of the given substances. Compare the results and create a ratio. In the given question: 12.06g of H divided by 1.01 g/mol of H = 12.06oles of H 71.95g of C divided by 12g/mol of C = 5.9 moles of C 15.97g of O divided by 16g/mol of O = .9 moles of O The ratio is 6:12:1 C6 H12 O
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A "shell" consists of all orbitals having
A. The same value of n B. The same value of m and l C. The same value of n, l, and ml D. The same value of n, l, and ms E. The same value of all four quantum numbers |
A: Orbitals with the same value of "n" are said to be in the same "shell".
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Arrange the following elements from largest to smallest according to their atomic radius : K, Mg, Ca, Cl, F
A. K, Ca, Mg, Cl, F B. K, Mg, Ca, F, Cl C. F, Cl, Mg, Ca, K D. Ca, K, Mg, F, Cl E. Ca, Mg, K, Cl, F |
A: When determining Atomic Radius, a simple rule may be followed. Atomic Radius will increase as you move from right to left on the periodic table. It will also increase as you move from top to bottom. So the periodic table will give us the answer.
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Balance the equation Zn + HCl -> ZnCl2 + H2
A. 2Zn + HCl -> ZnCl2 + H2 B. Zn + 2HCl -> ZnCl2 + H2 C. 2Zn + 2HCl -> ZnCl2 + H2 D. Zn + 2HCl -> 2ZnCl2 + H2 E. Zn + 2HCl -> ZnCl2 + 2H2 |
B: Make sure that there are the same number of atoms of each element on both the left and right sides of the arrow.
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How many ml of gas are enclosed in a cylinder under a pressure of 760 Torr if the final volume of the gas is 50 ml at 1550 Torr?
A. (50 X 1550)/(760) B. (760 X 50)/(1550) C. (760)/(1550 X 50) D. (1550)/(760 X 50) E. (1550 X 760)/(50) |
A: Boyle's Law is used therefore P1V1=P2V2. You are give everything but V1 so you need to solve for it.
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If 3.24 g of hydrogen occupies 44.2 liters, calculate the density of hydrogen gas in grams per millileter.
A. (3.24)(44.2) B. (3.24)/(44,200) C. (44.2)/(3.24) D. (3.24)/(44.2) E. (44,200)/(3.24) |
B: Density = (weight/Volume) You must make sure your units are in g/ml or you will get the wrong answer.
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Which element is most reactive with Cl?
A. Ca B. F C. Al D. Sr E. Br |
D: The most reactive will be an elements that has the greatest differences in electronegativity. Which in this case will be found in the lower left area of the periodic table.
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Which element is the most electronegative?
A. C B. F C. Cl D. Br E. I |
B: When determining electronegativity, a simple rule may be followed. Electronegativity will increase as you move from left to right on the periodic table. It will also increase as you move from bottom to the top. As F is in the upper right of the table, it is the most electronegative.
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The air in a compartment has a pressure of 700 mm of Hg at 25 degrees C. When placed in the sun the temperature rose to 50 degrees C. what was the pressure in the tank at that time?
A. (323 X 700)/(298) B. (323)/(700 X 298) C. (700)/(323 X 298) D. (298 X 323)/(700) E. (298)/(700 X 323) |
A: The Gay-Lussac Law states that (P1/T1)=(P2/T2)
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An element that looses _______, will maintain the same atomic number.
A. Proton B. Electron C. Neutron D. Alpha particle E. Beta particle |
C: Atomic number is the number of protons in an element. The loss of a proton will by definition change the atomic number. Electrons and beta particles are the same thing and when they are lost, the atomic number will increase by 1. The loss of an alpha particle will decrease the atomic number by 2.
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If you change only the number of electrons in an atom you create a(n)
A. Different element B. Different atomic number C. Ion D. Isotope E. Different atomic weight |
C: By adding another electron in an atom, you create an anion. By removing an electron from an atom, you create a cation.
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Which of the following compounds are held together by ionic bonds? H2O, KBr, MgO.
A. H2O only B. KBr only C. H2O and KBr D. KBr and MgO E. None of the above |
D: Ionic bonds are found between non-metals and metals. Hydrogen and Oxygen are both non metals. K and Mg are metals and Br and O are non-metals.
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Which of the following is NOT true of Ionization energy
A. Ionization energy is the amount of energy required to remove an electron from an atom B. Group I elements have low ionization energy. C. 2nd ionization energy is the energy required to remove a 2nd electron from an atom D. The 2nd ionization energy is always higher than the first. E. The removal of the an electron is exothermic |
E: Group I elements will be very stable with the loss of an electron. It is much more difficult to remove an electron from a positive atom. The 2nd ionization energy is trying to remove an electron from a positve atom is will always be greater than the first ion energy. The removal of an electron requires an input of energy and therefore is Endothermic.
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How many moles are in 137 g of H2SO4?
A. 98 x 137 B. 137/96 C. 137/98 D. 96/137 E. 96 x 137 |
C: In order to calculate the number of moles of a compound, you must divide the number of grams of the compound by the molecular weight of that compound. So you must first calculate the molecular weight of H2SO4 which gives you 98 g/mol. Now divide 137 g by 98 g/mol and this gives you your answer.
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set up units first:
g/(g/mol) = moles because grams cancel, leaving moles^-1 which is moles. |
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If a mark on a piece of paper has a mass of 1 x 10-5g, and you assume that it is carbon, how many atoms are required to make this dot?
A. (1 x 10-5/14 g/mol) 6.02 x 1023 B. (12 g/mol/1 x 10-5) 6.02 x 1023 C. (1 x 10-5/6.02 x 1023) 12 g/mol D. (1 x 10-5/12 g/mol) 6.02 x 1023 E. (1 x 10-5/6.02 x 1023) 14 g/mol |
D: You need two facts to answer this question. You need to know the number of moles in the carbon dot and also the number of atoms in a mole of any substance, 6.02 x 1023(Avagadro's number). 1 x 10-5/12 g/mol gives you the number of moles in the carbon dot, then you multiply it by Avagadro's number to give you the answer.
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How many molecules are in a drop of water that weighs 0.36 g?
A. (0.30 moles/18 g)(6.02 X 1023) B. (0.36 g/18 g)(6.02 X 1023) C. (0.02 moles/18 g)(6.02 X 1023) D. (18 g/0.02 moles)(6.02 X 1023) E. (18 g/0.36 g)(6.02 X 1023) |
B: In order to find the number of molecules in a drop of water, you need to know the number of moles making that drop. This is done by dividing the weight of the drop by the molecular weight of the drop of water (18g).
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According to VESPR theory, the H-Te-H bond angle in H2Te is closest to?
A. 90 degrees B. 120 degrees C. 109.5 degrees D. 149.5 degrees E. 180 degrees |
C: Te has 6 valence electrons. Two are bonded to both Hydrogens and 2 lone pairs. Therefore, the geometry is tetrahedral, which is 109.5 degrees. The VESPR theory predicts H2O and H2S, CH4, and NH3; all close to 109.5 degrees
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What are the oxidation numbers of NaMnO4?
A. Na +1, Mn +8, O4 -8 B. Na -1, Mn +8, O4 -7 C. Na +2, Mn +8, O4 -10 D. Na +1, Mn +7, O4 -8 E. Na +2, Mn +7, O4 -9 |
D: Rules of oxidation. 1 For atoms, molecules and neutral formulas, the total oxidation number is 0. Oxidation numbers are equal to the charge of the ion. 2 Group IA metals have an oxidation number of +1 and Group IIA metals have +2. 3 In compounds, the oxidation number of flourine is -1. 4 In compounds, the oxidation number of hydrogen is +1. 5 In most of its compounds, oxygen has an oxidation number of -2. 6 In two-element compounds with metals, Group 7A elements have an oxidation number of -2, and Group 5A elements have an oxidation number of -3.
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A mixture of nitrogen and oxygen gas is collected by displacement of water at 25 degrees C and 600 torr Hg pressure. If the partial pressure of nitrogen is 500 torr Hg, what is the partial pressure of oxygen? (Vapor pressure of water at 25 degrees C = 30 torr)
A. 30 torr B. 120 torr C. 490 torr D. 70 torr E. 110 torr |
D: Ptotal = p1 + p2 +...pn where Ptotal is the total pressure and p1 + p2 +...pn are the parial pressures of the gases present. So the equation for the problem can be written as Ptotal = poxygen + pnitrogen + pwater. Water is calculated because the pressure of nitrogen and oxygen are collected over water meaning that its vapor is present in the mixture. Therefore, Ptotal = 600, poxygen = ?, pnitrogen = 500, and pwater = 30. Solve for poxygen.
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Which will be followed by an increase in entropy?
A. Na(s) + H2O(l) --> NaOH(aq) + H2(g) B. I2(g) --> I2(s) C. H2SO4(aq) + Ba(OH)2(aq) --> BaSO4(s) + H2O(l) D. H2(g) +1/2O2(g) --> H2O(l) E. None of the above |
A: Entropy is the measure of disorganization. The order of increasing entropy = Solid-->liquid-->ions in solution-->gas. Look at the product and see it's phase. A gas product will always have the greatest entropy.
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An atom with the atomic number of 20 is of the element
A. Br B. Mg C. Ca D. Rb E. K |
C: Atomic number is the number of protons in the nucleus. Mass number is the sum of protons and neutrons.
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Consider the following equation which is not balanced: KO2(s) + H2O(l) -> KOH(s) + O2(g) How many moles of KO2 are needed to react with 4 moles of H2O?
A. 2 B. 4 C. 6 D. 8 E. 10 |
D: The first thing that needs to be done is to balance the equation. Do this by counting up the number of each element on each side. Find a common denominator to make sure there are the same number of elements on each side. The balanced equation for this problem is 4KO2 + 2H2O --> 4KOH + 3O2. The question asks for the relation ship between the number of moles of KO2 and H2O. The question doubles the number of moles of H2O, so the number of moles of KO2 must be doubles as well.
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Which of the following will NOT affect the rate of a reaction?
A. Gibbs Free Energy B. Temperature C. Concentration of reactant D. Presence of a catalyst E. pH |
A: Gibbs free Energy (Δ G) is related to the spontaneity of a reaction. It has nothing to do with the rate at which the reaction occurs.
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Which of the following is the best explanation of why PCl5 has a trigonal bipyramidal configuration?
A. Has 2 electron pairs B. Has 5 electron pairs C. Has 3 electron pairs D. Has 4 electron pairs E. Has 1 electron pair |
B: When figuring out the geometry of a molecule, you must first write out the Lewis dot structure. Remember that electrons repel each other and want to be as far away from each other as possible. So for this molecule P is the central atom and it has 3 electron pairs connecting Cl in a triangle. The remaining 2 pairs connecting P to the 2 Cl atoms are in a perpendicular line to the triangle. This is a trigonal bipyramidal molecule.
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For a certain reaction, 0.050 moles of HCl is needed. What volume of 0.888 M HCl(aq) is needed?
A. 37.2 mL B. 12.9 mL C. 29.7 mL D. 56.3 mL E. 61.2 mL |
D: Molarity is the number of moles of a compound in 1 liter of solution. Figure out which is the solution and which is the compound and do the necessary calculations to get mol/L.
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.050mol / .888mol/liter = .0563 liters
set up units first, moles cancel, liter^-1 becomes liters. |
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Using Avragado's number find the number of molecules in 36 g of SO4?
A. (36g)(96g)/(6.02 x 10^23) B. (36g)/(6.02 x 10^23) x (96) C. [(36g)/ (96g)] x (6.02 x 10^23) D. [(96g)/ (6.02 x 10^23)] x (36g) E. [(96g)(6.02 x 10^23)]/(36g) |
C: This is a simple stoichiometry problem. The final answer needs to be put into molecules. First, convert the mass of SO4 to moles by taking 36g and dividing it by 96g (mass number of SO4. Multiply this by Avagadro's number and you get your answer.
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Where is the triple point located?
(see graph) |
A: You should first know the 3 phase areas on this graph. Above B-A-D is the soild phase. Below B-A-C is the gas phase. Within D-A-C is the liquid phase. Point A is the triple point where soild, liquid and gas can coexist. Point C is the critical point where liquid and gas are indistinguishable.
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If 1 mole of H2 and 1 mole of O2 are mixed and allowed to react according to the equation 2H2 + O2 → 2H2O, the maximum number of moles of H2O that could be produced is
A. 2/3 B. 3/2 C. 2/1 D. 1/2 E. 1/1 |
E.
This problem has two main steps. First, we need to determine which of the reactants will limit the amount of products; then we can use the balanced equation to calculate how much product is formed. We are told that we begin with one mole of each of the reactants. When we divide this by the coefficients of the reactants in the balanced equation, we find that hydrogen is the limiting reactant. For every two moles of hydrogen, we produce two moles of water, so using the one mole of hydrogen we're given, we can produce one mole of water. |
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For the equilibrium (NH4) 3PO43NH4+ + PO43- the solubility product expression (Ksp) is
A. 3[NH4+][PO43-] B. [NH4+]3[PO43-] C. [NH4+][PO43-] D. [3NH4+][PO43-] E. [NH4+]3 [PO43-]/[(NH4)3PO4] |
B.
Recall that the solubility product expression is written for species in solution. Therefore, the solid ammonium sulfate will not be a part of the equation. In addition to this, we must keep in mind that the concentration of each species will be raised to the power of its coefficient. Since there are three ammonium ions per dissolved ammonium sulfate unit, the ammonium ion concentration will be raised to a power of three. Overall, this gives: [NH4+]3[PO43-]. |
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Which of the following compounds contains 42% chlorine and 57% oxygen by weight?
A. HCl B. HClO C. ClO2 D. HClO3 E. HClO4 |
D.
When we are asked to determine an empirical formula from percentages by weight, it is usually most convenient to think of a 100-g sample. Of course, any sample would give the correct answer, but using a 100-g sample, the percentages translate directly into weights (in g). In a 100-g sample of the compound, we therefore have 42 g chlorine and 57 g oxygen. To deduce the empirical formula, we need to find out the number of moles of Cl and O atoms to which these values correspond. The atomic weight of chlorine is 35.5 g/mol, so 42 g corresponds to 42/35.5, or 1.2 mol, and the atomic weight of oxygen is 16 g/mol, which corresponds to 57/16, or 3.6. Out next step is to divide both of these values by the smaller of the values. From this we obtain a ratio of 3 oxygen atoms to 1 chlorine atom. The only compound in which the ratio of oxygen to chlorine is 3:1 is choice D. Had you been really stuck, you could have eliminated choice C because it is the only compound of the five that contains only oxygen and chlorine, and the given weight percentages did not add up to 100, so there must have been another element in the correct compound. |
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Consider this reaction: A + B → C. Given the following information about the rate of this reaction, what is the rate expression?
(see chart) A. k[A][B] B. k[A]2[B] C. k[A][B]2 D. k[A]2[B]2 E. k[A]2[B]4 |
C.
We know that the general rate expression for a bimolecular reaction like the one given in this problem is k[A]x[B]y. We need to use the rate data given to determine the values of x and y. First, take a look at the first and second experiments. The concentration of A was varied, while B was held constant. [A] doubled, and so did the rate. This represents a first-order dependence on [A], so x=1. Now take a look at the first and third experiments. The concentration of B doubled, while [A] was held constant, and the rate increased fourfold. This represents a second-order dependence on [B], so y=2. Therefore, the rate expression is k[A][B]2. |
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When empty, the mass of a beaker is 115 g. When filled with 300 mL of an unknown liquid, the combined mass is 427 g. What volume (in milliliters) would 1 kg of the liquid occupy?
A. (300)(1000)/(427-115) B. (300)(1000)(427-115) C. (427-115)(1000)/(300) D. 300/427 E. 300/(427-115) |
A.
Although it isn't stated explicitly in the problem, this question asks you to determine the density of the unknown liquid and use this to find a volume. Using dimensional analysis will be very helpful here: the answer must be in milliliters, so we've got to get rid of g and kg. We also don't want to forget that the mass of the beaker is given as 115 g- we'll want to determine what the actual mass of the liquid in the beaker is by subtracting one from the other. This quantity will be (427-115), so eliminate any answer choices that don't contain that difference- that takes care of (D). Since we're trying to cancel the mass of the liquid with the mass of the given quantity, one must appear in the denominator and the other in the numerator. Finally, multiply this by the 300 mL of the original unknown experiment, and choice (A) is clearly correct. |
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A pot containing an unknown liquid is boiling on a stove. The temperature of the liquid is 102º C. Which of the following is NOT a possibility?
A. The liquid is a solution of water and table salt. B. The liquid is pure water and the stove is below sea level. C. The liquid is pure water and the stove is above sea level. D. The liquid is a solution of water and sugar. E. The liquid is a mixture of water and antifreeze. |
C.
This question tests your knowledge of colligative properties. The correct answer will be the one that doesn't allow the temperature of the liquid to rise above 100 degrees Celsius under the given conditions. Let's work by process of elimination. Choice (A) suggests that the liquid is a solution of water and table salt. Dissolved particles cause the boiling point of liquids to rise, so (A) is incorrect; so are choice (D) and (E). Now we're left with choices (B) and (C). We're assuming the liquid is pure water, and the atmospheric pressure is varied because of the altitude. Which one will cause the boiling point of water to rise? To answer this, think back to the definition of boiling point: the point at which the vapor pressure of a liquid is equal to the atmospheric pressure. This will happen at a lower temperature for lower pressures, which corresponds to higher altitudes, so choice (C) is correct. |
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Which of the following is NOT a good conductor of electricity?
A. Fe (s) B. NaCl (aq) C. NaCl (l) D. NaCl (s) E. Cu (s) |
D.
Metals are good conductors of electricity. From this, we know that we can eliminate choices (A) and (E) because both are metals. Now we are left with three different forms of sodium chloride. What are the differences between them? Well, the aqueous solution of sodium chloride will be a superb conductor of electricity; ionic solutions are well-known conductors. The molten form will also be a good conductor. Only the solid form will be an insulator, so it is the correct choice. |
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Concentrated sulfuric acid is 18 M. What volume of water would be needed to dilute 25 mL of concentrated H2SO4 solution to 3 M ?
A. 25 mL B. 50 mL C. 75 mL D. 125 mL E. 150 mL |
D.
For dilution problems like this one, a helpful formula will simplify your calculations: MiVi = MfVf, where Mi and Mf are the initial and final molarity of the solution respectively, and Vi and Vf are the initial and final volume of the solution respectively. To determine the amount of water needed to perform the dilution, our first step is to determine the final volume of the solution (the volume after dilution). (18 M)(0.025 L) = (3 M)Vf Vf = 0.150 L = 150 mL The amount of water we need to add is therefore the difference between the final and initial volumes, or (150 – 25) mL = 125 mL. |
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Which one of the following processes is accompanied by an increase in entropy?
A. Sorting groceries by food group in the refrigerator B. Filtering a solid from a liquid suspension C. Condensation of water D. Evaporation of rubbing alcohol E. Deposition of carbon dioxide |
D.
Recall that the second law of thermodynamics states that in a real, spontaneous process the entropy of the universe (the system being considered plus its surroundings) must increase. We can think of entropy as disorder or randomness, and it is helpful to keep a few typical entropically-favored processes in mind: processes such as boiling, sublimation, melting and mixing. Take a look at the answer choices. Choice (A), sorting, will be the opposite of mixing, so it is accompanied by a decrease in entropy. The same goes for choice (B), which separates two things by filtration. Condensation is the opposite of boiling, so choice (C) is incorrect, and choice (E), deposition, is also incorrect because it is the opposite of sublimation. This leaves choice (D), evaporation. Evaporation is entropically equivalent to boiling, so (D) is the correct answer. |
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Hydrogen, deuterium and tritium are
A. allotropes. B. isomers. C. isotopes D. identical except for their nuclear charge. E. resonance structures. |
C.
This is an example of a question for which choosing the correct answer depends solely on your knowledge of general chemistry. How are hydrogen, deuterium and tritium related? They have identical structures except that deuterium has an extra neutron and tritium has two extra neutrons in their nuclei. Therefore, they are isotopes. Allotropes are different forms of elements, isomers are molecules with the same formula and different structures and resonance structures are different ways of representing the electron density in a molecule. The three isotopes given are not identical- they have different nuclear structures, though they are all electrically neutral (each contains one electron per atom.) |
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Which of the following can be classified as a Brønsted acid?
A. CH4 B. OH– C. Li+ D. SnCl4 E. CH3COOH |
E.
A Brønsted acid (or Brønsted-Lowry acid) is a species that can donate one or more proton. Don't confuse this definition with other acid definitions, such as Arrhenius and Lewis. Choice (A) is not basic or acidic: most hydrocarbons are neutral. Choice (B) you should recognize as hydroxide ion, a strong base. Choice (C) is a Lewis acid but not a Brønsted acid, so it is incorrect. Choice (D) is a Lewis acid because it can accept an electron pair, but it does not have a hydrogen ion to donate. Therefore, the correct answer is choice (E), acetic acid, which can donate a proton in solution. |
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Which of the following pairs would give the bond with the most ionic character?
A. Rb and F B. Li and Cl C. K and Br D. Si and O E. Mg and P |
A.
The question stem gives a helpful clue- we're looking for a pair that has the most ionic character. What makes a bond ionic? The most ionic pair will be one in which the two ions most closely approximate point charges. This is possible when the positive ion completely donates its electron to the negative ion. Another way of saying this is that the species involved in the bond differ greatly in electronegativity. So, we are looking for a positive ion that completely loses its valence electron and a very electronegative element for the anion. Which of the anions is most electronegative? F, which appears in choice (A). However, we must also consider the cation. The cation which most closely approximates a point positive charge will be one that has the most shielding between the nucleus (the positive charge) and the valence electron to be lost. This will be Rb, so choice (A) is correct. Choices (B) and (C) will have significant ionic character as well, but not to the extent that (A) does. Choices (D) and (E) both contain elements from the center of the periodic table: Si and P, so it will be more difficult to pair them with elements of differing electronegativity. Therefore, these combinations will not have high ionic character when bonded. |
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The equilibrium for the reaction below can be shifted to the right by
Cl2 (g) + PCl3 PCl5(g) ΔH = -88 kJ/mol A. decreasing the volume. B. decreasing the pressure. C. increasing the temperature. D. adding a catalyst. E. removing chlorine from the container. |
A.
This question calls for an application of Le Châtelier's principle to the given system. Le Châtelier's principle states that when a system is in equilibrium, any stress applied to it will cause the equilibrium to shift in a direction that alleviates the stress. The stress can be in the form of a change in temperature, pressure, volume, or concentration of any of the species in the system. For the system given in the question stem, we have one mole of PCl5 in equilibrium with one mole of PCl3 and one mole Cl2. All species are in the gas phase. If we decrease the volume after the system has reached equilibrium, the equilibrium will shift towards the side with the smaller number of moles of gases, i.e., to the right. Therefore, choice (A) is correct. Choice (B) is incorrect because decreasing the pressure would have the opposite effect. It would cause the equilibrium to shift to the side with the larger number of moles of gas, in this case the reactant side. Choice (C) is incorrect. Since the reaction is exothermic ( H < 0), increasing the temperature would favor the reactants. Choice (D) is incorrect because the addition of a catalyst does not affect the thermodynamics of a reaction; it increases both the forward and reverse rates of reaction, so it won't affect the equilibrium. Removing chlorine from the container will shift the reaction left to compensate, so choice (E) is incorrect. |
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The electronic configuration of a particular neutral atom is 1s22s22p63s23p63d64s2. The number of unpaired electrons in this atom is
A. 0 B. 1 C. 2 D. 4 E. 6 |
D.
The best thing to do here is to examine the electron configuration given and determine which electrons are the valence electrons. Clearly, the n=1 and n=2 levels are filled, so only n=3 must be considered. The s subshell is full (2 electrons), the p subshell is full (6 electrons) and this leaves the d orbitals. Remember that electrons prefer to be unpaired if they can be, so place each electron in an empty orbital. This works for the first five electrons; then there is one additional electron. Place it in one of the half-full orbitals, and you'll see that you have four unpaired electrons left. This is choice (D). |
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Which of the following compounds is not planar?
A. NO3- B. BF3 C. AlCl3 D. PBr3 E. SO3 |
D.
At its core, this is a Lewis dot structure/ VSEPR question. In order to determine which of the compounds is not planar, we must calculate the number of valence electrons for each. The nitrate ion has 24 electrons; 6 from each of the three oxygen atoms, five from the nitrogen atom and one from the negative charge. Boron trifluoride also has 24 electrons – three from boron and seven from each of the three fluorine atoms. SO3 has 24 electrons- six from each of the oxygens and six from the sulfur atom. Aluminum trichloride has 24 electrons; 3 from the aluminum atom, 21 total from the chlorines. Phosporus tribromide differs from the other four compounds because it has 26 electrons; five from the phosphorus, and 21 total from the three bromine atoms. Since it has an extra pair of valence electrons, the three substituents around the central phosphorus atom will be pushed out of the plane. Instead of having a trigonal planar shape, this compound will have a trigonal pyramidal shape. |
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When an oxygen nucleus containing nine neutrons emits an alpha particle, the other product of the balanced reaction
17O -->4He + ? is: 8 2 A. 21 Ne 10 B. 13C 6 C. 13Al 6 D. 12C 6 E. 13Ne 10 |
B.
alpha particle is 4He 2 The key to answering this question correctly is to balance the equation given, then choose the corresponding element. The mass numbers and the atomic numbers must be balanced in order for the equation to be valid. Only choices (B) and (C) satisfy this requirement. Choice (C) is incorrect because it denotes the wrong element; the subscript (6, in this case) is the atomic number of the species in question, whereas choice (C) uses the mass number, the superscript to determine the identity of the element. Choice (A) adds the masses and atomic numbers instead of balancing them. |
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Which of the following increases down a column in the periodic table?
A. Electronegativity B. Electrical conductivity C. Ductility D. Ionization energy E. Atomic radius |
E.
Electronegativity and ionization energy decrease down a column in the periodic table, so choices (A) and (D) are incorrect. Ductility (the ability of a substance to be pulled into wires) and electrical conductivity vary by substance and don't demonstrate a clear trend based on the position within a column of the periodic table, so choices (B) and (C) are incorrect. Only (E), atomic radius, increases as we move down a column. Since there is more shielding due to additional filled orbitals of electrons, the valence electrons are not as tightly held and the atomic radius increases. Choice (E) is correct. |
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A certain radioactive compound has a half-life of 3.0 minutes. One-eighth of the compound will remain after
A. 3/8 of one minute. B. 6.0 minutes. C. 9.0 minutes. D. 18.0 minutes. E. 27.0 minutes. |
C.
The half-life of a radioactive substance is the time it takes for half of the sample to decompose. Therefore, after three minutes, this sample will have one-half of its original activity. After an additional three minutes (6.0 total) the sample will have one-fourth of its original activity. After another three minutes, 9.0 minutes total, the sample will have one-eighth of its original activity. Therefore, the answer is choice (C). Choice (A) is a trap- just multiplying 3.0 minutes by one-eighth won't give the correct answer. |
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How many moles of AgIO3 (Ksp = 3.1 x 10-8) will dissolve in one liter of a 10-5 M solution of NaIO3?
A. sqrt(Ksp) B. sqrt(Ksp) / (10^-5) C. sqrt(ksp) * 10^-5 D. 10^-5 - sqrt(Ksp) E. sqrt(ksp) - 10^-5 |
E.
Answering this question correctly requires an understanding of the common ion effect. We know that if a solution already contains some iodate ions, less silver iodate will dissolve; Le Chatelier's principle tells us that the equilibrium will lie to the left. Therefore, our task is to determine how the common ion effect will operate in this case. The simplest way to approach the calculation is to think of how much silver iodate would dissolve if the liter contained only water. This is just the square root of the solubility product constant given. Then, we can subtract the ions that are already there from the sodium iodate, and we have choice (E). |
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Given the following series of reactions and their standard reduction potentials, which one will most readily occur as an oxidation reaction?
A. Al3+(aq) + 3e– → Al(s) Eº = –1.706 V B. Cd2+(aq) + 2e– → Cd(s) Eº = –0.4026 V C. Pb2+(aq) + 2e– → Pb(s) Eº = –0.1263 V D. I2(s) + 2e– → 2I– (aq) Eº = +0.535 V E. H2O2 + 2H3O+ + 2e– → 4H2O Eº = +1.776 V |
A.
Reduction leads to the gain of electrons, while oxidation removes electrons. So we can rewrite all the equations in oxidation form, i.e., reverse them. At the same time, we need to reverse the sign for the standard reduction potentials to convert them into oxidation potentials. Recall that Δ G = –nFE. Thermodynamically, the reaction with the most negative Δ G is the most spontaneous, so the most positive Eox is the most spontaneous. Choose Eox = +1.706 V, which is choice (A) after it has been converted into an oxidation half-reaction. Don't be misled by choice (E), which at 1.776 has a greater magnitude. |
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In the following reaction,
3CuS + 8HNO3 -->3CuSO4 + 8NO + 4H2O the oxidation number of nitrogen changes A. from -1 to +2. B. from +1 to +3. C. from +5 to +2 D. from +2 to +3 E. from +5 to -2. |
C.
It isn't necessary to assign oxidation numbers to all species in this reaction; just focus on the species that contain nitrogen; nitric acid on the left and NO gas on the right. Determining the oxidation number of nitrogen in NO is relatively easy since the compound contains only two atoms. From this, we'll be able to eliminate some of the wrong answer choices and move on to nitric acid. NO is neutral, so the oxidation numbers of nitrogen and oxygen will have to add up to zero. The oxidation number of oxygen is –2 except in rare cases, so the oxidation number of nitrogen is +2. Armed with this information, we can eliminate all the choices except (A) and (C). Now, let's examine the oxidation number of nitrogen in nitric acid. Each of the oxygen atoms contributes a –2 charge, which gives –6 total. Hydrogen will be +1, leaving a +5 oxidation number for the overall neutral compound. Therefore, the correct answer is choice (C). |
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Why does raising the temperature cause an increase in the rate of a chemical reaction?
A. As the concentration of reactants increases, the number of collisions increases. B. A greater proportion of reactants has sufficient kinetic energy to overcome the activation barrier. C. The geometry of the reactants shifts slightly to make collisions more effective. D. The reactants move more slowly, making each collision last longer, which in turn gives the reactants a better chance of resulting in a reaction. E. Most reactions are exothermic, so increasing the temperature actually hinders the reaction rate. |
B.
This question is a tough one, so predicting the answer is a good plan, but you'll still need to check all the answer choices since they are complex. You don't want to miss any subtleties! Raising the temperature is a way of saying "adding kinetic energy" to a system. Therefore, when we raise the temperature of a reaction system, we'll be imparting kinetic energy to the reactants (and products, if there are any already). Why will this increase the rate of reaction? Well, now the reactants have more energy, so when they collide with one another, there is a greater chance that they will have sufficient energy to overcome the activation energy barrier. Keep in mind that not all collisions between reactant species will result in reactions. If the molecules have the wrong spatial orientation or if they collide with insufficient force, no reaction will occur and the reactants will leave the collision exactly as they entered it. Choice (B) is correct. Choice (A) is not untrue; as concentrations of reactants increase, the reaction rate will also increase, but this is irrelevant since the question asks about temperature. Although geometry is key to a successful reaction collision, choice (C) is incorrect because it is not a rise in temperature that causes the correct spatial orientation to occur. Choice (D) is the exact opposite of the correct reasoning; if the reactant particles are moving more slowly, they will be less likely to collide effectively since the collision won't be energetic enough. Choice (E) is incorrect because it brings in irrelevant information – we don't know whether most reactions are exothermic, so we can't make a generalization about that. |
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Which of the following is the electronic configuration of Mn2+?
A. [Ar]4s23d5 B. [Ar]3d5 C. [Ar]4s23d3 D. [Ar]3d3 E. [Ar]4s23d7 |
B.
Assigning an electron configuration to an ion is a matter of determining the number of electrons surrounding that ion, then using the Aufbau (building-up) principle to place the electrons in their proper orbitals. Manganese has seven valence electrons, so when it loses two to achieve a +2 charge, it will be left with five. Don't be thrown off by choice (E), which suggests that the +2 charge represents an additional two electrons! Since we know we have five electrons, we can eliminate choices (A) and (D). This leaves us with (B) and (C). Since the Aufbau series tells us that electrons will be removed from a 4s orbital before a 3d orbital, the answer is choice (B). |
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Which of the following is the correct Lewis structure for NOF?
A. N and F each with one bond to O B. N and O each with one bond to F C. N with 2 bonds to F O with one bond to F D. N and F each with 2 bonds to O E. N with 2 bonds to O F with one bond to O |
E.
There are a couple of important things to keep in mind when constructing a proper Lewis structure. First, the number of valence electrons in the molecule must match the structure. Then, the structure with the least separation of charge will be preferable to the others. Therefore, our first step is to determine the number of valence electrons in the correct molecular structure. Nitrogen contributes five valence electrons, oxygen contributes six and fluorine seven. This gives a total of eighteen electrons. Armed with this information, we can eliminate choices (A), (B), and (D), since those choices contain 20, 20, and 16 electrons respectively. Choices (C) and (E) both contain the correct number of electrons, so we'll need to consider the positions of the atoms in order to determine which is the correct structure. Choice (C) shows fluorine as the central atom. This is not possible, since fluorine will have a positive charge and the negative charge will lie on the oxygen. Fluorine is more electronegative than oxygen, so the opposite will be true. Therefore, oxygen is the central atom, and choice (E) is correct. |
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What is the standard enthalpy change for the following reaction?
2 SO2 + O2 → 2 SO3 SO2 -297 O2 0 SO3 -396 A. +99 kJ/mol B. -99 kJ/mol C. -198 kJ/mol D. -495 kJ/mol E. +495 kJ/mol |
C.
Use an effective Kaplan strategy here: estimate! The value given for sulfur trioxide is very close to -400, and that given for sulfur dioxide is close to -300, so use that to your advantage instead of keeping track of lots of numbers- the answer choices are quite far apart, so there's no risk that you'll go astray. The enthalpy change for the reaction is simply the enthalpy of the products minus the reactants: in this case, 2(-400)-2(-300), or about –200. The only answer choice that comes close is choice (C). |
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Which of the following compounds is NOT capable of hydrogen bonding?
A. C4H11N B. C4H9NO C. C4H7F3 D. C4H8O2 E. C4H10O |
C.
A compound can hydrogen bond if it has an electronegative element (N, O or F) bonded to a hydrogen atom. In order to determine which of the listed compounds does not meet this requirement, draw the structures as carbon chains. Choice (C) cannot hydrogen bond because the fluorine atoms are bonded to a carbon atom. We know that they can't be bonded to hydrogen atoms because fluorine only forms one bond per atom, and the compound contains more than just HF. |
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A sample of gas is expanded isothermally from a volume of 2 L to 3.5 L. If the final pressure is 0.4 atm, what was the initial pressure of the system?
A. 7 atm B. (7/0.4) atm C. (3.5/0.4) atm D. 0.7 atm E. (2/0.4) atm |
D.
This is a fairly straightforward Boyle's law question. We know that the expansion was effected isothermally (at a constant temperature) so we can use P1V1=P2V2 to determine the fourth variable when we have the other three. In this case, that is (2 L)(P1) = (3.5 L)(0.4 atm), or P1 = 0.7 atm. |
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Which of the following describes the thermodynamics and electrochemical properties of a battery?
A. positive Gibbs free energy; spontaneous; oxidation at cathode B. positive Gibbs free energy; nonspontaneous; oxidation at cathode C. negative Gibbs free energy; spontaneous; oxidation at cathode D. negative Gibbs free energy; nonspontaneous; oxidation at anode E. negative Gibbs free energy; spontaneous; oxidation at anode |
E.
This question might seem daunting since it asks you to put a battery into three categories: positive or negative Gibbs free energy change, spontaneity and electrode character. In addition to that, you might be wondering, what kind of cell is a battery? The most helpful thing here is to take a step back and ask yourself what you already know about batteries. The main distinction between galvanic (or voltaic) cells and electrolytic cells is that one is used to do work, while the other requires electrical energy to induce the reaction. A battery most certainly falls into the first category – we use batteries all the time to run small appliances. Therefore, you know that the correct answer will be one that contains the attribute "spontaneous." This eliminates choices (B) and (D). The next step is to consider how spontaneity is related to the change in Gibbs free energy. When a reaction is spontaneous, it has a negative change in Gibbs free energy. This allows us to eliminate choice (A). Finally, consider what is true for all kinds of cells: the oxidation half-reaction occurs at the anode, bringing us to choice (E). |
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The pH of a solution of HNO3 is 3. What is its hydrogen ion concentration?
A. 3 M B. 10-3 M C. 103 M D. 1.5 M E. 3-10 M |
B.
The pH and the proton (hydrogen ion) concentration of an aqueous solution are related in the following way: pH = -log[H+]. We know the pH, so we must work backwards. Luckily, we're given a number that is relatively easy to work with. If the pH is 3, the [H+] is 10-3 M. |
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