MCAT Gen Chem Problems & Facts

What is the concentration of Cl- ions in a 0.1M solution of calcium chloride?

A: 0.2M The formula for calcium chloride is CaCl2 and it produces twice as many Cl- ions as Ca2+ ions in solution when it dissolves

Which cations is most likely to be found in place of Fe(II) in the square planar binding domain of hemoglobin?

A: Co2+ (Picked Mg2+) Co2+ is closely related to Fe2+ as a transition metal and support a square planar coordination environment

Which of the following properties is associated with the existence of glycine as a dipolar ion in aqueous solution?

A: High dipole moment Polarity in neutral molecules results from an uneven distribution of electron density, which can arise from separation of unlike charges (this occurs in zwitterions and ylides). Additionally, molecules containing strongly electron-withdrawing or donating substituents are highly polar and possess high dipole moments

Which of the following is most likely to undergo positive beta decay? A. 14C B. 13C C. 17O D. 22Na

Answer: D Positive beta decay, known as positron emission, occurs when the proton-to-neutron ratio is too high. Of the options given, 22Na has the highest ratio as it has 11 protons and 11 neutrons (1:1)

How many ppm is 1%

A: 10000 ppm stands for parts per million. Since 1$ is one part in 100 parts. The question becomes 1 is to 100 as * is to 100000 where x is unknown. Thus, x = 10000

The addition of which of the following compounds to water would reduce the solubility of ferrous (II) hydroxide? A. Ammonia B. Ethanol C. Acetic acid D. Trichloroethane

A: Ammonia This is a common ion effect question. The presence of a common ion in the solution prior to dissolving will cause the reaction to reach equilibrium faster (i.e. at a lower concentration of reactants). Because ferrous (II) hydroxide produces hydroxide ions when it dissolves, the equilibrium will be changed by changes in pH. More specifically, an increase in pH indicates there are fewer H+ ions and more OH- ions in solution. Thus, the concentration of hydroxide ions will increase and the solubility of ferrous (II) hydroxide will be reduced. Ammonia is the only base of those listed.

A student in an organic chemistry lab grabs a bottle that supposedly contains an aqueous solution of ammonium phosphate. Unfortunately, he notices thick precipitate lining the sides of the container. Which solution was most likely contaminated with? A. sodium carbonate. B. barium nitrate. C. a small amount of additional phosphate ion

A: Barium nitrate Ammonium always forms soluble salts. Because we are looking for a substance that forms a precipitate with phosphate, barium cation is a perfect example as it would form barium phosphate, an insoluble compound.

As increasing amounts of NaCl are added to water

A: Boiling point will increase, melting point will decrease, and vapor pressure will decrease Colligative properties include boiling point elevation, melting (freezing) point depression, and vapor pressure depression. As more solute is added, these properties are seen with an increasing effect.

Which of these options correctly list five elements in order of decreasing reactivity with water?

A: Cs, K, Ca, Mg, Be Reactivity with water is heavily based on ionization energy. Elements with low ionization energy values tend to lose electrons easily, allowing them to rapidly form oxides in water. The list above clearly has the five elements from lowest to highest ionization energy.

Which of the following acids is expected to generate the strongest hydrogen bonding? A. HI B. HF C. HCl D. HBr

A: HF Hydrogen bond is an intermolecular force that occurs when H is bonded to a very electronegative elements like F, O, or N.

O2 dissociates more readily from Hb in an acidic environment. This dissociation will therefore occur most readily when the PCO2 is

A: High, because the equation (CO2+H2O -> H2CO3 -> H+ + HCO3-) will proceed to the right. (I picked "High, because the equation will proceed to the left" because CO2 is acidic) Analysis of equation 1 leads to the conclusion that increasing concentration of carbon dioxide will result in higher levels of H+ (lower pH). The lower pH increases the dissociation of oxygen

H2O is liquid at room temperature, whereas H2S, H2Se, and H2Te are all gases. Which of the following best explains why H2O is liquid at room temperature?

A: Hydrogen bonds form between H2O molecules H2O is capable of forming intermolecular hydrogen bonds, but the rest are not

While solving a chemistry problem, a student uses the equation ΔG = -nFE to determine that the ΔG = -178 kJ/mol. Which of the following must be true about this reaction? I. The reaction is spontaneous. II. The reaction is exothermic. III. The reaction increases entropy.

A: I The change in Gibbs free energy determines the spontaneity of the reaction. A negative G indicates a spontaneous reaction. Whether a reaction is endo or exothermic is determined by enthalpy (H) not G. Whether the entropy of reaction increases or decreases is determined by the change in entropy (S) not G

In an isolated system, entropy is maximized when: I. the system is at equilibrium. II. the system is far from equilibrium. III. the system is unable to perform work.

A: I and III Recall at equilibrium, with no free energy change, the system is unable to perform work. At equilibrium, there are no energy gradients within the isolated system, so energy is maximally dispersed, resulting in maximal entropy - A system far from equilibrium constant contains large energy gradients which are able to perform work. The formation of large energy gradients requires localization/concentration of energy, which results in decreased entropy

The term "ideal gas" refers to a gas for which certain assumptions have been made. Which one of the following assumptions applies to an ideal gas?

A: Individual molecular volume and intermolecular forces are negligible It's assumed that the molecules in an ideal gas have no volume and no interactions between the molecules

What are the units for k in the following rate law: rate = k[A]2[B]? Note that the concentration unit is mol/L.

A: L2mol-2s-1 We can see that the reaction is third order overall, and reaction rates are measured in M/s, where M = mol/L. Thus, we can set up our equation: M/s = k (mol/L)3 = k (mol3/L3) k = (M/s) (L3/mol3) = (mol/L•s) (L3/mol3) k = L2/s•mol2 = L2mol-2s-1

Suppose that at the end of reaction 1 the level of the aqueous solution were 26cm higher inside the buret then outside. Compared to ambient pressure, the pressure of the gas inside the buret would be

A: Lower Because the level of the liquid inside is higher than outside which results from the fact that the air pressure outside is higher than the pressure inside. Pressure builds on the inside by raising the liquid level. When the sum of the inside air pressure plus liquid pressure equals the outside air pressure, the liquid level stops rising (attains equilibrium)

Which experimental techniques was most likely used by the students to determine the rate of reaction?

A: Monitor the increase in absorbance of the solution at 360nm The experiment describes that the substrate was chosen based n the fact that it produced a yellow colored product (compound 2). The complementary color to yellow is purple, which is ~360nm.

Which of these choices correctly places three nitrogen-nitrogen bonds in order of increasing bond energy?

A: N-N, N=N, N(triple)N As bond order increases, bond length shortens and bond energy increases.

Two additional compounds were studied: NO2(OH) dissolved in water and produced an acidic solution, and Ni(OH)2 dissolved only in an acidic solution. What type of compounds were these?

A: NO2(OH) was an oxyacid and Ni(OH)2 was a base The first substance is nitric acid (HNO3). Since this dissolves to generate an acidic solution, the bond between O and H in NO2OH breaks when the substance dissolves, making it an oxyacid. Ni(OH)2 is insoluble in neutral but will dissolve if the solution is acidic. This is typical of substances featuring basic anions. The bond between Ni and O breaks when Ni(OH)2 dissolves. The hydroxide ion that is produced quickly reacts with proteins in solution and can't react with Ni2+ to form a precipitate

What is the best explanation for the fact that a solution of NaNO2(aq) is basic?

A: NO2- is hydrolyzed with the formation of OH- (aq) ions NO2- reacts with water, forming OH- ions

Which of the following ingredients would not aid a multivitamin in decreasing stomach acidity? A. Mg(OH)2 B. Al(OH)3 C. C2H5OH D. NaCN

Answer: C A compound that would help decrease stomach acidity would be an antacid. Antacids are basic compounds that react with and neutralize stomach acids. We are looking for an option that is NOT a base (an acid or any non-base). Choice C is ethanol, which suits what we are looking for

Tissue engineers have been developing a synthetic mineral apatite to use in prosthetics that is comprised of repeating molybdenum hexacarbonyl [Mo(CO)6] subunits. This bonding is likely to involve which of the following orbitals? A. sp3 B. sp3d2 C. d2sp3 D. dsp3

Answer: C Here, the central molybdenum atom is bound to six substituents. In such a configuration, the molecule takes the shape of an octahedron. The number of substituents (bonding regions) around the atom is equal to the number of orbitals that can hybridize: here, there are two 4d, one 5s, and three 5p atomic orbitals, for an overall hybridization of d^2sp^3. - B is false; in sp^3 d^2 hybridization, all of the orbitals have the same principle quantum number. In d^2 sp^3 hybridization, the principal quantum number of the d orbitals is one less than the principal quantum umbers of the s and p orbitals. We see d^2 sp^3 hybridization in the transition metals and sp^3 d^2 hybridization in the nonmetals. - dsp^3 hybridization would be expected if the central atom had only 5 attached substituents. This would yield a trigonal bipyramidal rather than an octahedral geometry

Subsequent solvation studies employing other haloalcohols revealed the strongest interaction between Trp19 in monomeric melittin and the hydrophobic component of what solvent molecule? A. 2,2-difluoroethanol B. 2-fluoroethanol C. 2,2-dichloroethanol D. 2-chloroethanol

Answer: D The four molecules differ in the number and strength of electron-withdrawing halogen atoms present. Specifically, as the number and electronegativity of the halogens present decreases, the polarity of the molecule will decrease as well, increasing its hydrophobicity. This will strengthen the interaction of the hydrophobic component of the molecule with Trp19. - Fluorine is more electronegative than chlorine - The addition of a second chlorine atom increases the electronegativity and decreases the hydrophobicity

What is the net charge of both pantothenate and phosphopantothenate in aqueous solution at pH 7?

A: -1 for pantothenate and -3 for phosphopantothenate In water at pH 7, the carboxylic acid of pantothenate will lost a hydrogen ion to become a carboxylate (-1) for a net charge of -1 on pantothenate. The phosphate groups of phosphopantothenate will lost two hydrogen ions (-2) and the carboxylic acid will lose one hydrogen ion (-1) for a net charge of -3 on the phosphopantothenate.

What is the normality of a 0.015M solution of phosphoric acid?

A: 0.045N The chemical formula of phosphoric acid is H3PO4. NORMALITY refers to the number of moles of PROTONS per liter of solution ("molarity of proton"). Normality can be calculated by multiplying the molarity of the solution by the number of protons per molecule of acid (here, 3) (0.015M solution)*(3 protons per molecule)=0.045N

Ten moles of the monoprotic, weakly acidic medication aspirin were added to water to make one liter of solution. If the pH of the resulting solution was 5.9, what is the approximate Kb for the non-diffusible form of aspirin?

A: 0.1 Since we are given pH in the question stem, we will not be able to find Kb immediately. Instead, we need to calculate Ka and solve for Kb from that value. The Ka for the dissociation of a generic acid HA can be written as Ka = [H+][A-]/[HA], where all concentrations are measured at equilibrium. In the solution of aspirin described, the initial concentration of drug is 10 M. Since only a small amount of this weak acid will dissociate, this value is a good approximation for our final equilibrium [HA]. Next, we must find the proton concentration. Remember, [H+] = 10-pH. Here, the pH of the solution is 5.9, so [H+] = 10-5.9 M ~ 10-6 M. Since each HA molecule dissociates into equal parts [H+] and [A-], our value for [A-] must be 10-6 M as well. Returning to the Ka expression, Ka = [(10-6 M )(10-6 M)] / (10 - 10-6 M). Remember, we can estimate that [HA] = 10 M, even though its true final value is 10 M - 10-6 M! [(10-6 M )( 10-6 M)] / (10 M) = 10-12 M / 10 M = 10-13 M In water at 25°C, Ka•Kb = 10-14. Given this, Kb = 10-14/Ka = 10 -14/10-13 = 10-1 = 0.1

An artificial vesicle containing a 1 M glucose solution is composed of a phospholipid bilayer lacking any protein components other than aquaporin channels. Assuming an ideal solution, what is the ratio of the osmotic pressure measured immediately after immersion of the vesicle in de-ionized water to the osmotic pressure measured immediately after immersion of an identical vesicle containing the original volume of 1 M glucose solution added to an equal volume of 1 M KCl solution in deionized water?

A: 0.67 The situation described in the question is analogous to a semipermeable membrane. Water is able to pass through aquaporin channels present in the liposome, but large uncharged particles (glucose) and ions (K+ and Cl -) are impermeable and will remain trapped within the liposome. If assumed to be ideal, the osmotic pressure, π, exerted by the solution due to molarity differences across the membrane is defined as π = iMRT, where i is the van't Hoff factor, M is the molarity of the solution, R is the universal gas constant, and T is the absolute temperature of the solution. A change in osmotic pressure at constant temperature is due to changes in iM, a term that is equivalent to the concentration of dissolved particles produced by solute in solution. When compared to the original volume of 1 M glucose, the new combined solution has twice the volume and three times the number of dissolved particles (1 M KCl, a strongly electrolytic solution, produces 1 M concentrations of both K+ and Cl- in solution), or an increase in the concentration of dissolved particles by a factor of 1.5. This is equivalent to a combined molarity of dissolved particles of 1.5 M. The ratio of osmotic pressure is then [1 M dissolved glucose] / [1.5 M dissolved glucose + KCl] = 0.67

Given the role of the reaction in cellular respiration, what is the most likely Eº value for the reduction of oxygen to water?

A: 0.82V Positive Eº values correspond to compounds that are easily reduced. As the final electron acceptor in the ETC, oxygen must be prone to reduction than any of the other electron carriers. Positive Eº means that the associated reaction is spontaneous. Since metabolic respiration is a biological process to produce energy for the cell, the reactions that drive it must be net energy producers

The half-life of 18F is 110 minutes. If 5 grams of FDG remain after 5 hours and 30 minutes, how much energy was emitted from the patient's body in the form of gamma rays from radioactive decay of the FDG?

A: 1.19 x 1026 keV This question requires us to determine how many half-lives have occurred after 5.5 hours. 5.5 hr x (60 min/1 hr) = 330 minutes, or 3 half-lives. Working backwards from the final amount of 5 g we see that we must have started with 40 g of of the original sample: 5 g x 2 (first half-life) x 2 (second half-life) x 2 (third half-life) = 40 g original sample Thus 35 grams of FDG must have undergone positive beta decay. 35 grams x (1 mol/181 grams) = about 1/6 mol of FDG, which means that 1/3 moles of gamma rays were emitted. (1/3) x (6.022 x 1023) * 511 keV = approximately 1 x 1026 keV.

In a certain kinetic experiment, the enzymatically catalyzed hydrolysis of ATP proceeds at a constant rate of 2μMs^-1. If the volume of solution is 1mL, what is the total number of ATP molecules that hydrolyzed after 1 min?

A: 1.2*10^-7 mol The total number of molecules that were hydrolyzed can be calculated by multiplying the rate in μMs^-1 by the time (in second) and volume of the solution (in L): 2*10^-6 mol*L^-1 * s^-1 * 60s * 1 * 10^-3 L = 1.2 * 10^-7 mol

The transition from N to O offers an exception to the trend for first ionization energy due to Hund's rules of spin pairing. If this spin pairing wasn't present, what would be the expected first ionization energy for the p-orbital on an oxygen atom?

A: 1700 kJ/mol In the figure from the passage, we see a dip between group 5 (N) and group 6 (O). If we examine the trend for the first 6 elements, IE goes from 500 (Li) to 1500 (N), giving a slope of about 200kJ/mol per. This trend would have continued, if spin pairing DID NOT occur, giving oxygen the IE of 1700. In actuality, the value shown is about 1300 kJ/mol, meaning spin paring allows for an IE difference, which represents the amount of destabilization that occurs for the p-type orbital containing two electrons of opposite spin.

In one trial of this experiment, significant impurities were detected in the extracted caffeine. Based on the results of the experiment, which of the following would be the expected melting point range for this extracted sample?

A: 195 - 220 celcius (I picked: 245-267) A melting point range includes the temperature when the first crystal of a compound starts to melt and the temperature when the compound is entirely melted. For a pure compound, the melting point range is narrow (this is how we use melting point determination to identify an unknown pure compound). The presence of impurities in a compound LOWERS and BROADENS the melting point. According to the passage, the melting point of pure caffeine is 235 celcius

What is the molar solubility of ferrous (II) hydroxide in water at 25°C?

A: 2.1*10^-5 The passage states that the Ksp of ferrous (II) hydroxide is 3.2 x 10-14. When Ksp is known, we can determine molar solubility from the dissociation reaction. Fe(OH)2 (aq) + H2O (l) → Fe2+(aq) + 2 OH- (aq) Note that Fe(OH)2 dissociates into three ions (one Fe2+ and 2 OH-). Given this 2:1 ratio, Ksp = [Fe2+][OH-]2 = [x][2x]2 = 4x3, where x represents the molar solubility. Next, we must divide Ksp by 4, then take its cube root to solve for x. Dividing 3.2 by 4 is more difficult than dividing 32 by 4, so we can manipulate scientific notation and rewrite Ksp in an easier format. Ksp = 4x3 = 3.2 x 10-14 = 32 x 10-15 8 x 10-15 = x3 2 x 10-5 = x

What is the ratio of neutrons in radon-202 to that in radon-206?

A: 29:30 The number of neutrons for each isotope can be found by subtracting the atomic number of radon (86) from the relevant mass number. Because radon-202 contains (202-86) = 116 neutrons, this gives us a ration of 116:120, which simplifies to 29:30

A student sought to remove the H2O and CO2 gas emitted by the animals in separate chambers instead of using silica gel and soda lime. 224 mL of each gas was collected at a pressure of 1 atm. What will be the expected ratio of the mass of CO2 to that of H2O if a valve between the two chambers was opened, allowing the gasses to freely mingle with one another? (Note: the temperatures of all chambers were held constant.)

A: 2:1 In this, as the two chambers are connected, the pressure and volume for each gas must change while the temperature is held constant. For both gases, the new volume is the total volume of both chambers. Given that the volumes of the two gases are equal (224mL each) and that they have the same initial pressure, they will have the same partial pressure in the new chamber. For CO2: PiVi = PfVf: (1 atm)(224 mL) = Pf (448 mL) Pf = 0.5 atm For H2O: PiVi = PfVf: (1 atm)(224 mL) = Pf (448 mL) Pf = 0.5 atm Next, use the partial pressure formula to find moles of each gas: Ptotal = (mol CO2)/(total mol) x Ptotal Ptotal = (mol H2O)/(total mol) x Ptotal Since both gases have the same partial pressure, and since the total number of moles and pressure are equal for both gases, they have the same number of moles. Thus, the ratio of the masses of the two gases is the ratio of their molar masses; (Molar mass of CO2) / (molar mass of H2O) = 44 g / 18 g ~ 2.

In a gas experiment, a student wished to measure the deviation of non-ideal xenon gas at various temperatures and pressures. For one mole of non-ideal xenon gas at a pressure of 1000 atm and a temperature of 100K, what is the most likely value for the quantity PV/RT?

A: 3 For one mole of an ideal gas (n=1) the quantity PV/RT equals 1 for all pressures. Deviations from the ideal gas law occur at high pressures and low temperatures. Here, the main assumption that is broken is that gas particles have no volume. Recall that as P increases, V (volume of a container) decreases. At a pressure as high as 1000atm, the gas must be compressed into a very small volume. Under these, the volumes of the gas particles will begin to be substantial in comparison to the overall volume, causing the real volume to increase (Vreal > Videal). If this is still confusing, imagine pressure increasing further, approaching infinite pressure. According to PV = nRT, volume would approach zero. However, at a certain point, volume will not be able to decrease, because the particle volume cannot be compressed. As a result, the higher the pressure, the more Vreal will deviate from Videal, or Vpredicted. If Vreal is significantly higher than expected, the PV term in PV/nRT will also be high, exceeding 1.

During the reduction of a mole of oxygen to water shown in Figure 1, how much charge is transferred?

A: 4*10^5 C First, we need to balance the reaction: O2 (g) + 4 H+ → 2 H2O (l) This reaction is now balanced in terms of atoms, but as a redox reaction, it must also be balanced with regard to charge. We currently have a +4 charge on the left and a 0 charge on the right. To balance, we must add 4 electrons to the left side of the reaction. O2 (g) + 4 H+ + 4 e- → 2 H2O (l) Now, we can see that 4 moles of electrons are transferred per mole of O2. Faraday's constant tells us that approximately 105 coulombs are present per mole of electrons, so: 4 mol e- x 105 C/mol e- = 4 x 10^5 C

Under what set of conditions would an air embolism be MOST likely to form in blood?

A: 42°C and PO2 = 120 mmHg According to the first and second paragraphs, an air embolism is caused when a bubble of air is formed in the blood and blocks a vessel. If a gas is insoluble in a fluid, it can escape as a bubble (think of warm soda going flat). Gases generally have greater solubility in cooler liquids. Also, the solubility of a gas increases as the partial pressure of the gas increases. This is primarily due to the hydrogen bonding effects of aqueous fluids that produce cavities within the fluid; these cavities can accommodate the gas molecules. Thus, we would expect an air embolism to form when gas is insoluble, which occurs in high-temperature, low-pressure systems.

Following the procedure of the experiment, how much total caffeine could you expect to extract?

A: 9.75g Caffeine is largely nonpolar, so it will be more soluble in dichloromethane than in water. Therefore, it will dissolve more in the dichloromethane layer. Using the partition coefficient equations and the caffeine properties provided by the passage, we must add up the amount of caffeine dissolved in each layer in both the first extraction and the second extraction. First extraction: Ψ ≅ solubility of solute in organic layer / solubility of solute in aqueous layer: Ψ = (140 g / 1000 mL) / (20 g / 1000 mL) = 7 Three tablets, each weighing 3.3 g, were used. Thus, the total amount of sample is around 10 g. If we assume that X grams dissolve in dichloromethane, then (10-X) g dissolve in the aqueous layer. Therefore: Ψ = (X g / 100 mL of CH2Cl2) / ([10-X] g/ 100 mL of H2O) = 7 X = 70 - 7X 8X = 70 X = 70/8 = 8.75 g dissolved in dichloromethane The second extraction was done using the aqueous layer, which holds 1.25 g caffeine initially. Ψ = (X g / 100 mL of CH2Cl2) / ([1.25-X] g/ 100 mL of H2O) = 7 X = 8.75 - 7X X ≅ 1 g Total caffeine extracted = 8.75 g + 1 g = 9.75 g

An additional experiment on the bovine extracts using a known ADAMTS inhibitor would serve as

A: A positive control A positive control is a control group not exposed to the experimental treatment but exposed to some other treatment that is KNOWN to produce the expected effect. In this passage, the scientists wished to test the effect of the analogs on enzyme activity. Using a KNOWN inhibitor is an example of comparing the expected effects of this inhibitor to those of the analog treatment, which is a positive control. - A negative control group is a group that's not exposed to the experimental treatment or any other treatment that is expected to have an effect. - In randomized controls, the groups that receive different experimental treatments are determined randomly - A false negative is when a test result appears negative when it should have been positive

The principal quantum number is a measure of which of the following?

A: Approximate radial size of an electron cloud The principal quantum number n is most closely associated with the potential energy of the electron. Since potential energy is proportional to the square of the distance of two oppositely charged particles by Coulombs's, n is associated with the radial size of the electron cloud.

Besides directly killing bacteria, how does pasteurization with hot water treat IPB populations?

A: By decreasing the solubility of oxygen, preventing further bacterial growth Oxygen, like all gases, is more soluble in cold water than warm water. Thus, the process of heating water (pasteurization) would lower solubility of oxygen and reduce the amount available to IPB. - Only gases are less soluble in warm than cold solutions. For instance, ferrous (II) hydroxide would be more soluble in warm water than cold water

What is the most likely formula for the ingredient shown on the label in Figure 1, called dibasic calcium phosphate?

A: CaHPO4 Phosphoric acid is H3PO4, and all three of its hydrogen atoms are acidic. As a result, there are three associated anions (PO43-, HPO42-, and H2PO4-), which are typically named phosphate, hydrogen phosphate, and dihydrogen phosphate, respectively. Alternatively, we can refer to these anions in terms of their ability to act as Brønsted-Lowry bases. For example, PO43- has the ability to gain three protons, so it can be called tribasic phosphate. Calcium ion is Ca2+, and when combined with the dibasic anion, HPO42-, the resulting formula is that in choice C.

For the student cell that produced the highest positive voltage, which of the following changes would produce an even higher positive voltage?

A: Decreasing the concentration of the Mg2+solution and increasing the concentration of the Cu2+ solution The overall net ionic redox reaction for this cell is: Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) The Nernst equation, E = E° - 0.059/n log Q, can be used to predict cell voltages under non-standard conditions. The reaction quotient, Q, takes the same form as the equilibrium expression, which in this case is Q = [Mg2+]/[Cu2+]. Here, we want our E value to be as positive as possible, so we need log Q to be as negative as possible (since subtracting a negative quantity from E° is the same as adding a positive quantity). For this to occur, the reaction quotient (Q) needs to be very small. Therefore, in order to produce the highest possible voltage, we want the concentration of magnesium ion to be essentially zero and the Cu2+concentration to be as high as possible. Choice D is the correct answer.

he reduction potential for Zn2+ is -0.66 V. Combined with the data in Table 1, which of the following most likely represents the standard reduction potentials for the half-reactions shown below? Cu2+ (1.0 M aq) + 2 e- + Cu (s) E1° = Ni2+ (1.0 M aq) + 2 e- + Ni (s) E2° = Mg2+ (1.0 M aq) + 2 e- + Mg (s) E3° =

A: E1 = 0.34V, E2 = -0.23V, E3 = -2.18V For a galvanic cell, the total cell potential is the sum of the reduction potential of the cathode and the oxidation potential of the anode. Since we're looking for reduction potentials, and since reduction occurs at the cathode, we are looking for E°cat. Let's begin with Cu2+ and consider the Zn (s)/Zn2+ (1.0 M aq)//Cu2+ (1.0 M aq)/Cu (s) cell from the data for Group 1. In the first paragraph of the passage, it states that zinc is the anode and copper is the cathode for this cell. Therefore, we need to reverse the sign of the reduction potential for zinc to turn it into the oxidation potential. E° = E°cat + E°an 0.99 V = E°cat + 0.66 V E°cat = +0.33 V Note that the passage states, "In all cases, the voltage readings were slightly less than the theoretical values for these cells." Therefore, our calculated value of +0.33 V is reasonably close to the theoretical +0.34 V given in answer options A, B, and C. Choice D can be eliminated. Now, if we use the oxidation potential for zinc (+0.66 V) in the Zn (s)/Zn2+ (1.0 M aq)//Ni2+ (1.0 M aq)/Ni (s) cell in a similar analysis using 0.43 V as the total cell potential, we get: E° = E°cat + E°an 0.43 V = E°cat + 0.66 V E° cat = -0.23 V This value is shown only in choice C, so we can choose that option and move on. However, if you had time and wanted to be absolutely certain, you could check the reduction potential for magnesium. Unfortunately, we do not know whether zinc is oxidized or reduced in its reaction with magnesium. If we assume that zinc is oxidized, and if we use the observed total potential from Table 1 for Mg (s)/Mg2+(1.0 M aq)//Zn2+ (1.0 M aq)/Zn (s), we get: E° = E°cat + E°an 1.51 V = E°cat + 0.66 V E°cat = +0.85 V This is not close to the value in choice C, but we still need to consider the possibility that magnesium is oxidized (at the anode) and zinc is reduced (at the cathode). We can assume that this is true, solve for the oxidation potential of magnesium, and then switch the sign to obtain the reduction potential (which the question asks for). Using the reduction potential for zinc gives: E° = E°cat + E°an 1.51 V = -0.66 V + E°an E°an = +2.17 V E°cat, then, must be equal to -2.17 V. This is close to the value for magnesium given in choice C. Similar analyses could be done with any of the other combinations of cathodes and anodes

The crystal precipitate formed in collected effluent was composed principally of calcium carbonate, with strontium carbonate inclusions. The large majority of the crystal units did NOT include strontium because the:

A: Effluent strontium ion concentration was much smaller than the effluent calcium ion concentration (I picked: "solubility constant of strontium carbonate is less than that of calcium carbonate") Precipitate forms when the ion product, Q, for a compound exceeds the Ksp for the same compound. Both Q and Ksp take the same form, but Q includes current ion concentration while Ksp must use equilibrium concentrations. The Ksp values given in the passage for the carbonates of calcium and strontium are very similar. However, the influent and effluent concentrations of calcium ion are much greater than strontium ion, suggesting that the Q of calcium carbonate is much more likely to exceed its Ksp. Side note: pure liquids (l) and solids do not appear in equilibrium constant expressions such as Keq, Ksp, and Ka. Only gaseous and aqueous (aq) components should be included

What is the primary process responsible for the loss of latent heat and entropy from the ocean at the air-sea interface

A: Evaporation A negative latent heat implies that the phase change that is occurring is consuming energy. Therefore, the phase change that is occurring is melting, evaporation or sublimation. The fact that this process is occurring at the air sea interface means that the phase change must involve the gas phase. That fact rules out melting. Since sublimation is not an option, will go with evaporation. It makes sense that with evaporation, the ocean is losing entropy at the expense of the atmosphere which is gaining gas particles. - An example of a positive latent heat for ocean is condensation as it's a process that releases energy

A certain type of tissue is sensitive to radiation with the damage the tissue receives being directly proportional to the charge on the irradiating particle. Which of the following radiation types will cause the least damage?

A: Gamma Gamma particles have no charge as they are simply high-energy photons. Since the question says that the damage is proportional to charge, a particle with no charge would cause the least damage - Alpha particles have a +2 charge and would be the most damaging

Which of the following statements correctly describe the methods used in the caffeine experiment? I. The retention factor in a TLC procedure depends on the solvent system, temperature, and the adsorbent. II. A polar compound will exhibit a smaller retention factor on a TLC plate than a less polar compound. III. Anhydrous methanol has a greater eluting strength than pentane when used as solvents in a TLC procedure.

A: I, II, and III I is true; TLC depends on the different affinity of a compound for the stationary vs. mobile phase. Therefore, depending on how polar the solvent and adsorbent are, a compound moves on a TLC plate at a certain rate. Temperature also affects the rate of movement II. A polar compound will be attracted to the adsorbent (via dipole-dipole) and move slower on a TLC whereas a nonpolar have more affinity for the mobile phase and move faster III. The eluting strength depends on how strongly a compound ADSORBS onto the adsorbent. Since typical adsorbents are highly polar, eluting strength increases with increasing solvent polarity.

Which of the following is (are) present in a mixture of NH3 and BF3? I. FLP II. Brønsted-Lowry acid III. Brønsted-Lowry base

A: III BL bases are species that accept protons, and BL acids are species that donate protons. NH3 has a lone pair on the N that accepts protons to form NH4+, or ammonium. Thus, it is a BL base and RN III is correct. I: The passage states that FLP are mixtures or compounds that contain Lewis acids and Lewis bases that are sterically hindered, such as Compound 1. Though NH3 and BF3 are Lewis bases and acids, they are not sterically hindered, so they are not FLP. II: NH3 is a BL base, but BF3 is not a BL acid because it does not have a proton to donate.

A researcher carries out a column chromatography at physiological pH, using a stationary medium with a net positive charge. If a solution containing the following oligopeptides is poured into the column, which oligopeptide will most likely be found in the first fraction collected?

A: KRVV If the stationary phase has a net positive charge, then oligopeptides with negative charges will be attracted to the stationary phase and will move more slowly through the column. KRVV has a charge of +2; it would not bind the stationary phase and elute quickly. Thus, it would move the quickest through the column and be found in the first fraction collected

A patient presents in the emergency department having ingested a large quantity of tolbutamide. Intravenous administration which of the following compounds is most likely to increase the rate of urinary excretion of the drug? A. KCl B. NaHCO3 C. NH4ClO4 D. NaCl

A: NAHCO3 To increase the percentage of drug excreted in the urine, we must decrease the fraction of tolbutamide capable of reabsorption, or diffusion out of the lumen of the nephron. Passage indicates that for weakly acidic drugs, the uncharged state is capable of diffusion through membranes much more than the charged form. Thus, we must maximize the charged form of tolbutamide to prevent reabsorption. We do this to a weak acid by deprotonating the drug via administration of a base. This will increase blood and urinary pH and increase the fraction of ionized drug present. Urinary alkalization can be accomplished by administration of basic salt like NaHCO3.

Turbulent flow in humans is a major risk factor for atherosclerosis, the buildup of plaque on arterial endothelium. Ignoring any potential effects of turbulence, what effect would atherosclerosis have on blood flow?

A: Narrowing of the artery causes the velocity to increase and the hydrostatic pressure to decrease The buildup of plaque in the lumen of the vessel would cause a narrowing of the vessel. The question then implies that we must consider the Venturi effect, which states that a fluid's velocity must increase as it passes through a constriction, based on the continuity (A1v1 = A2v2) principle. According to the Bernoulli principle, then, the hydrostatic pressure must decrease to conserve energy. - The osmotic pressure (determined by the solute composition of the fluid) will not be affected by this narrowing

A student places a vial of gas at the bottom of a graduated cylinder. A liquid sample is suspended in a chamber at the top of the cylinder. The vial is broken and the gas is allowed to diffuse throughout the cylinder. When the two compounds mix, a black precipitate is formed. Which gas should the student choose to minimize the time required for the precipitate to form?

A: Neon Rate of effusion is inversely related to the square root of molecular weight: rate ∝ 1/√ (MW). Thus, the smaller the molecular weight, the higher the rate of effusion. Examining the periodic table, we can see that neon has the lowest molecular weight.

What was the most likely purpose of adding bovine serum albumin to the kinetics experiments in the passage? Bovine serum albumin:

A: Prevents the esterase from adhering to the walls of the vessel Passage states that albumin is a protein that mobilizes proteins and lipids in serum. In the context of a kinetics experiment, it's logical to assume that albumin is added to maintain homogeneity and prevent the enzyme from adhering to walls and other surfaces which would inhibit its activity

What of the following types of electromagnetic radiation would have the shortest wavelength?

A: Radiation that ejects an electron from an sp orbital Shorter-wavelength EMR (like gamma rays) carries much more energy than longer wavelength. Thus, we should look for the highest-energy EMR. The closer an electron is to the nucleus, the harder it is to eject. Therefore, the most s character of all of the answer choices contains the electrons hardest to eject.

Given the following elementary reaction, what is the rate law? 3A+B -> 2C

A: Rate = k[A]^3[B] For elementary reactions, stoichiometric coefficients can be used to write the rate law. These coefficients become exponents according to the following theoretical example: for the reaction aA + bB → cC, rate = k[A]^a[B]^b.

During strenuous exercise, lactic acid buildup in cells causes the creation of a hydronium complex known as the Eigen cation (H9O4+). If water molecules then experience hydrogen bond attractions to the Eigen cation, this attractive force:

A: Results in a semi-stable shell of water molecules around the hydronium Process known as "hydration" or "solvation" occurs when the attractive force of an ion molecule causes a think shell of water molecules to surround it. For hydronium, each of the H attracts the O in H2O due to hydrogen bonding. These H2O molecules cause a "shell" of water molecules to surround the hydronium. - Regular H+ ions almost never exist in H2O solutions. Instead, they combine with an H2O to form H3O+.

Which of the following formulas represents a general structure of a fatty acid salt produced in reaction 1 (saponification)

A: Rn-CO2- Na+ A fatty acid has the general formula Rn-CO2H. The corresponding fatty acid salt has the general formula Rn-CO2- Na+

Under anaerobic conditions, bacteria can sometimes derive energy from the oxidation of sulfur-containing species. Which of the following can NOT be oxidized by anaerobic bacteria?

A: SO42- (aq) Of the answer choices, the oxidation number for sulfur would need to be in a reduced form. The highest oxidation state for sulfur is +6, corresponding to the loss of all of its valence electrons. The oxidation number for sulfur (x) in sulfate, SO42-, can be determined by assuming that the oxygen atoms are oxides having a -2 charge and that the sum of the oxidation numbers must equal the overall charge of the ion, giving -2 = x + 4(-2) -2 = x - 8 +6 = x The oxidation number of sulfur in sulfate is +6, and choice B is the best answer. The oxidation number of sulfur in hydrogen sulfide (H2S), thiosulfate (S2O32-) and yellow elemental sulfur (S8) are -2, +2 and 0, respectively, all of which could be oxidized and be a source of energy for anaerobic bacteria.

Efforts to treat lactic acid buildup in muscles were attempted using dissected muscle specimens in the laboratory. One of these experiments involved ammonium formation from dissolved ammonia. Under conditions of excessive lactic acid:

A: The final concentration of ammonium will be higher than otherwise due to lower pH In acidic environment, a base such as ammonia (NH3) dissolves into ints conjugate acid ammonium (NH4+) to a greater extend that would have been in a neutral/basic environment.

Pulmonologists use BCG to measure respiratory rate during exercise by having patients exhale into an aqueous solution containing the indicator. Which of the following would be observed in the final solution?

A: The formation of carbonic acid and the solution turning yellow The patient's exhalation will cause CO2 from his breath to be absorbed into the solution. As we've seen in the blood, when carbon dioxide mixes with water, it forms carbonic acid. CO2 + H2O -> H2CO3 -> H+ + HCO3-. As shown, when this carbonic acid dissociates, it will cause the water to go from neutral to acidic, so the BCG will end up in its acidic form and turn yellow.

When aqueous solutions of the various anions and cations were mixed, precipitates formed because

A: The solubility product of a compound was exceeded The amount of a substance that will DISSOLVE IN WATER is described by the Ksp. If the amount of the compound present is in excess of the Ksp, a precipitate would form to maintain the Ksp

The ATP-dependent phosphorylation of a protein target is catalyzed by which class of enzyme?

A: Transferase Kinases catalyze the transfer phosphate groups from ATP to target proteins and are classified as transferases.

Absorption of radiation corresponding to what portion of the electromagnetic spectrum is most likely to excite Aequorea fluorophore, a green fluorescent protein used in biological experiments?

A: Ultraviolet Fluorescence is the emission of a lower energy photon from a fluorophore excited by the prior absorption of a higher-energy photon. Thus, the Aequorea fluorophore must be excited by the absorption of a photon of higher energy (and frequency) than the associated with green visible light.

The unknown compound was probably a

A: Weak acid (Picked "strong acid") The passage states: "The compound completely dissolve in water and weakly conducted electricity. The hydrogen ion concentration of the unknown aqueous solution was 1*10^-5M"; picked my answer because "completely dissolve". However, because the hydrogen ion concentration for this solution is greater than 1*10^-7 but much less than the concentration of the compound itself, the compound only partially ionized to form hydrogen ions, which is a behavior of a weak acid.

In comparison to the cohesive forces between water molecules of the protein solution droplet, how can the strength of interaction between water and oil molecules at an oil-water interface be characterized?

A: Weaker, because they are forces creased by induced polarity in nonpolar molecules The Van der Waals forces existing between molecules of water and of oil are predominately of the induced dipole-dipole. Here, a small, temporary dipole is induced in molecules of oil by the permanent dipole of water, resulting in a weak attraction between the molecules. This is a weaker interaction than the hydrogen bonding predominating the interaction between water molecules.

Would deviations from the ideal gas law be observed for gaseous nitrogen at 180GPa and room temperature?

A: Yes, because at this pressure, molecular volumes and intermolecular forces become significant According to the passage, the solid form of nitrogen is stable above 65 GPa. The ideal gas law assumes that molecular volumes and intermolecular attractive forces are negligible. At a pressure at which solid nitrogen is predicted to be stable, these assumptions are invalid.

Before coordination to protoporphyrin, what is the ground state electron configuration of Co2+?

A: [Ar]3d^7 During the ionization of transition metals, electrons from 4s subshell orbitals are generally removed before those from 3d subshell orbitals. This is because electrons of 4s subshell orbitals are higher energy than those in 3d subshell orbitals

What is the electronic configuration of the Co(II) center found in vitamin B12?

A: [Ar]3d^7 (I picked: [Ar]4s^2 3d^7) CO(II) is a diction formed from the atomic element by the loss of two 4s electrons (outer shell). Thus, only 7 3d electrons remain in the valence shell.

According to the description in the passage, Earth is:

A: closed system Earth satisfies both conditions of a closed system. There IS energy exchange but not mass exchange with surroundings. - An isolated system exchanges NEITHER energy nor mass with its surroundings.

Assume the hydrolysis of ATP proceeds with G= -30kJ/mol: ATP + H2O -> ADP + Pi Which expression gives the ratio of ADP to ATP at equilibrium if the [Pi]=1M? (RT=2.5KJ/mol)

A: e^12 The free energy of the reaction G is related to the equilibrium constant Keq = [ADP][[Pi]/[ATP] as G= -RTln(Keq). Thus, by plugging in, [ADP/ATP] = e^12

More complete fractionation of proteins using an Size-exclusion chromatography (SEC) column could be achieve by using a

A: longer column Like other chromatographies, increasing the column length enhances the resolution of the column, leading to more completion fraction by SEC by providing the physical means of separating the proteins.

Which of the following electronic transitions for a hydrogen atom would result in the emission of a photon that would be visible to the human eye?

A: n=4 to n=2 The visible spectrum contains electromagnetic signals with wavelengths ranging 400-700nm. The wavelength of light emitted during a particular electronic transition determined by the energy difference between the initial and final energy levels. For light to be emitted AT ALL, an electron must travel from a higher to a lower level to release energy.

The combustion of ethanol forms carbon dioxide and water. If 2 moles of ethanol and 4 moles of oxygen are combined in a combustion, the limiting reagent is

A: oxygen It is not ethanol because we are NOT given a balanced reaction. Once we write a balanced equation, C2H5OH + 3 O2 → 2 CO2 + 3 H2O; we see that reacting two moles of ethanol requires 6 moles of oxygen.

Aluminum belongs to what block of elements in the periodic table?

A: p Aluminum is found in group 13 (3A) of the periodic table; it is a p-block element

Consider the chemical reaction NH4+(aq) + NO3- (aq) → N2(g) + 2H2O(l). Trial [NH4+] [NO3-] [Reaction Rate] 1 0.010M 0.02M 0.20M/s 2 0.015M 0.02M 0.30M/s 3 0.010M 0.01M 0.05M/s

A: rate = k[NH4+][NO3-]2 With rate law problems, we should first identify two trials where only one of the reactants changes concentration. Between trials 1 and 3, we can see that the concentration of nitrate doubles while the rate quadruples. This means that the reaction is second order with respect to nitrate (eliminate choices A and C). Between trials 1 and 2, the ammonium concentration stays the same while the nitrate concentration increases by a factor of 1.5. We can see that this causes the reaction rate to increase by a factor of 1.5, indicating that the reaction is first order with respect to ammonium

In locations with very low dissolved O2 concentrations in drinking water, the observed Ksp of ferrous (II) hydroxide will be

A: the same Equilibrium constants, including Ksp, are not affected by concentration. These constants are ONLY altered by changing temperature

In which of the following aqueous carbohydrate solutions would a given vesicle have the greatest buoyancy? A. 0.15 M sucrose B. 0.15 M glucose C. 0.20 M fructose D. 0.25 M galactose

Answer: A Buoyancy is the force that results from the displacement of fluid when an object is submerged. The magnitude of the buoyant force is equal to the weight of the fluid displaced. Therefore, the greater the density of the solution, the larger the buoyant force. Glucose, fructose, and galactose are all monosaccharides with the molecular formula C6H12O6, so their relative densities will depend on their concentrations. The most dense of those three choices, then, is choice D. However, choice A is sucrose, a disaccharide with the formula C12H22O11. Since sucrose has a molecular weight almost twice that of galactose, a 0.15 M sucrose solution would have a density approximately equivalent to that of a 0.3 M galactose solution. Choice D is not this highly concentrated, so we can eliminate it and choose option A.

#7. Which of the following has the smallest atomic/ionic radius? A. K+ B. Ca2+ C. Ar D. Cl-

Answer: B All four answer choices are isoelectronic: they have the exact same configuration and number of electrons. The main difference between them is the number of protons in their nucleic. Protons, which are positively charged, attract electrons. Calcium has the highest atomic number (the most protons) so its electrons will be pulled closest to the nucleus. - As a general rule, cations have the smallest radii and anions have the largest radii

Which of the following phosphate ions is amphiprotic? A. PO43- B. HPO42- C. P2O74- D. H3PO4

Answer: B An amphiprotic species is one that can act both as an acid and a base. Hydrogen phosphate HPO42- can dissociate a hydrogen ion to form PO43- and therefore act as an acid. It can also accept a hydrogen ion from water to form H2PO4- and act as a base

If the average bone mineral density is 3.88 g/cm3, which of the following is a reasonable estimation of the density of the average human body? A. 0.001 g/cm3 B. 1 g/cm3 C. 3.88 g/cm3 D. 4.88 g/cm3

Answer: B Bones are the most dense part of the human body so the overall body should have a density lower than 3.88g/cm^3. We can also consider the fact that humans typically float right at the surface of water, implying that the density of the human body is approximately equal to the density of water.

Hydrogen bonds vary in strength depending upon the donor and acceptor atoms. Which of the following hydrogen bonds has the lowest bond dissociation enthalpy? (A dashed line indicates the hydrogen bond in question.) A. F-H - - - :O B. N-H - - - :O C. O-H - - - :O D. P-H - - - :O

Answer: D This question is asking for the weakest bond. Since the acceptor atom is oxygen in all four choices, we need only consider the effect of the electronegativity of the donor atom on the strength of the bond. Fluorine is the most electronegative atom on the list, followed by oxygen, then nitrogen, and lastly phosphorus.

What happens when HCl reacts with CO3^2-?

When HCl is added and reacts with CO3^2-, it forms CO2(g)

What is the net charge on a phenylalanine molecule at pH 1?

A: +1 Being an amino acid, phenylalanine has an acidic carboxy group that will be protonated at a pH of 1 (remember, such a pH is highly acidic). Additionally, it has a basic amino group that will also be protonated. Finally, it contains a neutral toluene side chain. In total, the charge will be (0 from the carboxy group) + (+1 from the amino group) + (0 from the side chain) = +1. *When the carboxy group is protonated, it's neutralized with a charge 0 not +1*

Of the isomers below, that with the highest boiling point is: A. 2-methyl-2-butene. B. trans-2-pentene. C. 3-methyl-2-butene. D. cis-2-pentene.

Trans-2-pentene Boiling point is affected mostly by intermolecular forces and molecular weight. However, these are all hydrocarbon chains. For a hydrocarbon chain, the less branching resembles a more straight change which will "stack" atop like and have higher boiling points.

The heat of vaporization of water is 2256kJ/kg, while the heat of fusion is 334 kJ/kg. What amount of heat is required to fully convert 120g of ice at 0 degree into water, also at 0 degree?

a: 40080J Here, we want to use the heat of fusion not vaporization. Note that Q=McT only works when temperature is changing; thus, for phase transitions, we use Q=mL where L is the latent heat. In this, we have (0.12Kg)(334KJ/Kg), which equals 40000J

According to equation 1, the concentration of the polymer with respect to [HPO4^-2] is

A: (1/n)[HPO4^2-] Equation 1: nCDP -> (CP)n + nHPO4^2- According to the balanced coefficients, each time one molecule of the polymer (CP)n is formed, there also are n hydrogen phosphate ions HPO4^-2 produced. Therefore, the concentration of (CP)n must be 1/n times the concentration of HPO4^-2 Q2. The pKa for the dissociation of H2PO4- to HPO4^2- is 6.7. What is the initial ratio of [HPO4^2-]:[H2PO4-] in the buffer solution of experiment 1? The acid dissociation reaction referred to in the question is: H2PO4- (aq) -> HPO4^-2 + H+ The equilibrium constant for this reaction is the acid dissociation constant Ka: [HPO4^-2][H+]/[H2PO4^-]. The acid dissociation constant is often presented as the pKa where pKa = -log10Ka. At a given temperature, the relative concentrations of the weak acid H2PO4^- and its conjugate base HPO4^-2 will depend on [H+] and Ka of the acid. Since the solution was buffered to pH = 8.7, [H+] is held constant. Since Ka is also a constant, the ratio of the concentrations of weak acid and conjugate base must be fixed. Thus, we can simply solve: pH = pKa + log([conj. base]/[conj. acid]) = pKa + log([HPO4^-2]/[H2PO4^-]) where log([HPO4^-2]/[H2PO4-]) = 2. Taking the antilog, 10^2 = [HPO4^-2]/[H2PO4-] = 100:1

The table below gives the reduction potentials of two metals. Ag+ + e- -> Ag (E = 0.8V) Sn2+ + 2e- -> Sn (E = -0.14V) A system is constructed with two electrodes, one made of solid silver and the other composed of solid tin. Current from an external generator is then applied to power a reaction. If the species above are the only ones that react in the solution, the E°cell for this system could be

A: -0.94V The stem mentions the use of external current to promote this reaction, indicating that it must be nonspontaneous (electrolytic). Thus, the tin cation would reduce (E° = -0.14 V) while the silver metal would oxidize (E° = -0.80 V). (-0.14) + (-0.80) = -0.94. - If the answer was 0.94V, it indicates a galvanic cell due to the lack of a negative sign.

What is the approximate pH of a saturated aqueous solution of hydrochloric acid whose molarity is 10.6M?

A: -1 Hydrochloric acid is a strong acid that completely dissociates in aqueous solution. In this problem, the hydronium ion concentration is 10.6M. The pH is the -log of the hydronium ion concentration: -log[10] = -1. - Note that it's possible to have a negative pH and pH value greater than 14

If one mole of sucrose depresses the freezing point of a certain volume of water by 3.9 degree, what would be the freezing point of the same volume of water when one mole of malonic acid is added?

A: -11.7 degree Because freezing point depression is a colligate property, it depends on the number of particles in solution. Sucrose is organic and doesn't dissolve in water but carboxylic acids lost a proton at pH ~7; thus, malonic acid dissociates into three particles: its deprotonated form and two free protons. Thus, one mole will depress this volume by 3 times the amount the sucrose affected it, -11.8 degree.

1.25 mol of Ca(NO3)2 is added to exactly 0.25 kg of water. If the Kf of water is 1.86 °C ∙ kg / mol, what will the solution's new freezing point be?

A: -28 celcius Freezing point depression can be calculated using ΔTf = Kfmi, where Kf is a constant, m is molality, and i is the Van't Hoff factor, or number of dissolved particles per molecule. Here, m is (1.25 mol) / (0.25 kg) = 5 m, while i = 3. ΔTf = Kfmi = (1.86 °C ∙ kg / mol)(5 m)(3) = approximately 28°C. Since water normally freezes at 0°C, its new freezing point will be -28°C.

When 2 moles of hydrofluoric acid are added to 100mL of water, the resulting solution has a pH ~4. What is the percent dissociation of HF?

A: 0.0005% Remember that HF is a weak acid that doesn't fully dissociate in water. This question gives pH: pH = -log[H+]; therefore, [H+] = 10^-pH where [H+] = 10^-4 M. "Percent dissociation" indicates the percent of the original acid concentration that has dissociated into H+ and F- ions; this value equals to [H+]/[HF] * 100%. Original [HF] = 2moles/0.1L = 20M Thus, % dissociation = (10^-4 M H+)/(20M HF) = 0.0005%

What is the molar concentration of Na+ (aq) in a solution prepared by mixing 10mL of 0.01M NaHCO3 (aq) solution with 10mL of 0.01M Na2CO3(aq) solution?

A: 0.015 mole/L Since 1 equivalent NaHCO3 provides 1 equivalent of Na+, the molar concentration of Na+(aq) in 0.01M NaHCO3(aq) solution is also 0.01M = 0.01 mole/L. The molar concentration of Na+ (aq) in 0.01M Na2CO3 provides 2 equivalents of Na+. When equal volumes of these two solutions are mixed, the resulting molar concentration is equal to the average, (0.01mol/L + 0.02mol/L)/2

What is the approximate partial pressure of oxygen at 1500m?

A: 0.18 atm Table in the passage shows the barometric pressure at 1500m = 641 mmHg or 85 kPa. *Remember the conversion* 1 atm = 760 mmHg = 10^5 Pa; thus, we conclude that the pressure at 1500m is 0.84atm. The passage also states that air is 21% oxygen, meaning ~21% of the 0.84atm will be due to oxygen. According to Dalton's law of partial pressures, 0.84 atm * 0.21 = 0.18 atm.

How much heat is produced from the complete combustion of 30.0 g of methane, if the enthalpy of reaction is -890 kJ/mol?

A: 1.7 * 10^6 J Methane has a molecular formula of CH4 with its molecular weight 16g/mol. 30g CH4 * (1 mol/16g) * (890kJ/mol) = 1.7*10^6 J

Which of the following relationships must hold? (note 1m^3 = 1000L)

A: 100 kPa is equivalent to 100J/L This is a unit conversion problem! 100 kPa = 100,000 Pa = 100,000 N/m2 = 105 N/m2 100 J/L x (1000 L/m3) = 105 J/m3 = 105 Nm/m3 = 105 N/m2

What would be the approximate ratio of lactate to lactic acid in the final lactated ringer's solution after the initial preparation by Student A?

A: 1000:1 The Henderson-Hasselbalch equation is used to solve for the ratio of the conjugate base (lactate) to the acid (lactic acid) in a buffer solution of known pH. According to the equation, pH = pka + log[A-]/log[HA] where [A-]/[HA] = 10^(pH-pKa). pH is given as 6.6 so we only need pKa. It's given that Kb for lactate is 2.5*10^-11; Ka is found using Ka*Kb = 10^-14 at 25 celsius. Ka = (1*10^-14)/(2.5*10^-11) = 4*10^-4. pKa = -logKa = -log(4*10^-4) We use our shortcut: -log[C*10^-E] = (E-1).(10-C); thus, we find that -log(4*10^-4) = 3.6. [A-]/[HA] = 10^(pH-pKa) = 10^(6.6-3.6) = 10^3 = 1000:1

What is the sum of the protons, neutrons, and electron in strontium-90?

A: 128 In a neutral atom, the number of electrons is equal to the number of protons. The number of neutrons can be found by subtracting the atomic number from the mass number. The atomic number of strontium (Sr) is 38; thus, the number of neutrons is 52. The sum of protons, neutrons, and electrons in strontium is therefore 38+38+52=128

A hospital purchases brand-new GKS-Co and GKS-X machines. Five years after installation, what is the expected ratio of the total atomic mass of radioactive material (both before and after decay) in the Co machine to that in the X machine, assuming both machines start with the same mass of radioactive material?

A: 1:1 This question is NOT asking about the percentage of an isotope left after radioactive decay. Instead, it is asking about atomic mass. B-decay does cause a nuclear transmutation of protons to neutrons or vice versa, the atomic mass lost in these processes is negligible; thus, the atomic mass will be the same in both samples.

If an archaebacteria species lives in a pool that is 0.01M HCl(aq), what is the pH of the water?

A: 2 The pH of a solution is -log[H+]. The strong acid HCl completely dissociates in water; thus, an HCl concentration of 0.01M means that the H+ concentration is 0.01M (10^-2 M); pH is 2 Q2. What is the pH of a 0.001M NaOH solution? A: 11 A 0.001M NaOH solution has a pOH(-log[OH-]) of 3. The pH of the solution is therefore 14-3=11

1.5M tartaric acid (a diprotic acid) is titrated with 0.5M KOH to generate a diprotic titration. If the initial volume of tartaric acid was 750mL, what volume of KOH is required to reach point 2 (1st equivalent point) on the curve?

A: 2.25L Point 2 is the first equivalence point where we need to add enough KOH to neutralize the *first proton* (*1/2* of the amount required to fully neutralize tartaric acid). We use N1V1=N2V2:(3N tartaric acid)(0.75L) = (0.5M KOH)(x L). Solving for x, we get a volume of 4.5L. However, this amount gives us the volume of base needed to fully neutralize *both* protons. Dividing it by two, 2.25L

What is the pH of a 0.10 M aqueous solution of acetylsalicylic acid?

A: 2.3 Acetylsalicylic acid is a weak acid with pKa 3.5. Thus, the pH of this solution must be less than the pKa because the compound is in its acid form and pH of 3.5 would mean that the concentration of weak acid and conjugate base were equal (a buffer). Ka = [H+][A-]/[HA] 10^-3.5 = x^2/(0.1-x) Since x will be small, we can approximate 0.1-x ~ 0.1, giving: 10^-3.5 = x^2/(0.1) 10^-3.5(0.1) = x^2 10^-4.5 = x^2 [10-^2.25][10^-2.25] = ^ 10^-2.25 = x Since x equals the hydrogen ion concentration, taking the -log(10^-2.25) = 2.25.

The frequency used in ultrasound imaging must be greater than

A: 20 kHz Ultrasound is sound with a frequency above the human range of hearing (20Hz-20kHz); thus, anything greater than 20kHz will be ultrasound

Na2CO3 + HCl -> CO2 + H2O + NaCl Consider the above unbalanced equation. For this reaction, how many mL of a 2M solution of Na2CO3 are required to produce 11.2L of CO2 at STP?

A: 250mL First, we must balance the equation: Na2CO3 + 2HCl -> CO2 + H2O + 2NaCl Because one mole of an ideal gas at STP occupy a volume of 22.4L, 11.2L of CO2 gas at STP must represent 0.5 mole of CO2. Because of the mole ratio, one mole of CO2 is produced when one mole of Na2CO3 reacts. Thus, the amount of Na2CO3 required to produce 0.5 mole of CO2 gas must also be 0.5 mole. From its mole ratio with Na2CO3, we find that the answer is 250mL

Electrons in sample 2 (ultraviolet) could have made which of the following jumps, as indicated by their principal quantum numbers?

A: 4 to 1 To determine the final principal quantum number involved in the transition, we must be familiar with the different series of spectra. The Lyman series (ultraviolet rays) involves any emission with the ground state of electron n=1. Balmer series (visible rays) has the final state of n =2 Paschen series (infrared) contains emission to n = 3

A restaurant manager wants to melt 10 kg of ice at -8°C into water at 20°C. If his restaurant has an ambient pressure of 1 atm, how much heat will the entire process require? Note that cice = 2.03 J/(g∙°C), cwater = 4.18 J/(g∙°C), and Hfusion (water) = 334 J/g.

A: 4.3*10^6 j This process must be broken down into steps. First, ice at -8°C becomes ice at 0°C. Plugging known values in to Q = mcΔT yields Q = (1 × 104 g)(2.03 J/(g∙°C))(8°C) = 1.6 × 105 J. Next, ice at 0°C is melted to form water at the same temperature. Since this is a phase change, we must use the latent heat of fusion, giving us (1 × 104 g)(334 J/g), or 3.34 × 106 J. Finally, the temperature of water is increased from 0 to 20°C. This step requires (1 × 104 g)(4.18 J/(g∙°C))(20°C), or 8.4 × 105 J. In total, the sum of these values gives us 4.3 × 106 J.

The osmolarity of blood is approximately 298 mOsm. What is the difference in osmotic pressure between the blood and lactated Ringer's solution at standard temperature?

A: 50 Pa The osmotic pressure can be found by the equation n=MRT. To find the total molarity of lactated Ringer's solution, simply add up the concentrations of ALL ions shown (sodium + chloride + lactate + potassium + calcium) = 0.2725 Osm. We are given the osmolarity of blood as 0.298 Osm. Thus, △п = (0.298 - 0.2725)(8.314 x 273) = 0.025 x 8.314 x 273 = 54 Pa

Ignoring stereochemistry, how many different tripeptides may exist that contain the same three amino acids as the molecule shown below?

A: 6 The formula for the number of possible peptides that contain one each of n amino acid is *n!*. Given compound has n=3, thus, the answer is 6. For a tripeptide ABC, the following combinations are possible: ABC, ACB, BAC, BCA, CAB, and CBA

A student combines 364.6g of HCl with 80g NaOH in 5L of water. What additional H2O must be added to the mixture to yield a solution with a pH 1?

A: 75L From here we have 10 moles of HCl and 2 moles of the strong base. When mixed in solution, the base fully neutralizes 2 moles of HCl according to the 1:1 stoichiometric ratio between these reagents. 8 moles of HCl remains in 5L of water. However, we need a total H+ concentration of 0.1M to yield a pH of 1; which can be obtained by adding enough water to establish a total volume 80 L (MV=MV where 8=(0.1)(80))

Which of the following most closely approximates the pKa of phenolphthalein?

A: 9.3 Indicators undergo a color change near the desired *pH*; this typically occurs because of a reversible change in the protonation state of the indicator. It's desirable that the *pKa* of a chosen indicator is within +/- 1 of the target pH. For the titrations performed in the study of acetic acid (weak acid) and sodium hydroxide (strong base), the endpoint of the titration occurs at a pH greater than 7. Thus, only 9.3 will function well as an indicator for a pH in this range.

Al3+ has a reduction potential of -1.66 while Cd2+ has a reduction potential of -0.4. In an electrolytic cell at standard conditions

A: Al3+ will reduce at cathode and Cd will oxidize at anode. *Because Al3+ has a more negative reduction potential* than Cd2+, it is Al3+ that will reduce and Cd (s) that will oxidize. - More negative reduction potential, reduced!

At which electrode is aluminum produced in a galvanic cell and an electrolytic cell?

A: At the cathode in both cells Reduction of the Al3+ to form Al(s) occurs at the cathode both in a galvanic and an electrolytic cell

Ba2+ is an ion that is very toxic to mammals when taken internally. Which of the following compounds, mixed in water, would be the safest if accidentally swallowed?

A: BaSO4, Ksp = 1.1*10^-10 (smallest amount given) The lower Ksp, the lower the concentration of the cation and anion in an aqueous solution and the lower the solubility of the compound in water. If mixed with water and accidentally swallowed, the Ba salt with the lowest value of Ksp would be the safest

If trifluoroacetic acid is the reagent used in SP synthesis, which of the following would be the most effective for creating an effective support system? A.NH3 B.NaOH C.p-bromophenol D.Benzene

A: Benzene The passage indicates that an ideal support must be inert to all reagents and solvents used during SP synthesis. This tells you that we are using a weak acid; we want a molecule that will be inert to the actions of an acid. Benzene, an aromatic hydrocarbon, is very stable (inert) in the presence of acids/bases and won't react with the reagents. - Ammonia is weakly basic and will interact with the weak acid; sodium hydroxide is a strong base and will react with the acid. - p-bromophenol is a benzene ring with an electron-withdrawing bromine and hydroxyl group attached. The "p" indicates that the two substituents are located para to each other on the ring. This results in an acidic proton on the hydroxyl group, indicating that it can possibly react with the reagent mentioned by the question.

If a leak develops in the vacuum distillation apparatus, the boiling points of the two components of caraway seed oil will

A: Both increase The boiling point of a liquid is the temperature at which *the vapor pressure of the liquid equals the surface pressure*. The normal boiling point is measured at 1atm pressure. The vapor pressure of a liquid increases with increasing temperature. Thus, the boiling point of a liquid decreases as the pressure on the surface of the liquid is decreased. If a leak develops, the surface pressure increases, as will the boiling points of both liquids

Alcohols generally require acid catalysis in order to undergo substitution by nucleophiles. The acid catalyst enhances the reaction by

A: Creating a better leaving group Remember that the substitution reaction serves to replace the hydroxyl group and hydroxide ion is one of the worst leaving groups in substitution reactions. Under acidic conditions, the hydroxyl group is protonated such that the leaving group is now water, a superior leaving group (rather than hydroxide ion).

Student E accounted for the equivalent weight found for succinic acid by analyzing its titration with NaOH(aq) and concluding that it is

A: Diprotic and requires twice the number of moles of NaOH expected for a monoprotic acid Equivalent mass is the mass of an acid that yields one mole of hydrogen ions (= mass of base that reacts with one mole of hydrogen ions). Formula for succinic acid is H2C4H4O4, from which its molar mass is calculated as 118. Table 2 gives the equivalent weight of the acid as 59.1. Since the molar mass is two times the equivalent weight, the titration requires two moles of NaOH per mole of succinic acid (there are two titratable hydrogen atoms per molecule)

Which compound precipitated in trial 3 of experiment 2 from student A and failed to precipitate in trial 3 of experiment 2 from student B?

A: FeF2 To determine which compounds will precipitate, it's necessary to compare the *reaction quotient (=ion product) Q* to the Ksp. If Q < Ksp, no precipitate will form if Q = Ksp, the solution is saturated if Q > Ksp, precipitate will form Here, Fe(OH) 2 and FeF2 must be evaluated. The solution concentrations can be solved from the protocol in the passage: If 0.001 moles of each salt were added to 10 ml of water, then the initial concentration before further dilution = 0.001 mol / 0.01 L = 0.1 M. In trial 3 from Student A, 5 ml of each salt solution are mixed for a total volume of 10 ml, so the concentration of each metal should equal 0.1 M x 5 ml / 10 ml = 0.05 M. Evaluate Q vs Ksp for Fe(OH)2: The equilibrium expression is Fe(OH)2 ↔ Fe2+ + 2OH-. Solving for the Ksp gives: Ksp = [Fe2+][OH-]2. The [Fe2+] = 0.05 = 5 x 10-2 (shown above). [OH-] can be found from the pH. If the pH = 6.6, then the pOH = 14 - 6.6 = 7.4. Therefore, [OH-] = 10-7.4, which we can round to 10-7.5. Plugging in for Q gives: Q = (5 x 10-2)(1 x 10-7.5)2 = (5 x 10-2)(1 x 10-15) = 5 x 10-17. Therefore, Q < Ksp, as 5 x 10-17 < 8 x 10-11, and the Fe(OH)2 compound will not precipitate (eliminate choice A). Evaluate Q vs Ksp for FeF2: The equilibrium expression is FeF2 ↔ Fe2+ + 2F-. Solving for Ksp gives Ksp = [Fe2+][F-]2. [Fe2+] = 0.05 = 5 x 10-2 (shown above), and [F-] = 2[Fe2+] (based on the equilibrium reaction above). [F-] = 1 x 10-1. Plugging in for Q gives: Q = (5 x 10-2)(1 x 10-1)2 = (5 x 10-2)(1 x 10-2) = 5 x 10-4 Therefore, Q > Ksp, as 5 x 10-4 > 2.36 x 10-6, and the FeF2 compound will precipitate. From Figure 2, it is clear that no precipitate forms in trial 3 from Student B, so FeF2 will not precipitate under more dilute conditions.

For an arsenic atoms (contains 4 electrons for the outermost) , what is the nuclear charge on an n = 3 and n = 2 electrons?

A: For n =3, 23 & for n =2, 31 Effective nuclear charge is found by subtracting the total number of electrons in all shells preceding the one in question from the nuclear charge (# protons). Arsenic has an atomic number of 32, but we are given As+ with 32 electrons. For n=3, we only need to consider those in the first two shells; starting from the first ring, the numbers of electrons in those shells are 2 and 8. Therefore, the effective nuclear change when n = 3 is (33-10) = 23 For n=2, use the same logic, which results in (33-2) = 31

A particular element overwhelmingly prefers to form one bond and possess three pairs. Which group does this species likely belong in?

A: Group 17 This question is asking the element that has seven valence electrons: one from the bond and six from the lone pairs. Thus it's group 17 with 7 valence electrons.

Experiment 1 was repeated with 0.4g of calcium, and the gas that evolved was collected. The identity of the gas and its approximate volume at 1atm and 27 celcius were

A: H2, 250mL Calcium undergoes the following reaction with water: Ca(s) + 2H2O(l) -> Ca2+ + 2OH- + H2 The gas produced was H2. If 0.4g calcium reacted, the moles of calcium reacted was 0.01mol; the amount of H2 formed was also 0.01 mol. The volume of 0.01 mol H2 = (0.01mol)(0.0821L)(300K)/1atm = 0.246 L

Calcium oxalate, a salt found in large amounts in kidney stones, dissociates according to the equilibrium below. CaC2O4 (s) → Ca2+ + C2O42- Ksp = 2.7 × 10-9 A student places a chunk of solid calcium oxalate in 10 L of water and notices that a small amount dissolves. How can he best increase the concentration of C2O42- in his solution? A. He should add a dilute solution of calcium nitrate to the vessel. B. He should add an anion that forms a precipitate with calcium, such as hydroxide. C. He should add a cation that forms a precipitate with oxalate, such as magnesium. D. He should decrease the temperature of the solution.

A: He should add an anion that forms a precipitate with calcium such as hydroxide B is correct. For this question, consider Le Châtelier's principle. To increase the concentration of C2O42-, the student must exert a pressure on the system to shift the equilibrium to the right (in other words, toward the dissociated ions). Adding OH- will take calcium out of solution by way of precipitation, forcing the equilibrium in the direction that will regenerate more oxalate ion. A: According to the common ion effect, this would actually shift the equilibrium to the left (toward the undissociated solid). This is the opposite of the student's intention. C: While this will shift the reaction in the proper direction, it will also remove oxalate from solution and convert it to a solid salt. This option would be a better choice if the student wished to increase [Ca2+]. D: This change will likely decrease Ksp, causing the compound to become less soluble overall. - Ksp ONLY responds to changes in temperature; neither changes in ion concentration nor the common ion effect can alter the Ksp of a compound.

Which of the following experimental modifications will most likely improve the degree of separation between limonene and (+)-carvone?

A: Heating the distillation flask at a slower rate (I picked: "Using a vacuum source that can achieve a lower pressure inside the distillation apparatus") The separation of the two liquids takes place in the fractionating column as the two liquids vaporize and condense, with the lower-boiling liquid distilling first. If the fractionating column is shortened, the liquids will vaporize and condense fewer times (worsening the degree of separation). Cooling the condenser with ice water will have no effect on the degree of separation because condensation takes place after the separation has occurred. Creating a lower pressure inside the distilling apparatus will LOWER the boiling points of both liquids and NARROW their difference in boiling point (making it necessary to increase the fractionating column to achieve the same degree of separation). Heating the distillation flask at a slower rate will allow both liquids more time in the fractionating column (increasing the number of theoretical plates, allowing liquid and vapor to equilibrate).

The hemoglobin (Hb) dissociation curve at high altitude has a distinct sigmoidal shape from that at atmospheric pressure. Which of the following best explains this shape?

A: Homotropic regulation by oxygen occurs The sigmoidal shape of the curve implies that as each oxygen molecule binds to Hb, the affinity of Hb for oxygen goes up. Homotropic regulation is when a molecule serves as a substrate for its target enzyme as well as a regulatory molecule of the enzyme's activity. Oxygen is a homotropic allosteric modulator of hemoglobin. The four subunits of hemoglobin bind to oxygen cooperatively, meaning the binding of oxygen to one of the four subunits increases the likelihood that the remaining sites will bind with oxygen as well.

Which of the following changes in state function occurred during the dissolution shown in equation 1 (solid -> liquid): I. H < 0 II. G > 0 III. S > 0

A: I and III Because heat is liberated, the dissolution of solid NaOH is an exothermic process with enthalpy change H < 0. As NaOH dissolves, the system become more disordered, corresponding to a positive entropy change. The free energy is given as G=H-TS, thus it must be less than 0.

Properties of transition metals include: I. the presence of multiple stable oxidation states. II. the tendency to form brightly-colored gases. III. the ability to exist in many paramagnetic compounds due to unpaired d-shell electrons.

A: I and III Transition metals generally possess multiple distinct oxidation states (i.e Mg has oxidation numbers ranging from -3 to +7). They are present in the d block of the periodic table, meaning that many of them do have unpaired d-shell electrons where they can exist in paramagnetic states. While transition metals are typically associated with bright colors, they typically do not enter the gas phase.

A researcher attempts to maximize the yield of the following reaction: A (g) + B (g) → C (s) + D (g) (ΔH = -92 kJ/mol) Which of these changes will help accomplish his goal? I. Increasing the pressure II. Decreasing the temperature III. Removing product D as it forms IV. Lowering the concentration of reactant A

A: I, II, III I is true because an increase in pressure shifts the reaction to the right because it decreases the number of moles of *gas* as it counteracts the higher pressure. II- the reaction is exothermic where heat can be written as a product; thus, a reduction in temperature (~eliminating one of the products "heat") provokes the reaction to counteract this change by shifting to the product side. III: Removing product as it forms is the classic way to increase yield as it disrupts equilibrium and shift it to re-form the product molecules.

Which of the following phase changes are exothermic processes? I. Liquid to gas II. Liquid to solid III. Gas to solid

A: II and III Gases have more heat energy than liquids and liquids have more heat than solids. Thus, II and III are exothermic. - I is endothermic (i.e need to invest heat energy to equilibrate and change liquid -> gas)

Which of these statements accurately describe the behavior of photoelectrons? I. A photoelectron's energy can exceed that of the incident photon. II. A photoelectron's energy cannot equal the energy of the incident photon. III. The threshold frequency and the likelihood of electron ejection are proportional. IV. The intensity of incident light is unrelated to the energy of an ejected photoelectron.

A: II and IV From conservation of energy, the energy of a photoelectron may only be less than the incident photon due to the existence of the work function (energy required for an electron to be released). Because some of the incident photon's energy is devoted to overcoming this threshold, not all of it is converted to KE for the ejected electron. The intensity of the incident ray determines the number of ejected photoelectrons, not the individual energies. III is wrong because the threshold frequency is the frequency to be surpassed for a photoelectron to be emitted; it is the energy of the incident photon that determines ejection.

The Nernst equation: Vm=61/z * log(ion outside/ion inside) Given that Vm represents the equilibrium membrane potential in mV for a ion and z represents the valence of the ion. In neurons, the overall equilibrium potential is determined almost solely by potassium cations. Which set of conditions will yield a membrane potential closest to its actual resting value in humans?

A: K+ outside = 10mM, K+ inside = 149mM In humans, the resting membrane potential of a typical neuron is approximately -70 mV. For this reason, and since K+ has a valence of +1, we know that the log term in the above equation must be negative. Therefore, [K+inside] needs to be larger than [K+outside]. Furthermore, if the internal potassium concentration were exactly ten times the external [K+], the value for Vm would be -61 mV. As we are looking for a potential that is even more negative, [K+inside] must be relatively larger. Choice A displays a [K+inside] value that is around 15 times greater than [K+outside], which resembles values found in vivo.

Air is bubbled through distilled water. The solution will have a pH

A: Less than 7, because carbon dioxide undergoes hydrolysis Air contains elemental nitrogen as its major component, but elemental nitrogen is essentially INERT and is NEITHER an acid/base. On the other hand, carbon dioxide undergoes a hydrolysis reaction in liquid water to produce carbonic acid, a weak acid that produces a solution whose pH is less than the pH of water.

In the reaction shown in equation 1, Al(OH)3 acts as what kind of acid or base?

A: Lewis acid In the reaction the Al in Al(OH)3 accepts an electron pair from the oH- of NaOH and acting as a lewis acid

In aqueous solution, copper ion can react with water molecules to form a vibrant blue complex. This takes place according to the reaction below: Cu^2+ + 6 H2O → [Cu(H2O)6]2+ In this reaction, H2O serves as:

A: Lewis base Since this reaction involves the formation of a bond, we are dealing with the transfer of electrons not the movement of a proton from one molecule to another. Water (with its two available lone pairs) is able to donate its electrons to copper (which, as a cation, accepts electrons); thus it's lewis base

If sodium sulfate was added to the mixture containing silver ions and the yellow precipitate, in theory it could lead to the precipitation of silver sulfate. Which of the following might correctly explain what might be observed after a significant amount of time elapsed?

A: Little to no silver sulfate formation because the Ksp of silver sulfate is very large compared to the Ksp of the yellow precipitate The Ksp, or solubility product, of a substance is defined as the product of each of the substance's dissolved ion concentration raised to the power of its stoichiometric coefficient. For silver sulfate, the dissolution reaction is: Ag2SO4 (s) → 2 Ag+ (aq) + SO42- (aq) which means that the Ksp equation is: Ksp = [Ag+]2[SO42-] A large Ksp suggests that a substance is more soluble than a substance with a lower Ksp.

To a first approximation, the ionization constant of H2S is

A: Much less than 1 H2S is a weak acid that dissociates in water but only to a very small extent; this is much less than one (is non-zero)

All of the below statements are true regarding the relationship between oxygen and nitrogen except that

A: Nitrogen has a lower ionization energy than oxygen Usually, ionization energy increases as it moves towards noble gases; however, the nitrogen-containing group forms an exception to this rule. Nitrogen has an electron configuration of [He]2s^2 2p^3, meaning that removal of another electron would disrupt *half-filled p* and entirely full s orbitals. In contrast, oxygen has an electron configuration of [He]2s^2 2p^4. Ionization of this species results in half-filled p, which is a highly energetically favorable state; therefore, oxygen will be easier to ionize.

The Ksp of lead iodide is 7.1 × 10^-9. A chemical engineer adds 0.0025 mol of KI to a solution of 0.00004 mol Pb(NO3)2 in 500 mL of water. Should the engineer expect to see a solid precipitate?

A: No, because the ion product is less than the Ksp of lead iodide. Lead diode is PbI2 where Ksp = [Pb2+][I-]^2 with lead and iodide concentrations of 0.00008M (=8*10^-5) and 0.005M (5*10^-3). Therefore, (8 × 10-5)(5 × 10-3)2 = (8 × 10-5)(25 × 10-6). The product of these terms is 200 × 10-11, or 2.0 × 10-9. As this quantity is smaller than the Ksp, this substance will not precipitate.

Free radicals from ionizing radiation are highly unstable and have carcinogenic effects that are likely result from damage to

A: Nucleic acids Free radicals can cause cancer, which is a result of poorly regulated cell growth and division. Cancer is generally a product of mutations in DNA that disrupt these processes. Cancer is always related to GENETIC MUTATION

Which of the following is the most plausible explanation for the fact that the saponification of the triacylglycerol in the passage resulted in 4 different fatty acid salts?

A: One of the fatty acid salts was unsaturated, and a small percentage isomerized under the reaction conditions (I picked: "one of the fatty acid salts was unsaturated, and it completely isomerized under the reaction conditions") If one of the R groups in the triacylglycerol contained a C=C and if isomerization of the double bond occurred during the saponification, 4 fatty acids would be obtained instead of 3.

Which of the following compounds contains a weak Bronsted-Lowry base? A. Ammonium cation B. Lithium hydroxide C. Sodium amide (NaNH2) D. Potassium acetate

A: Potassium acetate The acetate anion (C2H3O2-) is the conjugate base of acetic acid as it readily gains a proton to form acetic acid in solution. - Both B and C are strongly basic - Ammonium (NH4+) is acidic.. it is ammonia (NH3) that is a classic weak base

Which of the following compounds is amphiprotic?

A: Sodium bicarbonate, NaHCO3 The prefix "amphi" means "both"; an amphiprotic species is one that can act as both an acid and a base. Sodium bicarbonate dissolves in aqueous solution to produce sodium ions and bicarbonate ions. While sodium ion is neither acidic/basic, the bicarbonate ion (HCO3-) is a Bronsted-Lowey acid by loss of a hydrogen and can act as a B-L base by accepting a hydrogen ion to form carbonic acid (H2CO3). - Acetic acid (HC2H3O2) acts as a weak acid by losing a hydrogen ion but is incapable of accepting a hydrogen ion to any significant extent. - Sodium acetate (NaC2H3O2) dissolves in aqueous solution to form sodium and acetate ions (acetate ions can accept hydrogen ions to form acetic acid and act as a base). However, hydrogen atoms in acetate (C2H3O2-) are covalently bonded to a carbon atom that don't ionize to any extent in aqueous solution (thus, acetate doesn't act as an acid) - Sodium carbonate (Na2CO3) dissociates into sodium ions and carbonate ions in aqueous solution. Carbonate ions can accept hydrogen ions to form bicarbonate (carbonate acts as a base that doesn't have hydrogen atoms to act as an acid)

Which of the following best explains why arginine is more basic than lysine?

A: The electron-donating groups around the basic nitrogen on arginine make its conjugate acid more stable Since, in its protonated form, arginine has electron-donating groups via resonance with other nitrogens; it is a more stable conjugate acid.

A student notices that a flask containing several reagents forms ice crystals on its outer surface as they react. He places the flask in a warm bath to prevent further crystal formation. How would this bath affect the new concentration of product at equilibrium?

A: The reaction is endothermic; therefore, product concentration should increase The ice crystal formation indicates that this reaction requires heat energy to proceed; therefore, it must be endothermic. heat can be written as a reactant in this process where addition of additional reactant shifts the reaction to the product side to reestablish equilibrium.

3.75 moles of nitrogen gas is held in a rubber ballon with a current volume of 1L. The balloon is placed in a room with an initial pressure of 1 atm and temperature 25 celcius. If the pressure drops to 0.75 atm, which change in volume is necessary if the temperature is to remain the same?

A: The volume must be increased to 1.33L We can use Boyle's law P1V1=P2V2 given (1atm)(1L)=(0.75atm)(X), the new volume must be 1.33L

Fluid pressure changes with depth are assumed to be linear. Which statement best explains which this does not hold true for atmospheric pressure changes?

A: The volume of a mass of air is not constant Hydrostatic pressure for liquids is linear because as depth changes, the density of liquid remains constant. Gases, however, are compressible and have densities that change according to the forces applied to them.

Which of the following statements most accurately describes the solubility properties of fatty acid salts?

A: They can partially dissolve in both polar and nonpolar media (I picked: "They are soluble in polar media only") The fatty acid salt contains a long hydrocarbon chain, which is soluble in nonpolar solvents. The salt contains the charged group -CO2- NA+, which is soluble in polar solvents

How much sodium hydroxide is needed to completely saponify a triacylglycerol?

A: Three equivalents, because one OH- ion is required to saponify each of the three fatty acid groups One hydroxide ion is required to hydrolyze one ester linkage of a triacylglycerol molecule. Because there are three ester linkages in a triacylglycerol, three equivalents of sodium hydroxide are needed to completely saponify.

If red litmus paper is dipped into the Na2CO3 solution, it will

A: Turn blue, because carbonate reacts with water to produce OH- In water, carbonate will undergo the following reaction: CO3^2- + H2O -> HCO3 + OH-. Red litmus paper will turn blue in a base!

In the galvanic cell, if the electrode on the left side is the cathode (Ag2SO4 - anode is ZnSO4), which species undergoes the most spontaneous oxidation?

A: Zn(s) The silver electrode is the cathode. Because the cathode is always the site of reduction, Ag+ reduce to Ag during process; thus, the only oxidation reaction must be Zn(s) -> Zn2+ + 2e-

Zn has an oxidation potential of +0.76, while Al has an oxidation potential of +1.66. Which metal serves as the better oxidizing agent?

A: Zn2+ because it is more prone to gaining electrons Here, Zn has a less positive oxidation potential; thus, it is less likely to oxidize than Al; however, we switch the signs of the reactions to find their reduction potentials. Zn2+ has a reduction potential of -0.76 while Al has a reduction potential of -1.66; thus, Zn2+ is better oxidizing agent. - Recall that reduction potential is the opposite of the oxidation potential

What is the electron configuration of Os3+?

A: [Xe]4f^14 5d^5 When determining the electron configuration of a cation, remember that the *highest-energy electrons* should be removed first. Because the 6s subshell possesses the highest principal quantum number, its electrons must also have the most energy involved. Therefore, two electrons in the 6s shell are removed first; then, a single electron from the 5d shell must be taken off because it has higher energy than the 4f electrons.

In equation 1, HPO4^2- is the conjugate

A: base of H2PO4- A base differs from its conjugate acid only in the lack of a single H+ Q2. What is the conjugate base of chlorobenzoic acid? A: C7H4ClO2- Conjugate acids and bases differ only by the presence/absence of a proton H+. The conjugate base of an acid is formed from the acid by the loss of a proton H+.

Which of the following electronic transitions for hydrogen would result in the emission of a quantized amount of energy?

A: n=5 -> n=4 In order to emit a PHOTON of energy (QUANTIZED), an electron must go from a higher energy level to a lower energy level

Which of the bromomethanes is least polar A. CBr4 B. CHBr3 C. CH2Br2 D. CH3Br

Answer: A (I picked D) The polarity of each of the bromomethanes is determined by the vector sum of all of the individual bond polarities within a given molecule of the compound. Carbon is sp3 hybridized and tetrahedral in all of the bromomethanes. Thus, the vector sum of the bond polarities in CBr4 is zero whereas the vector sum is greater than zero in other compounds

Chloroacetic acid has a pKa of 2.85, while formic acid has a pKa of 3.75. Which of the following statements is true? A. The Ka of formic acid is larger than the Ka of chloroacetic acid. B. The Kb of the formate anion is larger than the Kb of the chloroacetate anion. C. The pKb of the chloroacetate anion is smaller than the pKb of the formate anion. D. The Ka of formic acid falls between 10-2 and 10-3.

Answer: B- The Kb of the formate anion is larger than the kb of the chloroacetate anion The higher the pKa, the weaker the acid; thus, formic acid is less acidic than chloroacetic acid. Note that weak acids have relatively stronger conjugate bases, and the stronger the bases, the larger its Kb value. A: This is backwards; because Formic acid is less acidic, it has a smaller Ka than chloroacetic acid D: Because Formic acid has a pKa between 3-4, its Ka falls between 10^-3 and 10^-4

All of the following statements about Ka and Kb are true except: A. if the Ka for Acid A is greater than the Ka for Acid B, the pKb of Acid A's conjugate base will be greater than the pKb of Acid B's conjugate. B. the product of the Ka of HPO42- and the Kb of HPO42- is equal to the value of Kw at that temperature. C. the product of the Ka of H2SO4 and the Kb of HSO4- is equal to 10-14 at 298 K. D. the Ka value for HCO3- and the Kb value for CO32- can multiply to equal a number other than 10-14.

Answer: B- The product of the Ka of HPO42- and the kb is equal to the value of Kw at that temperature When using Ka*Kb = Kw, remember that this is the Ka of the acid with Kb of its conjugate base (NOT the Ka and Kb of the same species). Thus, it is the peofuct of the ka of H2PO4- and the Kb of HPO4^2- that will be equal to the value of Kw at the relevant temperature. A: This is a true statement. If Acid A has a larger Ka than Acid B, Acid A must be more strongly acidic. A comparatively stronger acid will always have a comparatively weaker conjugate base, or one with a larger pKb. C: This is also true. The product of the Ka of an acid and the Kb of its conjugate will always be equal to Kw. At 25°C (or 298 K), Kw = 10-14. D: Remember, Kw is only equal to 10-14 at 25°C.

The auto-ionization of water is an endothermic process. If the temperature of a beaker of pure water is increased, A. the water will become acidic. B. the pH of the water will decrease. C. Kw will remain unchanged. D. the pOH of the water will increase.

Answer: B- the pH of the water will decrease The auto-ionization of water refers to its dissociation into hydronium and hydroxide ions. Auto-ionization is: heat + 2H2O -> H3O+ + OH-. The reaction shifts towards the product in response to increased temperature, resulting both [H+] and [OH-] to rise and decrease the pH and pOH to the same extent. A: Although pH decreases, pOH also decreases. As proton and hydroxide concentrations are equal, the solution is neutral. C: Kw, like other equilibrium constants, is temperature dependent!

Which of the following molecules has the highest predicted melting point? A. H2O B. CaCl2 C. Ne D. CH3COCH3

CaCl2 We know that the greater the intermolecular attractions between molecules, the higher the melting point of a sample of those molecules. Given that calcium chloride is a salt that participates in ionic bonds, they have high melting points. Water has hydrogen bonds, which are strong intermolecular forces but they don't have ionic bonds so less strong. Ne is a noble gas that only exert London force, which is extremely weak Acetone exert dipole-dipole, which are also weak.

What does the solution containing nickel(II) ions green colored?

The color arises because nickel(II) ion has partially filled d orbitals and the electrons in the lower energy d orbitals absorb visible light to move to the higher energy d orbitals