NS PRACTICE TEST 3
If a scientist chooses not to employ a SEC column, but wishes to fractionate a protein sample on the basis of only molecular weight, what technique would be most appropriate?
C is correct. SDS-PAGE is an electrophoretic technique which involves the binding of the anionic detergent SDS to a polypeptide chain. SDS binding denatures and imparts an even distribution of charge per unit mass to the protein, resulting in fractionation by approximate size alone during electrophoresis. A: Affinity chromatography is used to separate molecules, including proteins, based upon their specific interaction with the stationary phase. During protein purification, this is often an antigen-antibody, receptor-ligand, or enzyme-substrate interaction. B: Ion-exchange chromatography separates charged particles based on their affinity for charged elements of the ion exchange column. D: NMR spectroscopy is not a separatory technique. *SDS-PAGE
Selenocysteine is a non-standard amino acid that is present in all domains of life. It has the structure of cysteine, but the sulfur atom is replaced with selenium. The structure of selenocysteine is shown below: What is the absolute configuration of the alpha carbon in selenocysteine?
C is correct. When assigning absolute configuration, imagine the molecule so that the lowest-priority group is facing away. In this case, it is the implicit hydrogen, which by the terahedral geomtry of the alpha carbon already happens to be facing away. Then, assign priority based on molecular weight of the attached atoms. Nitrogen > C-Se > C-O > H. Follow those atoms in a clockwise circle to assign the R configuration. A, B, D: D and L are relative configurations, not absolute configurations. S is the opposite of the correct configuration. *R
What is the exclusion limit of the SEC column used to create the calibration curve shown in Figure 2?
D is correct. Figure 2 indicates that the exclusion limit of the SEC column is at log (MW) = 6. If true, the molecular weight can be found by taking the inverse log of 6, 106 D = 1000 kD.
How much heat is produced from the complete combustion of 30.0 g of methane, if the enthalpy of reaction is -890 kJ/mol?
B is correct. First, note that methane has a molecular formula of CH4. Thus, its molecular weight is approximately 12 + 4(1) = 16 g/mol. 30 g CH4 x (1 mol/16 g) x (890 kJ/mol) = 1.8 x 103 kJ (if we round 890 to 900 kJ/mol) Note that the answer options have units of joules (J), not kilojoules (kJ), and choice A should be eliminated. Converting the units gives: 1.8 x 103 kJ x (103 J /1 kJ) = 1.8 x 106 J * B. 1.7 x 106 J
What is the relationship of these molecules to each other? A. Meso compounds B. Enantiomers C. Diastereomers D. Same compound
B is correct. The configuration of the chiral compounds in each molecule show that the first molecule has a configuration of (S,S) while that of the second compound is (R,R), thus the molecules are mirror images of each other and are enantiomers. A: Meso compounds have an internal plane of symmetry, which is not present here. C: The compounds differ at both chiral centers, not just one. D: The compounds do not share the same configurations at the same chiral centers. *enatiomers
A balloon has a volume of 3.0 L at 25°C. What is the approximate volume of the balloon at 50°C?
C is correct. Charles' law states that there is a direct relationship between the volume of an ideal gas and its temperature, when pressure is constant. Note that the temperature must be in Kelvin! We can approximate the initial temperature as 300 K and the final temperature as 320 K. (3.0 L)/(300 K) = V2/(320 K) [(3.0)(320)] / (300) = (3.0)(1.1) = 3.3 L D: This answer would be obtained by incorrectly using Celsius temperature instead of converting to Kelvin.
What is the standard cell potential for the starch battery created?
C is correct. The passage states that the cell is meant to act like a galvanic cell, meaning that it proceeds in a spontaneous fashion. Since galvanic cells always have cell potentials that are greater than 0, we can eliminate any negative options (choices A and B) immediately. To decide between C and D, we need more information, namely the reduction potentials of the species involved. Paragraph 1 gives E°reduction of O2 as 0.816 V and E°reduction of AQDS as -0.10 V. Since both species cannot reduce at once, we must reverse the sign of one of the reduction potentials to find the oxidation potential of the relevant reaction. For galvanic cells, reverse the less positive reduction potential, which is -0.10 V here. The associated oxidation potential, then, is +0.10 V. Finally, since we now have a reduction and an oxidation, we can simply add the values directly to yield 0.816 V + 0.10 V = 0.916 V.
What is the net charge on a phenylalanine molecule at pH 1?
D is correct. This question is asking us to remember what factors would contribute to the net charge on a phenylalanine molecule. 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.
Phosphorous acid
H3PO3
Which of the following best explains why arginine is more basic than lysine?
B is correct. This question is asking us to determine why arginine is more basic than lysine. The reason must be related to how arginine is better able to handle being protonated, as this is the essence of being a base. Since, in its protonated form, arginine has electron-donating groups via resonance with other nitrogens, it is a more stable conjugate acid. A: The stability of arginine's conjugate base is not in question. C: A lack of electron-donating groups does not stabilize a conjugate acid. D: The stability of lysine's conjugate base is not in question. *The electron-donating groups around the basic nitrogen on arginine make its conjugate acid more stable.
Which of the following alkyl chlorides is most likely to undergo an SN1-type reaction?
D is correct. An SN1 reaction proceeds by way of a mechanism in which the leaving group (in this case a chloride) dissociates in a kinetically slow step, producing a planar carbocation which is then rapidly attacked by a nucleophile. Steric congestion tends to promote SN1-type mechanisms, so a tertiary alkyl halide, like t-butyl chloride, would favor pushing the leaving group off to form the carbocation. A, B: Chloromethane and chloroethane are primary alkyl halides and are not very sterically congested. Therefore, they would probably favor an SN2-type substitution mechanism. C: While the leaving group in 2-chloropropane is on a secondary carbon, which is somewhat sterically congested, it is not as crowded as in the tertiary carbon of 2-chloro-2-methylpropane. *2-chloro-2-methylpropane
Which of the following particles is expected to have the LEAST mass?
*A gamma particle D is correct. A gamma particle is a photon of electromagnetic energy, which does not have mass. A: An alpha particle consists of two protons and two neutrons, having a mass of 4 amu. B: A beta particle is the nuclear equivalent of an electron, which has a mass of approximately 1/1800 of a proton. C: A positron is the antiparticle of an electron and has its same mass.
A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push causes the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters?
*D. 5 m/s D is correct. Since the crate is initially at rest, the Vi = 0 m/s, the acceleration a = 2.0 m/s2, and the displacement = 6 m. The appropriate kinematic equation is vf2 = vi2 + 2ad. Substituting the variables gives: vf2 = 02 + 2(2)(6) vf2 = 24 vf = ~5 m/s
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 is correct. A process known as "hydration" or "solvation" occurs when the attractive force of an ion molecule causes a thin shell of water molecules to surround it. In the case of hydronium (H3O+), each of the H atoms attracts the O atom in an H2O molecule due to hydrogen bonding. These H2O molecules cause a "shell" of water molecules to surround the hydronium. B: This is misleading. Regular H+ ions almost never exist in H2O solutions. Instead, they combine with an H2O molecule to form H3O+. Bases are more than capable of neutralizing H3O+ ions. C: Although this describes an actual biological process in muscle cells, it has no connection to the hydration shell around H3O+ molecules. D: The effects of hydration shells have nothing to do with enhancing the mobility of hydronium. *results in a semi-stable shell of water molecules around the hydronium.
Atrial fibrillation (AF) is the most common type of heart arrhythmia, causing palpitations, fainting, chest pain and congestive heart failure. According to the information in the passage, what is most likely to be observed in the ECG of a person with AF?
A is correct. An arrhythmia is an irregular time period between heartbeats. Since the question asks about atrial fibrillation, and the P wave is attributed to contraction of the atria, this answer is most consistent with AF. This does not mean that the atria are not contracting at all. It is possible that the voltage differences are less than the electrical noise naturally produced by an ECG measurement and that blood is not efficiently being pumped from the atria to the ventricles. B, D: These choices incorrectly associate arrythmia with a regular heart rate. C: Absence of an R wave is much more dangerous problem than AF, since the R wave is a strong, sharp electrical signal coming from the ventricles. Absence of an R wave suggests ventricular fibrillation at best, which often is life-threatening and is usually associated with myocardial infarction, or what is commonly known as a "heart attack." *An absence of the P wave and an irregular rate
Salicylic acid can be considered a benzoic acid derivative. It can also be considered a derivative of which of the following?
A is correct. Salicylic acid not only has a carboxylic acid functional group, but a hydroxyl group and is an alcohol. Phenol is an alcohol where the alkyl group is a benzene ring. D: Note that phenyl ester would best describe acetylsalicylic acid (aspirin) but the question asks about salicylic acid. To deduce the structure of salicylic acid, look to the figure in the passage and de-acetylate it (remove the acetyl ester). *phenol
A 12 V battery is used to charge a 20 μF capacitor in a defibrillator. How much charge is stored on the plates of the capacitor?
A is correct. The capacitance (C) is the amount of charge stored per volt, C = q/V which means q = CV. The capacitance is given as 20 microfarads, which is 20 x 10-6 F. Substituting the voltage (12 V) and capacitance into the equation in scientific notation gives us the charge in coulombs. (20 x 10-6 F)(1.2 x 101 V) = 24 x 10-5 C = 2.4 x 10-4 C Converting the answers unit to millicoulombs (mC) gives the correct answer. Remember when making the exponent larger, we must make the coefficient smaller by the same factor. 2.4 x 10-4 C = 0.24 x 10-3 C = 0.24 mC
Which of the following types of electromagnetic radiation would have the shortest wavelength?
A is correct. This question asks us about the wavelength of EMR that ejects an electron from an atom. Shorter-wavelength EMR (such as γ rays) carries much more energy than longer-wavelength EMR (such as radio waves). Therefore, we must look for the answer choice that involves the highest-energy EMR. The closer an electron is to the nucleus, the harder it is to eject. Because sp-hybridized orbitals have the most s character of all of the answer choices, they contain the electrons that are hardest to eject. B: sp2-hyrbridized orbitals have 33% s character, while sp-hybridized orbitals have 50% s character. C: sp3-hyrbridized orbitals have 25% s character, while sp-hybridized orbitals have 50% s character. D: Radiation that excites an electron but does not eject it is of lower energy than radiation that is able to eject an electron. *Radiation that ejects an electron from an sp orbital
The electrical resistance of dry skin can be as much as 100 kΩ, but can be reduced to about 20 Ω if electrode contact area is large and a suitably conductive gel is used. If the peak voltage produced by the AED is 500 V and lasts 0.01 s, what is the maximum current delivered to the patient during defibrillation?
A is correct. Using Ohm's Law, V = IR: I = V/R = (500 V) / (20 Ω) = (5 x 102) / (2 x 101) = 5/2 x 101 = 2.5 x 101 amps
What is the pH of a 0.10 M aqueous solution of acetylsalicylic acid?
B is correct. Acetylsalicylic acid is a weak acid, with a pKa of 3.5. Therefore, the pH of this solution must be less than the pKa, because the compound is primarily in its acid form and a pH of 3.5 would mean that the concentration of weak acid and conjugate base were equal (a buffer). Choices C and D can be eliminated. A pH of 1.0 for an acid whose concentration is 0.10 M would require complete dissociation (as in a strong acid), eliminating choice A. Alternatively, the pH can be determined from the equilibrium expression: Ka = [H+][A-]/[HA] 10-3.5 = x2/(0.1-x) Since x will be small, we can approximate 0.1-x ~ 0.1, giving: 10-3.5 = x2/(0.1) 10-3.5(0.1) = x 2 10-4.5 = x2 [10-2.25][10-2.25] = x2 10-2.25 = x Since x equals the hydrogen ion concentration, taking the -log(10-2.25) = 2.25. *2.25
In a follow-up study, the emission spectra of both bound states of Y32 were shifted to longer wavelengths than the excitation spectra shown in Figure 2. What is the most likely cause of this shift?
B is correct. Longer-wavelength light corresponds to lower-energy light. If the energy required to excite electrons present in the CPV fluorophore exceeded that which is emitted when the electrons relaxed, then energy must have been lost during the return to the ground state following excitation separately from the energy of the emission. Only choice B, which states that there is a loss of vibrational energy during the transition, provides a possible explanation consistent with this requirement. A: The intensity of the emission or excitation peak is not directly related to the wavelength of the transition. C, D: While these statements are true, they do not address the question of why the shift occurred. *There is a loss of vibrational energy when electrons move from the excited to the ground state.
A 60 kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturing the energy each time his feet touch down, there's an average 10% loss with each compression. What must the runner's additional power output be to account for just this loss, if he averages 0.8 s per stride?
B is correct. This is a multi-step question, though each step is relatively straightforward. The gravitational potential energy at the runner's height is: PE = (60 kg)(10 m/s2)(0.5 m) = 300 J Most of this energy is conserved as the runner hits the ground and his muscles capture the energy as spring potential energy, so the question only asks about the lost energy, amounting to 10%, or 30 J. Don't worry about the mechanics of which foot lands first and how much of the kinetic energy each one absorbs. The question stem doesn't give that kind of information, so there's nothing for it anyway. So, an additional 30 J is needed per stride, and a stride occurs every 0.8 s. Thus: P = (30 J)/(0.8 s) = 40 W This is between answer choices B and C. So is it really 40 W? Well, no, 8 goes into 30 less than 4 times, so the answer should be lower. Also, g is closer to 9.8 than 10, so that also caused the value of energy, used in the power calculation, to be overestimated. 37 W, then, must be the match. *37
A student determined that her yield of aspirin was 3.9 g. What was her percent yield?
B is correct. We first need to determine if there is a limiting reagent and then determine the theoretical yield. The moles of acetic anhydride can be determined by using the volume and density information in step 2 of the procedure and the molecular weight given in the passage. 7 mL x 1.08 g/mL x 1 mol/102 g ~ 7 x 10-2 mol We can also calculate the moles of salicylic acid from the number of grams used in step 1 and the molecular weight given in the passage. 5.0 g x 1mol/138 g ~ (5/1.4) x 10-2 ~ 3.5 x 10-2 mol Since the stoichiometry is 1:1 from Reaction 1, the acetic anhydride is in excess and the limiting reagent is the salicylic acid. The stoichiometric ratio between salicylic acid and the acetyl salicylic acid is also 1:1; therefore, the theoretical yield can be calculated using the molecular weight from the passage. 3.5 x 10-2 mol x 180 g/mol ~ 3.5 x 2 x 10-2 x 102 ~ 7 g Since the question indicates that the actual yield was 3.9 g, which is about 4 g, the percent yield is (4/7) x 100 ~ 60%.
Y32 is pH-sensitive. This can be a problem when employing it to monitor changes in the energy balance of a cell because:
C is correct. If the biosensor is pH-sensitive, than changes in cellular pH over the course of a measurement could confuse changes in the energy balance of the cell, as reflected by the cell's ATP-to-ADP ratio. A: While this is true, the question specifically asks about how the fact that the biosensor is pH-sensitive could affect measurements made within a single cell over time. B: While the ATP-to-ADP ratio in most cells does to some extent depend on pH, this does not address the question of how pH changes influence the utility of the biosensor in measuring a cell's ATP-to-ADP ratio. D: Decreasing pH would favor the protonated CPV A-state. *pH may change over time in the same cell.
A second student repeated the experiment using glucose and the equivalent enzymes of glycolysis instead of starches. How would his results compare to those shown in Figure 3? Solution Initial Absorbance Absorbance at 2 minutes Average Current Enzyme A 0.40 0.62 0.0 mA Enzyme B 0.39 0.36 10 mA Enzyme C 0.42 0.37 30 mA Enzyme D 0.41 0.79 0.1 mA
C is correct. The first difference between the first student and the second is the use of glucose, a monosaccharide, as the fuel. Paragraph 2 states that iodine binds selectively to linear-chain polysaccharides. As a result, the iodine will NOT bind the glucose, causing the solution to remain brown. We can eliminate choices A and B. Next, we must determine the effect of using glucose and the glycolytic enzymes on the current generated by the battery. From Figure 1, we can infer that the amount of current generated should be directly related to the production of NADH. Unlike in the biostarch battery, glycolysis does not require the breakdown of starch. Thus, NADH is generated directly and more quickly via the conversion of glyceraldehyde 3-phosphate (GAP) to 1,3-bisphosphoglycerate (1,3-BPG). The enzymes should produce 1 NADH per GAP molecule, or 2 NADH per glucose molecule (since each unit of glucose breaks down into two GAP molecules). The original experiment only produced 1 NADH per glucose conversion to phosphogluconate, so we can conclude that the second student should observe a higher initial current than the first. *Brownish color with higher initial current
If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a height of 0.5 m and its legs are compressed by 0.2 m?
C is correct. This is a tricky question, because the passage states that nature, biology has no perfectly elastic springs, which means energy cannot be completely conserved. This does not, however, mean that all the energy is lost, nor does it make sense for the potential energy stored in a spring to be negative. So, if energy were completely conserved, what would it be? Potential energy of the rabbit at the peak of its height is PE = (3)(10)(0.5) = 15 J, or, if we're more exact and use 9.81 for g, slightly less than that. 14.7 J seems about right. 13 J is clearly a little on the low side, but this is exactly what's needed for this question, since the question stem is dropping a big hint that some energy is lost. A: The spring wouldn't store negative energy. B: This would imply that all of the energy was lost and that no energy was stored in the spring (muscle) after landing. The passage tells us that the springs are imperfect, but not that they would dissipate all energy D: The rabbit only has ~14.7 J of energy at the top of the jump, so the springs can't store that exact amount of energy. That would mean they were perfect springs. *13
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?
C is correct. When reading questions, be careful not to read too quickly. In this case, fast but inefficient reading will lead us to assume that it is asking about the percentage of a certain isotope that is left after radioactive decay. However, the question is asking about atomic mass. While β-decay does cause a nuclear transmutation of protons to neutrons (β-plus) or neutrons to protons (β-minus), the atomic mass lost in these processes is negligible. This means that whether after one (Co) or five (X) half-lives, the atomic mass will be the same in both samples. A: This is the ratio of undecayed nuclei between the two samples. The GKS-Co machine has 1/2 of its nuclei undecayed and the GKS-X machine has 1/32 of its nuclei undecayed. The question, however, asks about all of the nuclei, both decayed and undecayed. *1:1
A student theorizes that the differences in ATP-to-ADP ratios between cells detected in the experiment could be due to variations in Y32 expression rather than differences in the cells' metabolic conditions. Does information presented in the passage support this possibility?
D is correct. According to the passage, the spectral ratio of the peaks of the excitation spectrum of a Y32-expressing cell directly reflects its cellular ATP-to-ADP ratio. The peak absorbance of the B-state, which is favored in the Mg2+-ATP bound conformation, is near 500 nm, while the peak absorbance of the A-state, which favors the ADP-bound conformation, is near 420 nm. Thus, the ratio of the fluorescence emission intensities when Y32 is excited at 500 nm versus when it is excited at 420 nm serves as a direct indicator of the cellular ATP-to-ADP ratio. If, as stated in the passage, the fluorescence intensity is proportional to the number of sensor molecules excited, then excitation at both 500 nm and 420 nm would change proportionally with changes in biosensor expression, and would have no impact on determining the ratio in which ATP and ADP are present in the cell. A: While this is true, it does not take into account the proportional increased fluorescence corresponding to both the ATP- and ADP-sensitive peaks. The ratio of ATP to ADP in the cell will not be affected. B: There is no indication in the passage that Y32 expression levels can influence the cellular ATP-to-ADP ratio. C: While this is true, it does not address the question posed in the question stem. *No, the spectral ratio intrinsically normalizes for the amount of biosensor.
More complete fractionation of proteins using an SEC column could be achieved by using a:
D is correct. As with other forms of chromatography, increasing the column length will enhance the resolution of the column, leading to more completion fraction by SEC. This is because the material of the matrix provides the physical means of separating the proteins. If the proteins come in contact with a longer length of matrix, the differences in retarding forces experienced by the proteins will have a greater cumulative influence on the migration of the proteins, lengthening the differences in their retention times. A: There is no passage information to suggest a relationship between solvent (buffer) polarity and the function of the SEC column. B: Applying a vacuum will increase the flow rate of buffer through the column. Increasing the flow rate will only increase the rate at which eluent is collected; it will not improve separation. C: Increasing the concentration of protein in the sample will likely decrease the resolution of the column, by increasing the chance of overlapping within the fractions of collected eluent. *longer column.
Colliding cells often meet on the near frictionless surface of vascular membranes. If Cell 1 collides into stationary Cell 2 on an arterial wall, which of the following describes what happens to Cell 2 after the collision?
D is correct. Cell 2 only accelerates while it is being pushed/in contact with cell 1, so statement I is false. Since the cells are on a frictionless surface, cell 2 will not decelerate after the collision; statement II is similarly incorrect. Once cell 2 is in motion, it will stay in motion with constant velocity, without changing direction. Therefore, it will have a constant speed (III) and a constant velocity (IV).
Which of the following types of proteins would be most efficiently fractionated by an SEC column?
D is correct. If SEC separates on the basis of differences in molecular weight, the technique's resolution should increase if the pair of proteins differ in their molecular weights to a greater extent. A, B, C: Resolution between separable proteins will be greatest if their molecular weights differ by the largest amount possible, so long as the molecular weights of both proteins are within the size selectivity range of the column. *Two proteins with substantially different molecular weights
Free radicals from ionizing radiation are highly unstable and have carcinogenic effects. These effects are most likely result from damage to:
D is correct. The question stem states that 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. Therefore, we must choose the answer that targets DNA, which is choice D. A, B, C: Damage to these structures/molecules would be unlikely to lead to uncontrolled cell growth (cancer). *D. nucleic acids.
An artificial leg designed for use by runners is spring-based, to mimic the compression required of a muscle during hard running. For safety reasons, it was determined that the leg should be able to absorb as much as 125 J of kinetic energy without compressing more than 10 cm, or the runner would be likely to stumble. What should the spring constant be?
D is correct. When the leg "absorbs" kinetic energy, it is converted to elastic potential energy. While this process involves the loss of some energy as heat, we can assume that it is a perfectly elastic process here for the sake of simplicity. Thus, the leg needs to hold up to 125 J of elastic PE. The formula for potential energy contained in a spring is PE = (1/2) kx2. k = 2 KE/x2 k = (2)(125 J) / (0.10 m)2 k = (250 J) / (.01 m2) k = 25,000 N/m
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 would correctly explain what might be observed after a significant amount of time elapsed?
Little to no silver sulfate formation, because the Ksp of silver sulfate is very large compared to the Ksp of the yellow precipitate. A is correct. This question is asking us to find a possible outcome and explanation for the outcome when adding sodium sulfate to the silver/yellow precipitate mix. This means we need to choose the answer that has a valid outcome and a proper explanation for that outcome. 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. It is not certain, given that the coefficients might be different and might affect the power to which the ion concentrations are raised. However, it provides a better possible explanation than the other answer choice options (and the phrasing of the question prompt, using the words "might explain", notes that there is some uncertainty involved). B: This answer cannot be correct as sodium sulfate will dissolve completely in water; this is actually true of all salts containing sodium. Even if it were a so-called "insoluble" salts, these will typically dissolve to a minute extent. (This is why they will have dissolved ion concentrations which can be multiplied to form a Ksp solubility product). C: Sodium ions do not catalyze the reaction of silver ions with sulfate ions; they are attracted to each other though electrostatic attraction (positive ions attracting negative ions). D: In the proposed reaction, sulfate ions constitute a reactant, not a catalyst. Remember that the definition of a catalyst is a molecule which speeds up a reaction rate, without itself being used up as the reaction progresses.
Rutherford observations
Most of volume of gold atom is empty space Alpha particle and gold nuclei are both positively charged Alpha particles are the nuclei of helium atoms, 4He2+. Since the vast majority of alpha particles pass straight on through the gold foil, without deflection or rebounding, they must not have interacted significantly with anything having significant mass or charge, making RN I a reasonable conclusion. In order for the alpha particles to rebound or be deflected, the nucleus of the gold atoms must have the same sign of electrical charge as the alpha particle, causing a repulsive electrical force that is dependent upon the distance between the two charged particles, according to Coulomb's law. Therefore, RN III is a reasonable conclusion as well.
The high-energy radiation produced by the γ rays has sufficient energy to:
radiation is strong enough to cause molecular electronic transitions by exciting electrons to higher energy levels in molecular orbitals. This indicates that that the radiation can either excite or eject electrons (depending on its energy) and can create free radicals (atoms with unpaired valence electrons). *generate free radicals. II. excite electrons to higher energy levels. III. eject electrons from molecular orbitals.