chem physics--aamc free full length exam

Would an increase in kcat be expected to affect Vmax?

. Vmax depends on the enzyme concentration, so if you double the amount of enzyme you double Vmax. Km and kcat are constants so changing the enzyme concentration will not change their value.

What number is STP

22.4 L/mol

What is a heterocycle?

A cyclic compound (like a ring) which contains at least one other element that is not carbon. Tryptophan and histidine are amino acid examples of this.

What amino acids are good at π-stacking

All the cyclic ones like tyrosine

What is the difference between an agonist and an antagonist?

An agonist activates a receptor to perform its biochemical function. An antagonist blocks the biochemical function of the protein it binds to.

How many stereoisomers of in a compound with 5 chiral centers ? A. 8 B. 16 C. 32 D. 64

C There are 5 chiral centers in a Compound. By application of the 2n rule, there are 32 possible stereoisomers.

what special group does arginine have

Guanidinium

What amino acid has a Imidazole ring?

Histidine.

Which values of km and kcat describe an enzyme with high catalytic efficiency?

Low Km high kcat describes an enzyme with high catalytic efficiency. Catalytic efficiency = kcat/km

How do you calculate Force

Mass Times Acceleration: Newton's Second Law.

can you use molecular weight to check that the protein you purified is still in its native state?

Molecular weight is a measure of the stability of the primary structure. Activity requires full conformational stability.

two open flasks I and II contain different volumes of the same liquid. Suppose that the pressure is measured at a point 10 cm below the surface of the liquid in each container. How will the pressures compare? A. The pressures will be equal. B. Pressure in flask I will be less. C. Pressure in flask II will be less. D. The pressures cannot be compared from the information given.

Solution: The correct answer is A. a) The pressure at a point 10 cm below the surface of the liquid is the same in both flasks because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. b) The pressure at a point 10 cm below the surface of the liquid in flask I is the same as the pressure in flask II at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. c) The pressure at a point 10 cm below the surface of the liquid in flask II is the same as the pressure in flask I at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. d) The pressures can be compared because both pressures are calculated according to the hydrostatic pressure formula p = ρgd, where ρ is the liquid density, g is the gravitational acceleration, and d is the depth where pressure is measured.

if all else is held constant, which of the following changes would NOT double the volume of a gas? A.Doubling the pressure B.Doubling the absolute temperature C.Halving the pressure D.Doubling the number of gas molecules

Solution: The correct answer is A. a) Based on Boyle's Law, P is inversely proportional to volume, thus, doubling the pressure of a gas sample will decrease, not increase, the volume. b) Doubling the absolute temperature of a gas sample will double the volume because T is directly proportional to V for an ideal gas according to Charles' Law. c) Halving the pressure of a gas sample will double its volume because P is inversely proportional to V for an ideal gas. d) Doubling the number of gas molecules present will double the volume according to Avogadro's Law.

What is a amphipathic chemical?

a chemical compound containing both polar (water-soluble) and nonpolar (not water-soluble) portions in its structure at the same time.

What does a high kcat/Km value mean?

a high kcat/Km ratio means the enzyme works well with not much substrate. This is called catalytic efficiency because if the enzyme is efficient, it means it doesn't need much substrate to achieve a high reaction rate

describe each of the below reactions A.RC(=O)R → RCH(OH)R B.ROPO32- → ROH + Pi C.RC(=O)NHR'→ RCOOH + R'NH2 D.RC(=O)OR'→ RCOOH + R'OH

a) This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed by Na+-NQR. b) This is the reaction catalyzed by a phosphatase. c )This is the reaction catalyzed by a protease or amidase. d) This is the reaction catalyzed by an esterase.

true or false: glycerol is a precursor for steroids

false Glycerol is the trihydroxyl backbone for triacyl glycerols, which are fats

true or false: Phenylalanine is precursor for steroids

false Phenylalanine is an amino acid, which are the components of proteins

true or false: fluorescence occur when the absorbed radiation has a photon energy smaller than the photon energy of the radiation emitted through florescence

false fluorescence can occur when the absorbed radiation has a photon energy larger than the photon energy of the radiation emitted through fluorescence.

true or false: Glucose is a precursor to steroids.

false glucose is a simple sugar it is not a steroid precursor

what special group does histidine have

imidazole

what special gorup does tryptophan have

indole

the native carbonic anhydrase has a met charge of -2.9 Like the unmodified enzyme, photochemically modified carbonic anhydrase at 20°C has full enzyme activity, which is the conversion of CO2 to H2CO3. Native carbonic anhydrase also has a net charge of -2.9 at pH 8.0. Which chromatographic technique would most likely separate a mixture of native carbonic anhydrase from carbonic anhydrase photochemically modified by CCl3CO2H? A. Anion-exchange chromatography B. Cation-exchange chromatography C. Gas-liquid chromatography D. Size-exclusion chromatography

Solution: The correct answer is A. a) Because the passage states that native carbonic anhydrase has a net charge of -2.9 and the modified enzyme would have greater negative charge, anion-exchange chromatography can separate them as this technique separates proteins with different negative charges. b)The native and modified Enzymes both have net negative charge. In cation-exchange chromatography, both Enzymes would elute together in the void volume. c)Proteins degrade before they would vaporize. d)The molecular weights of the native and modified proteins are too close (< 1 kDa difference) to allow separation by size-exclusion chromatography.

The second sequence consisted of six consecutive histidine residues (His)6. This sequence binds tightly to Ni2+ cations. In chromatography, (His)6 tag labeled proteins can be eluted from Ni2+-supported columns by adding a small molecule to the eluent that mimics the side chain of histidine. The second purification step is which type of chromatographic separation? A.Affinity B.Size exclusion C.Cation exchange D.Anion exchange

Solution: The correct answer is A. a) Displacement of the protein from the column in this step involved disrupting the binding of the (His)6tag to the column. This is a classic example of affinity chromatography. b)Size-exclusion chromatography separates proteins by molecular weight, not selective column binding. c)Cation-exchange chromatography separates proteins with different positive charges (or positive versus negative/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction. d) Anion-exchange chromatography separates proteins with different negative charges (or negative versus positive/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction.

Which structural change to Compound 1 would make it more water soluble? A.Replacing benzene CH with N in the ring B.Replacing C=O with C=CH2 C.Replacing N-N=N with CH-CH=CH D.Replacing NH with NCH3

Solution: The correct answer is A. a) Nitrogen in the benzene ring would have a lone pair that could accept a hydrogen bond from water, thus increasing the solubility of the compound. b) Replacing C=O with C=CH2 would decrease the polarity of the compound and make it less water soluble. c) Replacing N−N=N with CH−CH=CH would decrease the polarity of the compound and make it less water soluble. d) Replacing NH with NCH3 removes a hydrogen bond donor, thus decreasing water solubility.

What functional group transformation occurs in the product of the reaction catalyzed by Na+-NQR? A.RC(=O)R → RCH(OH)R B.ROPO32- → ROH + Pi C.RC(=O)NHR'→ RCOOH + R'NH2 D.RC(=O)OR'→ RCOOH + R'OH information from the passage The NADH:quinone oxidoreductase (Na+-NQR) is a transmembrane protein that catalyzes the reaction between NADH and ubiquinone coupled to the pumping of Na+ across the plasma membrane, resulting in a Na+ concentration gradient. The electron transport pathway in Na+-NQR is composed of four flavins (FAD, FMNc, FMNb, and riboflavin) and a [2Fe-2S] center, with electrons flowing in the direction: NADH → FAD → [2Fe-2S] → FMNc → FMNb → riboflavin → ubiquinone. Two electrons are transferred from NADH to FAD in the first step of the cycle, but all subsequent steps are one-electron transfers.

Solution: The correct answer is A. a) This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed by Na+-NQR. b) This is the reaction catalyzed by a phosphatase. c) This is the reaction catalyzed by a protease or amidase. d) This is the reaction catalyzed by an esterase. the only relavent information from the passage is that Na+-NQR is involved in oxidation (it is from the class of enzymes called oxidoreductase) which means it is involved in oxisation reduction reactions. choice a shows a carbonyl being reduced to an alcohol.

What is the identity of an atom that contains six protons and eight neutrons? A.Nitrogen B.Carbon C.Oxygen D.Silicon

Solution: The correct answer is B. a) The atom cannot be nitrogen because nitrogen contains seven protons. b)Carbon contains six protons because it also contains six electrons as a neutral atom. c)Oxygen contains eight protons because as a neutral atom it contains eight electrons. d) The silicon atom contains 14 protons.

A 2 kg mass and a 5 kg mass are connected by a massless cord suspended over a massless and frictionless pulley. If the acceleration due to gravity is g, what will be the acceleration of the masses after they are released from rest? A.2g/7 B.3g/7 C.5g/7 D.g

Solution: The correct answer is B. a) This is smaller than the magnitude of the acceleration of the masses after they are released from rest by a factor of 1.5. It implies that only the 2-kg mass is moving after the release. b) According to Newton's second law, the net force acting on the 5-kg mass is given by the expression Fnet= 5 kg × a1 = 5 kg × g - T, where a1 is the acceleration after the release and T is the tension in the cord. The net force acting on the 2-kg mass is given by the expression Fnet= 2 kg × a2 = 2 kg × g - T. Because the two masses move simultaneously but in opposite directions after they are released, a1 = -a2 = a. Substituting the expression T = 5 kg × (g - a) into the equation of motion of the 2-kg mass yields -2 kg × a = 2 kg × g - 5 kg × (g - a) = -3 kg × g + 5 kg × a. Then 7 kg × a = 3 kg × g, hence a = 3g/7. c) This is larger than the magnitude of the acceleration of the masses after they are released from rest by a factor of 1.7. It implies that only the 5-kg mass is moving after the release. d) This implies the two masses fall freely and separately without being connected by the cord.

If only [I] is increased, then [ESI] or [EI] increases. This is an example of: A.the Bose-Einstein Principle. B.the Heisenberg Uncertainty Principle. C.the Le Châtelier's Principle. D.the Pauli Exclusion Principle. given information from the passage: Compound 1 has been shown to inhibit HIV-1 protease with Ki = 60.3 μM (Table 1). Ki is the dissociation constant for the enzyme-bound inhibitor, which is either EI or ESI, depending on the type of inhibitor. The peptide substrate for HIV-1 protease is IRKILFLDG. HIV-1 protease hydrolyzes only the peptide bond after leucine and before phenylalanine Table 1 Kinetic Parameters of HIV-1 Protease with and without Compound 1 uM of compound 1 , KM(mM) kcat(s−1) KM(mM−1 s−1) Ki(μM 0 1.678 0.250 00.149 — 60.0 0.832 0.126 0.151 60.3 120 0.430 0.064 0.149 60.3 Note: HIV-1 protease follows Michaelis−Menten kinetics. A Lineweaver-Burk plot of Vo−1 versus [S]−1, both without Compound 1 and with different concentrations of Compound 1, generates a set of parallel straight lines with positive slopes.

Solution: The correct answer is C. a) The Bose-Einstein Principle states that a collection of atoms cooled close to absolute zero will coalesce into a single quantum state. b) The Heisenberg Uncertainty Principle states that one cannot know both the momentum and position of an object with absolute certainty. c) Le Châtelier's Principle states that in a reversible process, the application of stress to the system will cause the system to respond in a way that will relieve this stress. In this case, the reversible process is E + I forming EI or ES + I forming ESI. In either case, increasing [I] induces stress of the system, and the system relieves that stress by converting I to either more EI or ESI. d) The Pauli Exclusion Principle states that two or more identical fermions, e.g. electrons, cannot occupy the same quantum state.

Enantiomers can exhibit a difference in which chemical or physical property? A. Density B. Boiling point C. Smell D. IR spectrum

Solution: The correct answer is C. a) Enantiomers will not display different densities. b)The boiling points of enantiomers are identical, and they cannot be separated by distillation. c)Enantiomers have the same physical and chemical properties. They differ only in their three dimensional arrangement of atoms and their interactions with other chiral molecules. They can differ in their smell due to interacting differently with chiral odorant receptors. d) The IR spectra of enantiomers are identical when a normal light source is used. Circularly polarized light will potentially illustrate differences.

In μM•s-1 and μM, what should the approximate values of kcat/KM and Ki be, respectively, when [I] = 180 μM? A.33.5 and 15.7 B.75 and 30.1 C.150 and 60.3 D.300 and 120.6 given information from the passage: Compound 1 has been shown to inhibit HIV-1 protease with Ki = 60.3 μM (Table 1). Ki is the dissociation constant for the enzyme-bound inhibitor, which is either EI or ESI, depending on the type of inhibitor. The peptide substrate for HIV-1 protease is IRKILFLDG. HIV-1 protease hydrolyzes only the peptide bond after leucine and before phenylalanine Table 1 Kinetic Parameters of HIV-1 Protease with and without Compound 1 uM of compound 1 , KM(mM) kcat(s−1) KM(mM−1 s−1) Ki(μM 0 1.678 0.250 00.149 — 60.0 0.832 0.126 0.151 60.3 120 0.430 0.064 0.149 60.3 Note: HIV-1 protease follows Michaelis−Menten kinetics. A Lineweaver-Burk plot of Vo−1 versus [S]−1, both without Compound 1 and with different concentrations of Compound 1, generates a set of parallel straight lines with positive slopes.

Solution: The correct answer is C. a) These values are one-fourth of both the actual kcat/KM and the Ki. The latter is impossible as Ki , being an equilibrium constant, is unaffected by [I], and the former does not match the trend in Table 1. b) These values are half of both the actual kcat/KM and the Ki. The latter is impossible as Ki, being an equilibrium constant, is unaffected by [I], and the former does not match the trend in Table 1. c) Based on the data in Table 1, increasing [I] has no effect on kcat/KM, so it should remain at 150 μM•s-1. Also, Ki is an equilibrium constant, so it will not be affected by a change in [I]. d) These values are double both the actual kcat/KM and the Ki. The latter is impossible as Ki , being an equilibrium constant, is unaffected by [I], and the former does not match the trend in Table 1.

What is the number of neutrons in the nucleus of the atom (86 kr+) to produce laser radiations? A.48 B.49 C.50 D.51

Solution: The correct answer is C. a) This is less than the number of neutrons in the 8636Kr atom which contains 36 electrons, 36 protons, and 86 - 36 = 50 neutrons. b) This is one neutron less than the number of neutrons in the 8636Kr atom which contains 36 electrons, 36 protons, and 86 - 36 = 50 neutrons. c) The 8636Kr atom contains 36 electrons and 36 protons. Therefore, the number of protons is equal to 86 - 36 = 50 neutrons. d)This is one more than the number of neutrons in the 8636Kr atom which contains 36 electrons, 36 protons, and 86 - 36 = 50 neutrons.

The reaction between NADH and ubiquinone is exergonic, but the reaction, when catalyzed by Na+-NQR, does not generate much heat in vivo. What factor accounts for this difference? The reaction catalyzed by Na+-NQR in vivo: A.is more exothermic as a result of the lower activation energy. B.occurs sequentially in several small steps. C.maintains a large separation between the reacting centers. D.is coupled to the movement of a charged particle against a concentration gradient.

Solution: The correct answer is D. a) This is impossible. Even if it were true, this would make the heat generation larger, not smaller, for the catalyzed reaction. Catalysis does not change thermodynamics. b) By Hess's Law, the heat of reaction will sum and be the same. The fact that the reaction can be broken down into steps will not change the overall thermodynamics. c) This is also impossible. The reactants ultimately must be close together to react. d) The movement of a charged particle against its concentration gradient is energetically costly. Coupling the two processes: the redox reaction between NADH and ubiquinone and the movement of Na+ up a concentration gradient makes the overall process less exothermic.

Blood flows with a speed of 30 cm/s along a horizontal tube with a cross-section diameter of 1.6 cm. What is the blood flow speed in the part of the same tube that has a diameter of 0.8 cm? A. 7.5 cm/s B. 15 cm/s C. 60 cm/s D. 120 cm/s

Solution: The correct answer is D. a) A flow speed of 7.5 cm/s corresponds to a diameter of 3.2 cm, not 0.8 cm. b) A flow speed of 15 cm/s corresponds to a diameter of 2.25 cm, not 0.8 cm. c) A flow speed of 60 cm/s corresponds to a diameter of 1.125 cm, not 0.8 cm. d) The flow is characterized by the continuity equation because no amount of blood is lost between the two locations. The continuity equation is 30 cm/s × π × ((1.6 cm)/2)2 = v × π × ((0.8 cm)/2)2. Solving for v yields v = 4 × 30 cm/s = 120 cm/s.

What is the name of the ionic compound used to make Buffer B? A.Ammonium formate B.Ammonium carbonate C.Ammonium bicarbonate D.Ammonium acetate

Solution: The correct answer is D. a) Ammonium formate is NH4HCO2. b)Ammonium carbonate is (NH4)2CO3. c) Ammonium bicarbonate is NH4HCO3, a component of Buffer A. d) Ammonium acetate is NH4CH3CO2, a component of Buffer B.

A major obstacle to obtaining useful energy from a nuclear fusion reactor is containment of the fuel at the very high temperatures required for fusion. The reason such high temperatures are required is to: A. eliminate the strong nuclear force. B. remove electrical charge from reactants. C. decrease the density of the fuel. D. enable reactants to approach within range of the strong nuclear force.

Solution: The correct answer is D. a) The strong nuclear force cannot be eliminated by increasing the temperature because this force manifests whenever nucleons are present and within a range of picometers. b) The electrical charge is an intrinsic property of matter that is independent of temperature. c) Decreasing the density of the fuel is detrimental because the probability of fusion increases with the increase in fuel density because the attractive component of the strong nuclear force acts at short distances indicative of high density for the fuel material. d) The probability of fusion increases with the decrease in the average distance between fuel particles that enables attractive nuclear forces to overcome the repelling nuclear forces acting at medium and long distances. An increase in the temperature is equivalent to an increase in the root-mean-square speed of the fuel particles that will travel the average distance between fuel reactants in smaller times. The associated increase in the kinetic energy of the particles relative to the center of mass of the nuclear fuel system essentially correlates with a decrease in the electrostatic potential barrier that repels particles of the same electric charge. In turn, this increases the probability of particles to undergo the tunnel effect by penetrating the electrostatic barrier. These combined effects enable reactants to approach within range of the strong nuclear force.

4CO(g) → C3O2(g) + CO2(g) Reaction 1 Based on Reaction 1, when 1.0 atm of CO(g) completely reacts to form carbon suboxide at 550°C in a sealed container, what is the final pressure in the container? A.0.00 atm B.0.10 atm C.0.25 atm D.0.50 atm

Solution: The correct answer is D. a) There is still gas present in the container. Since pressure is directly proportional to moles of gas at constant V and T, this is a smaller pressure than the balanced equation allows. b) This result ignores the pressure contribution of one of the gases. c) Based on Reaction 1, 4 mol CO(g) forms 2 mol of gases. d) Because of the direct relationship between P and n at constant V and T, that means 1.0 atm CO(g) makes 0.50 atm of gases

How many moles of NaCl were contained in 500 mL of the buffer solution used to elute hMPRα? the buffer solution used to elute the protein contained 300 mM NaCl, 50 mM NaH2PO4 (pKa = 7.2), and various amounts of NaOH (MM = 40 g/mol) A. 7.5 x 10-3 B. 3.0 x 10-3 C. 3.0 x 10-2 D. 1.5 x10-1

Solution: The correct answer is D. a) This is the amount of NaCl found in only 7.5 mL of buffer. b) This is the amount of NaCl found in only 3.75 mL of buffer. c) This is the amount of NaCl found in 37.5 mL of buffer. d) The concentration of NaCl in the buffer solutions was 300 mM. The amount of NaCl contained in 500 mL can be calculated as: 300 mmol/L × 500 mL × 1L/1000 mL = 150 mmol = 150 × 10-3 mol = 1.5 × 10-1 mol.

Researchers performed in situ laser-induced fluorescence imaging and spectral analysis of different skin areas of patients with acne vulgaris. They used the fluorescence spectrometer depicted in Figure 2, which employs a 86Kr+ laser that simultaneously emits radiations of wavelengths 407 nm and 605 nm. Inside the laser, the noble gas is contained in a 11-cm3 tube. Two prisms were used to select the desired wavelength from the radiations emitted by the laser. A multi-fiber sensor assembly sent the 407-nm wavelength radiation to the skin sample and collected the fluorescence radiation emitted by the skin. Approximately how many moles of Kr+ are contained in the laser tube at 0°C and 1 atm? A.3 x 10-7 B.2 x 10-6 C.4 x 10-5 D.5 x 10-

Solution: The correct answer is D. a)This number of moles of gas occupies only 0.00672 mL at STP. b)This number of moles of gas occupies only 0.0448 mL at STP. c)This number of moles of gas only occupies 0.90 mL at STP. d) There is 1 mole of gas in a volume of 22.4 L = 2.24 x 10 cm3 at STP; there are approximately 5 x 10-4moles in 11 cm3.

what are steroids and where are they derived from?

Steroids are a class of lipids that are derived from cholesterol.

One function of the myelin sheath is to: A. insulate the axon from the surroundings. B. decrease the radius of the axon. C. produce Schwann cells. D. increase the capacitance of the axon.

The correct answer is A. A) The myelin sheath insulates the axon from the surroundings. This insulation acts to increase the rate of transmission of signals. B) The myelin sheath increases, not decreases, the radius of the axon. C) The Schwann cells form the myelin sheath, and not vice versa. D) The myelin sheath decreases, not increases, the capacitance of the axon.

in Circular dichroism (CD) , the signal measured arise from the chirality of A.α carbon. B.amide nitrogen. C.carbonyl carbon. D.β carbon.

The correct answer is A. a)In a protein, each amino acid residue, except Gly, has a chiral α carbon. b) Amide nitrogen is achiral. c) Carbonyl carbon is achiral. d) With the exception of Ile and Thr,the β carbon in amino acid residues is achiral. CD spectra arise from the massively greater presence of α carbon chirality.

Channel X transmits only the smallest substances dissolved in the extracellular fluid through the axon membrane. Which substance does Channel X transmit? A. Proteins B. Sodium ions C. Potassium ions D. Chloride ions

The correct answer is B. A) Proteins are very large in relation to cations. B) Sodium cations are small by virtue of their positive charge and low atomic number. Of the answer choices, sodium cations are the smallest. C) Potassium cations are larger than sodium cations because they are from the next period in the Periodic Table. D) Chloride ions are much larger than sodium cations. The two ions are isoelectronic, but sodium has two more protons in its nucleus. Because the "size" of an ion or atom is really dictated by the electron "cloud," sodium cations are smaller because of the greater electrostatic attraction of the more highly positively charged nucleus.

What amino acid has a Indole ring?

Tryptophan

What quantity of Compound 1 must be provided to prepare 100.00 mL of solution with a concentration equal to Ki? A.48.4 mg B.24.2 mg C.5.64 mg D.2.92 mg given information from the passage: - Compound 1 (molar mass: 483.5 g•mol−1) Table 1 Kinetic Parameters of HIV-1 Protease with and without Compound 1 uM compound 1 KM(mM) kcat(s−1) KM(mM−1 s−1) Ki(μM 0 1.678 0.250 00.149 — 60.0 0.832 0.126 0.151 60.3 120 0.430 0.064 0.149 60.3 Note: HIV-1 protease follows Michaelis−Menten kinetics. A Lineweaver-Burk plot of Vo−1 versus [S]−1, both without Compound 1 and with different concentrations of Compound 1, generates a set of parallel straight lines with positive slopes.

right answer is D.2.92 mg the question is asking: X has a molar mass of 483.5 g/mol. How much of X (in milli grams mg) is needed to make 100 mL of a 60.3 µM solution I recommend rewriting everything in the most useful units using scientific notation: 483.5 g/mol = 483.5 x 10****3 mg/mol 60.3 µM = 60.3 x 10****-6 mol/L 100 mL = 0.1 L Then it looks a bit more straightforward to calculate how much of Compound 1 we need: 60.3 x 10-6 mol/L = X/(0.1L) X = (60.3 x 10-6 mol/L)*(0.1 L) = 6.03 x 10-6 mol Now we just need to convert mol to mg: (6.03 x 10-6 mol)*(483.5 x 103 mg/mol) = 2.92 mg

Which of the following species has an electron configuration equivalent to that of a noble gas? A.Ca2+ B.Cu2+ C.O D.H

solution: The correct answer is A. a) Ca2+ has lost its valence electrons and thus attains the electronic configuration of the previous noble gas (Ar). b)Copper has 11 outer electrons. Loss of 2 electrons leaves the cation with a [Ar] 3d10 configuration. c)An oxygen atom is not a noble gas. It has a partially filled 2p subshell. d) A hydrogen atom is not a noble gas. It has a half-filled 1s subshell.

An ice cube at 0°C and 1 atm is heated to form steam at 100°C and 1 atm. Ignoring heat loss to the surroundings, what part of the process uses the most heat? (Note: Specific heat of water = 1 cal/g°C. Heat of fusion = 80 cal/g. Heat of vaporization = 540 cal/g.) A.Melting the ice cube B.Heating all the water from 0°C to 50°C C.Heating all the water from 50°C to 100°C D.Vaporizing all the water

solution: The correct answer is D. a) Melting the ice cube requires only 80 cal/g, which is less than the heat of vaporization. b) Heating all the water from 0°C to 50°C requires 50 cal/g, which is less than the heat of fusion. c) Heating all the water from 50°C to 100°C requires 50 cal/g, which is less than the heat of fusion and the heat of vaporization. d) Vaporizing all the water requires 540 cal/g, which is a greater heat requirement than specific heat or heat of fusion.

Which of the following substances is polar? A. NF3 B. CCl4 C. CO2 D. Li2

the correct answer is A. a)The geometry of trifluoroamine is impacted by the lone pair on nitrogen, making it trigonal pyramidal. No bond dipoles cancel; this results in a polar molecule. b)While each C-Cl bond in carbon tetrachloride is polar, the sum of the dipole moments cancel as a result of its tetrahedral molecular geometry. c)Carbon dioxide is a linear molecule. The bond dipole moments of each C=O bond cancel as they are in opposite directions. d)This molecule is necessarily non-polar as it is comprised of two identical atom

what is the relationship between the photon energy and the radiation wavelength?

the photon energy is inversely proportional to the radiation wavelength