Magoosh FL #1 Exam Review PDF

Magoosh FL #1 Exam Review 2/18/24 487 Section: Chem/Phys 1. What is the standard free energy (ΔG°) for reaction 3? a. What’s1 being tested: Equation that relates free energy to the cell potential: ΔG = -nFE, and being able to calculate for the number of electrons that were transferred during the rea...

Magoosh FL #1 Exam Review 2/18/24 487 Section: Chem/Phys 1. What is the standard free energy (ΔG°) for reaction 3? a. What’s1 being tested: Equation that relates free energy to the cell potential: ΔG = -nFE, and being able to calculate for the number of electrons that were transferred during the reaction. Equation that relates free energy to the cell potential: ΔG = -nFE ΔG ° E Change in free energy Standard electric potential for the redox reaction. Also known as the standard cell potential, is the voltage measured when the cell operates under standard conditions. F Faraday's constant (1x10^5 C/mol E) magnitude of charge that is carried by one mole of electrons n # of electrons that were transferred during the reaction 1. In the question, it shows that that everything is in standard condition, so there is nothing weird we must do there. From here, you would start by balancing the oxidation and reduction half-reactions. From here, we can see that the number of electrons is not balanced, so if you multiply the oxidation half reaction by 2 yields: 4Fe2+ -----> 2Fe2 O3 +4E3. From here, the rest of the reaction does not have to be balanced, we know that 4 electrons were transferred, and now we know that n=4. 4. Then, substituting into the equation gives: ΔG° = -nFE° = -(4) (1 x 105 )(-0.37) 2. Joules---> Kilojoules: divide by 1,000 3. Which of the following would you expect to be true regarding the following half reaction? Zn(s)---> Zn2+ +2eA. Here the mcat is testing understanding E and spontaneous vs. non-spontaneous. When to apply Nernst equation: The Nernst equation is used to calculate the potential (voltage) of a cell under nonstandard conditions, which means when the concentrations of the reactants and products are not at their standard state (1M for solutions and 1ATM for gases) or when the temperature is not at the standard temperature (25C or 298K) The Nernst equation is used to calculate the cell potential under non-standard conditions. The equation is: E=E°-RT/nFlnQ which can also be written as to E=Rt/zF ln (ion outside/ion inside) E° The standard cell potential R Gas constant 8.314 J/mol T Temperature in Kelvin n Number of moles of electrons transferred in the reaction F Faradays Constant 1x105 C/Mol Q The reaction quotient A. In the passage it said that zinc is preferentially oxidized over iron. Then it states; galvanization, a process where zinc is layered over steel, prevents rust formation by sacrificing zinc through oxidation instead of allowing iron to oxidize and form rust. This implies that the oxidation reaction of zinc-tozinc ions must have a more positive standard reduction potential compared to the oxidation of iron. a. In galvanization, zinc acts as the anode, which means it will undergo oxidation to protect the iron in the steel. The more positive standard reduction indicates the oxidation of zinc is more favorable or spontaneous compared to the oxidation of iron. 4. in reaction 2, the oxidation state of the oxygen atom in O2 AND H2 O are: a. here the mcat is testing oxidation state, along with understanding how a diatomic worked. Grou p 1 2 13 17 Oxyg en (O) Hydro gen (H) Elements Typical oxidation State Li, Na, K, etc. Be, Mg, Ca, etc. B, Al, Gam etc. F, Cl, Br, I, Etc. O +1 +2 +3 -1 -2 H +1 with non-mentals, -1 with metals Since O2 is a diatomic molecule, the atom shares the electrons equally, which results in no transfer of electrons between the atoms. If you were to have O 3 , it would be different because the electrons DO NOT share electrons equally. So, when calculating: - We know that oxygen has an oxidation state of –2, and since it is diatomic, it will have an overall oxidation state of –2. - Then, for H2 O we know that oxygen has a –2 charge and hydrogen have a +1. Since there are 2 hydrogen molecules, that will account for a +2 charge, since oxygen as a –2 charge, that will then have the overall net charge cancel out and be equal to 0. 6. A patient currently has a measured aortic valve area of 1.0cm2. By what % will the diameter of the valve have to decrease before stenosis can be considered severe? a. here the mcat is testing the relationship between diameter and area. In the passage, it stated that the valve area of a patient with severe aortic stenosis is 0.8cm2 Then, you must apply the following equation below which is used to calculate the area of a circle with the given diameter: A 1 = π(d1 /2)2 AND A 2 = π(d2 /2)2 o This equation will be used because we are calculating the area of a circle with a given diameter. ▪ A1=area of the circle (in this case A1 is the initial area of the aortic valve and A2 is the area after the diameter decreases). ▪ Π=represents the ratio of the circumference of the circle to its diameter. ▪ D1= the diameter of the circle ▪ (d1/d2)^2 calculates the radius of the circle by dividing the diameter by 2 and then squaring the results. The radius gives the area of the circle. Then, you the divide the equations: A 2 /A1 = (d 2 /d 1 )2 - We are trying to determine the change in the diameter of the aortic valve that corresponds to a change in the valve area. Taking the square root of both sides yields the following: d 2 /d 1 = √(A 2 /A 1 ) Then, we set A1 to 1.0 and A2 to 0.8 and you get the following: D2/d1=√ 0. 8= roughly 10% Short cut for finding square root of decimal: 7. During diastole, 50mL of blood accumulates in the left ventricle of a patient's heart. How much work must be done by the heart in order to pump this blood from the left ventricle to the patient's brain if they are standing upright and the distance from their heart to their head is 0.5 meters? (Note: assume the velocity of the blood in the heart and brain in negligible and that the density of blood is 1.06 g/cm3 ) Unit conversions: 1. Convert the volume to cubic meters: a. V=50mL =50*10^-6m^3 (1mL=1x10^-6) 2. Change the density of blood from g/cm^3 to kg/m^3 so that the units are all the same a. 1.06g/cm^3=1.06kg/L (1cm^3=1x10^-3L)=1060kg/m^3 (1L=1x10^-3) Work=fd Work=f(grav)d Work=m*g*d Work=(pv)*g*d W=(1060kg/m^3* 50*10^-6m^3)(10m/s^2)(0.5m) =(1.060x10^3)(50x10^-6) =(50x10^-3kg)(10m/s^2)(0.5m) =(0.05)(10)(0.5) =0.26J which is closest to 3x`0^-1 J 9. According to the experiment in the passage, how many moles of NaOH were added to the lactic acid solution at the equivalence point? Equivalence point: the # of moles of acid = the # of moles of added base To find equivalence point: 1. Find the # of moles of lactic acid originally present 2. (2grams of lactic acid) (1mol lactic acid/90grams) =1/45 moles =0.02 moles 11. According to the passage, what is the Ka of lactic acid? half equivalence point=pH=pKa (the flattest part on the titration curve) equivalence point=steepest part on the titration curve. it is where the amount of titrant added is jut enough to completely neutralize the analyte of the sol. in acidbase titration, moles of base=moles of acid. end point= right after the equivalence point. it is where the color changes and indicates the titration is over. pH=pKa at equivalence point pH=3.5 log(pka)=ka =10^-3.5 12. When the student repeats the experiment using an indicator to determine the equivalence point of the titration, which of the following indicators should he use? Using an acid-base indicator, marks the end point of a titration by changing color. The most common acid-base indicators are weak acids. They will exhibit a color when the proton is attached to the molecule and a different color when the proton is absent. crystal violet: Ka=1.3x10^-1 cresol red: Ka=1.3x10^-8 alizarin:Ka=2.0x10^-12 bromcressol green:Ka=2.5x10^-5 The equivalence point here is when the ph=8. So pka Therefore, you want an indicator with a pKa between 7-9. 13. Upon completing the experiment, the student discovered that the molarity of stock NaOH solution he used as a titrant was accurate, but it was contaminated with some additional NaOCl (bleach). How would this information affect the accuracy of his calculated value of K a for lactic acid? - when we look at bleach, we can break it apart: Na+ and OCl-. Na+ can dissolve in te solvent, but since is OCl- is a weak base, you have to think why would happen to the results if there was extra base in the NaOH solution. We know that the function of the base is to remove the acidic proton from in this case, the lactic acid molecule. Therefore, additional base would remove additional protons from the lactic acid molecule, which would make lactic acid stronger than it is, which means stronger acids lose protons more easily. Higher Ka values indicate stronger acids, so the student would have a higher Ka value. Ka Interpretation Acid/Base Behavior Value Ka > 1 Strong acid Completely dissociates in solution, forming many H+ ions Partially dissociates in solution, equilibrium favors Ka ≈ 1 Weak acid reactants Minimally dissociates in solution, equilibrium favors Ka < 1 Very weak acid reactants Extremely weak Ka ≈ 0 Virtually no dissociation in solution acid *14. A converging lens produces a real inverted image of a pencil located 0.15 m from the lens. If the image produced is located 2.5 cm from the lens, then the lens's focal length is: 1 1 1 = 𝑑𝑜 + 𝑑𝑖 𝑓 F=focal length of lens Do=object distance (distance from the object to the lens) Di=image distance (distance from the image to the lens) 1 1 1 = + = 15/7 𝑓 15 2.5 Purpose of equation: The lens equation is used to relate the focal length of a lens to the distances of an object and its image formed by the lens. It's a fundamental equation in optics and is used to calculate and predict the behavior of lenses in various optical systems. Focal length: units in M or cm - Distance from the lens at which parallel rays of light converge (converging lens) - Or appear to diverge from a diverging lens, after passing through the lens. Object distance units in M or cm - Distance from the object to the lens - For a real object: (+) - For a virtual object formed by a mirror lens(-) Image distance units in M or cm - Distance from the image to the lens. - For a real image formed by a lens: (+) - For a virtual image: (-) Converging optical elements focus light rays to a point, either real or virtual, after passing through or reflecting from the element. Diverging optical elements cause light rays to spread out or diverge after passing through or reflecting from the element. 15. Tension: T1=T2 in frictionless pulley system where the rope is massless and there is no friction. T=Mg+ma When a-=0, T=Mg T=Mg ,and since the 5kg mass descents with a constant velocity and the system is in equilibrium, the tension in the rope is equal to the equation given. M1=2kg, M2=5Kg M1a=T-fk Since mass 1 moves horizontally, with a constant velocity, a=0. Therefore: T=fk Since the tension in the rope is the same throughout: T=m2g M2g=fk Fn=m1g ( Fn is the force exerted by the surface on the object) Fk=uk*fn=uk*m1g M2g=uk*m1g Gravity cancels out and solve for uk: Uk=m2/m1 16. - first identify whether this is R or S: which it indeed is S. I was correct with identifying that, but I was wrong when identifying whether the R group was an ethanol or acetamide. Ethanol Acetamide Even if you did not know acetamide the main thing here was to identify that this was NOT ethanol, because ethanol should be a known structure. 17. All of the following are underlying assumptions/conclusions of the ideal gas law except: › “Intermolecular interactions are maximal” would be the correct answer because that assumption violates the assumption of the ideal gas law that says there are no intermolecular interactions. 18. i. is correct because in the last paragraph it states “the current flowing through the coil is directly proportional to the magnetic susceptibility.” ii. is false because the resistance will not change unless the shape or composition of the coil is physically changed. iii. Is not true because we can see that p is proportional to the square of the magnetic susceptibility and varies exponentially, not linearly. In electrical systems, where resistance is involved, the power can be expressed using ohms law: P=I^2R p=power I= current flowing through the resistor R=resistance of the resistor This equation appears to represent the power loss denoted by P in a magnetic material subjected to an alternating magnetic field: P=(KxB)^2R You can sub in the KxB for the I in the power equation. P: This represents the power loss in the magnetic material. Power loss refers to the energy lost as heat due to various factors like resistance, hysteresis, and eddy currents within the material. k: This is a constant related to the properties of the material, often associated with the material's magnetic permeability. χ: This represents the susceptibility of the material, which describes how it responds to an applied magnetic field. It's a dimensionless quantity. B: This represents the amplitude of the magnetic field to which the material is exposed. R: This represents the resistance of the material, often referring to the effective resistance that leads to power dissipation as heat. 20. If the radius of a current-carrying loop of wire is doubled, and the current is externally decreased by a factor of 2, what will be the net change in the magnetic dipole moment? In the passage they provided the following equation: μ = IA This equation is for a current flowing in a closed loop path. The magnetic dipole moment is μ, and can be expressed as the product of the current and the area of the loop. μ = IA will then be accommodated for the area of circle and will become: μ = I𝜋𝑟2 μ = IA μ 1= I𝜋𝑟2 μ 2= (I/2)𝜋(2𝑟2 ) μ is directly proportional to the current and directly proportional to the square of the radius (from the area of the loop) therefore if the radius was doubled then μ would double. 20. How much work is required to move an alpha particle from a 6-V potential region to an 8-V potential region? We know there is a difference of 2 volts between the 6-V and 8-V potential. Volts are J/C We need to calculate the energy (work) needed to move a charged particle through a given potential difference by doing what is presented on the right. Section: Biological and biochemical 1. “ A common method of immunocytochemical tagging involves the binding of protein A, derived from staphylococcus aureus to the Fc region of the antibodies. Gold particles couple to protein A, and the gold/protein A complexes present themselves as abnormal dark spots under the microscope.” - This should have been a hit for it to be lysosomes, which are found within macrophages, the abnormal dark spots. Macrophage: a type of white blood cell. They are a type of phagocyte, which means they engulf and digest foreign particles, microbes, and dead or dying cells. They are found in various tissues throughout the body. They also contribute to regulation of inflammation and tissue repair. Lysosome: membrane-bound organelle found in the cytoplasm of eukaryotic cells. Contain a variety of enzymes capable of breaking down various molecules including proteins, lipids, carbs, and nucleic acids. Play a critical role in cellular digestion, recycling cellular components, and degrading foreign martials that enter the cell through processes such as endocytosis and phagocytosis. They are referred to as the digestive organelles. Macrophages filled with lysosomes fulfills their primary function of phagocytosis. Erythrocyte: red blood cells which serve to transport oxygen and nutrients. You would not expect vesicles filled with proteolytic enzymes to present in erythrocytes. Helper t lymphocyte: primarily involved in the secretion of cytokines, which are signaling proteins, not proteolytic enzymes. B lymphocyte: primarily involved in the production and secretion of antibodies. They are NOT phagocytic, so you would not expect them to contain an abundance of lysosomes. Proteolytic enzymes, also known as proteases, are a class of enzymes that catalyze the hydrolysis of peptide bonds in proteins. In digestion, proteolytic enzymes such as pepsin (in the stomach) and trypsin, chymotrypsin, and carboxypeptidases (in the small intestine) break down dietary proteins into smaller peptides and amino acids, facilitating their absorption by the intestinal lining. These enzymes play a crucial role in the digestive process by enabling the body to access essential amino acids for protein synthesis and other metabolic functions. In cellular processes, proteolytic enzymes regulate protein turnover by degrading damaged or misfolded proteins. This helps maintain cellular homeostasis and ensures the removal of unwanted proteins. Proteolytic enzymes also participate in protein synthesis, signaling pathways, and various other cellular functions, contributing to the overall regulation and function of biological systems. 2. This question is simply asking, where are microfilaments located? Microfilaments: threadlike protein structures that give cells their shape. The two main microfilaments are actin and myosin in muscle cells. They are predominantly found in the cytoplasm of cells such as leukocytes. They are not found in the smooth ER, as this is where lipid synthesis and detoxification of poisons occurs. The Golgi apparatus is incorrect because this is where they package proteins in vesicles for storage or exocytosis(cells activity release molecules or particles from within the cell to the external environment. During this, vesicles, which are membrane bound sacs containing the material to be expelled, fuse with the cell membrane, leading to the discharge of the vesicle contents outside the cell. This is essential for various cellular functions such as secretion of hormones, neurotransmitters, enzymes, and waste products.). The rough ER is not correct because they are not present here and this is responsible for protein synthesis. 3. The purpose of protein A is to be the glue between the antibody and the particle of gold, which is what makes the particle visible in this procedure. If protein A were to bind to the antigen-binding portion of the antibody, you would see a diminished antigen-antibody interaction and the whole technique would be compromised. You do not want the protein binding to the antigen, rather you want the antigen-antibody interaction. For option A, it is true that the gold particles would still couple to protein A, but the protein being measured would not be able to bind to the antibody, because its binding site would be occupied. This would make the whole technique pointless. 4. A secondary immune response is primed with an infectious agent, memory b-cells will activate much more rapidly upon a subsequent exposure to an identical agent, resulting in prolific production of antibodies to combat the pathogen. Hence for the option choice of the same cohort of rabbits should be used over and over, because their immune systems will recognize the human cytokeratin more quickly due to the presence of memory cells developed after the first infection. The same cohort of rabbits will have immune systems that are able to recognize the cytokeratin and react more vigorously. Option B is incorrect because the amount of time required for rabbits to secrete large quantities of antibodies will DECREASE during each subsequent infection with human cytokeratin after the first. Option D is incorrect because macrophages ARE NOT directly involved in the secretion and production of the antibodies that will recognize the cytokeratin and facilitate a secondary immune response. 8. Start codon: AUG - remember, school STARTS in August Stop Codon: UGA - Remember, U Go Away Here, they are asking if we know what the initiation (start) codon is for the mRNA that enters the ribosome, leading with its 5’ end. Downstream of the 5’ UTR, you would expect the first codon to be there, which would be the start codon. 9. The duodenum is the first part of the small intestine, and it plays a crucial role in the digestion process. Bile and Lipase are secreted into the duodenum approximately midway through. It would be present in the distal duodenum but not the proximal duodenum. Insulin is an endocrine hormone and it is secreted into the blood and should not be significantly present anywhere in the lumen of the digestive tract. Chyme is a semi-liquid substance consisting of partially digested food and gastric juices. From the stomach, it enters the duodenum, where it encounters secretions from the gallbladder and pancreases, which raises pH from 2 to 7. You would expect this to be present in ALL parts of the duodenum. *10. MCAT seems to test the general function of an enzyme in one way or another. The 4 major points regarding the function of enzymes: 1. They increase the rate of the forward and reverse reaction. 2. They lower the activation energy of a reaction. 3. They do not alter the equilibrium concentrations of the reactants and products. 4. They are never consumed during a reaction. Characteristic Competitive Inhibitors Non-Competitive Inhibitors Binding Site Active site of the enzyme Allosteric site of the enzyme ( a site other than the enzymes active site, causing a change in enzymes shape) Interaction with Enzyme Reversible binding Reversible binding Effect on Substrate Competes with substrate for active site. Competitive inhibitors do not affect the max activity of an enzyme. Does not compete with substrate for active site, this results in a decrease in the enzyme molecules that can catalyze the reaction. Hence, Vmax is lowered. Substrate Increasing substrate concentration Increasing substrate concentration cannot Concentration can overcome inhibition overcome inhibition Reversibility Mode of Inhibition Inhibition can be overcome by adding more substrate Inhibition generally cannot be overcome by adding more substrate Directly blocks active site Alters enzyme's conformation, affecting its activity Change of Km Increased Unchanged 11. In the question it states “but that inhibition is overcome with the addition of excess substrate.” This indicates that it is a competitive inhibitor, because non-competitive inhibitors do not compete with the substrate for active site. The Vmax, is the max velocity of an enzymatic reaction when the enzyme is saturated with its substrate. For option A, it is incorrect because since this is a competitive inhibitor and a competitive inhibitor does not lower the Vmax of the enzyme. For option B, it states that it binds to the active site of the enzyme, which would indicate that it is a competitive inhibitor because they bind to active site of enzyme while non-competitive inhibitors bind to the allosteric site of the enzyme. For option C, it binds to the allosteric site of enzyme would describe a non-competitive inhibitor. For option D, it lowers the Km of the enzyme catalyzed reaction, the Km is the concentration at which we are at ½ of max velocity. A low Km indicates high affinity (the enzyme can achieve half of its max velocity at low substrate concentrations) while a high Km indicates a low affinity. The Km value is obtained as the substrate concentration at which the reaction rate is half of Vmax on the plot. Competitive inhibitors increase the Km, not decrease the Km, hence the inhibitor can be overcome with additional substrate. 12. The Michaelis-Menten equation describes the rate of an enzymatic reaction as a function of substrate concentration. The equation is: 𝑉= 𝑉𝑚𝑎𝑥 [𝑠] 𝐾𝑚+[𝑠] V=initial reaction rate Vmax= max reaction rate when the enzyme is saturated with substrate [s]=substrate concentration Km=Michaelis constant, which is equal to the substrate concentration at which the reaction rate is half of Vmax 1 𝑟𝑎𝑡𝑒 versus 1 [𝑠] , is defined as the Lineweaver- burke plot. This is a plot of the reciprocal of the Michaelis-Menten plot. This linearity is due to the reciprocal relationship between the reaction rate (v) and substrate concentration (s) within the Michaelis-Menten equation. This linearizes the hyperbolic curve of the Michaelis-Menten plot and allows for a more straightforward determination of Vmax and km. When we rearrange the Michaelis-Menten equation for the Lineweaver-burke plot, we get: 1 𝐾𝑚 + [𝑠] = 𝑣 𝑉𝑚𝑎𝑥[𝑠] 1 So, plotting 𝑟𝑎𝑡𝑒 versus 1 y-axis of 𝑉𝑚𝑎𝑥. 1 [𝑠] , will yield a straight line with a slope of 𝐾𝑚 𝑉𝑚𝑎𝑥 ⬚ and an intercept on the 13. A non- competitive inhibitor binds to an allosteric site of the enzyme, not the active site. Therefore, the flooding of an enzyme with substrate would not be able to overcome the effect of the non-competitive inhibitor. Increasing substrate concentration CANNOT overcome noncompetitive inhibition because the inhibitor binds to the enzyme regardless of whether the substrate is bound. On a graph, non-competitive inhibition results in a decrease in the Vmax value of the enzyme, while the Km value remains unchanged. As the concentration of the inhibitor increases, the Vmax decreases, but the Km value remains the same. This is because the inhibitor reduces enzymes catalytic efficiency without affecting its affinity for the substrate. A would not work because the concentration should still be increasing and the non-competitive inhibitor would not be affected by this. Also, the Vmax is the same here and noncompetitive inhibitors low the vmax. B is the correct answer because you have to look at a graph with decreased Vmax value, since non-competitive inhibitors lower Vmax. C is incorrect because the Vmax is the same. Analogy for understanding competitive vs non-competitive inhibitors Let's imagine you have a toy car (the enzyme) and colorful balls (the substrates) that you want the car to push around. When the car pushes the balls, it goes faster and faster until it reaches its maximum speed. Now, let's talk about two different kinds of friends who might come and play with you: 1. Competitive Friend: This friend loves playing with the same balls you have. So, when there are more balls around, your friend will grab some for themselves, leaving fewer for you to push. This friend competes with you for the balls. But, if there are lots and lots of balls, you can still find some to push even if your friend takes some. 2. Non-competitive Friend: This friend has a different game. They bring blocks and stack them on your car, making it harder for your car to move, even if there are plenty of balls around. Your car gets slower because of the blocks, no matter how many balls there are. In the case of the competitive friend, if there are many balls, you might still be able to push some, even if your friend takes a few. But with the non-competitive friend, even if there are many balls, your car struggles to move because of the blocks. Similarly, in biology, enzymes are like the car, substrates are like the balls, and inhibitors (competitive or non-competitive) are like the friends. Competitive inhibitors compete directly with the substrates, while non-competitive inhibitors hinder the enzyme's function regardless of how many substrates there are. 14. Passive diffusion: - movement of molecules from an area of HIGH concentration to an area of LOW concentration, driven solely by the concentration gradient. Requires spontaneously does not require the input of energy. - does NOT requires specific transport proteins but relies solely on the physical properties of the molecules and the characteristics of the membrane. Secondary active transport: - Coupled movement of 2 different molecules or ions across a membrane. The movement of one molecule against its concentration gradient is coupled with the movement of another molecule down its concentration gradient. - The energy required here is derived from the electrochemical gradient established by the primary active transport (sodium gradient established by the Na+/K+ ATPase pump). There are 2 different types of secondar active transport: o Symport: both molecules move in the same direction across the membrane o Antiport: molecules move in opposite directions across membrane. Active transport: - Active transport involves the movement of molecules against their concentration gradient, from an area of low concentration to an area of higher, which requires energy expenditure, usually in the form of ATP. Option A is incorrect because passive diffusion MAY play a role in the reabsorption of some small, non-polar molecules in the nephron, it is not the primary mechanism for the reabsorption of amino acids. Amino acids are relatively large and polar molecules that requires specific transport mechanisms or efficient reabsorption. Only small, uncharged molecules such as water and urea can be reabsorbed this way. Option B is correct because the proximal convoluted tubule is where sodium, glucose, and amino acid reabsorption occurs in the nephrons. These occur by active transport and require heavy energy expenditures. Option C is incorrect because hydrostatic pressure differential is the motivating force behind the filtration of fluid into the nephron, not the reabsorption of solutes out of the nephron. Option D is incorrect because the amino acid reabsorption in the nephron DOES NOT occur by passive diffusion. The idea of “active reabsorption” of water does not make sense. Water movement follows the active transport of solutes AGAINST their gradients. Property Energy Requirement Passive Diffusion Does not require energy input Active Transport Secondary Active Transport Requires energy input (usually ATP) Requires energy indirectly from an electrochemical gradient established by primary active transport Movement of Molecules Moves along concentration gradient Moves against concentration gradient One molecule moves against its concentration gradient, while another moves with its concentration gradient Specific Proteins Does not require specific transport proteins Requires specific transport proteins (pumps/carriers) Requires specific transport proteins (cotransporters) Role Plays a role in equilibrating concentrations across membranes Maintains concentration gradients, cellular homeostasis, and performs various physiological functions Couples’ uphill movement of one molecule with the downhill movement of another molecule Examples Movement of small, nonpolar molecules like Sodium-potassium pump oxygen and carbon (Na+/K+ ATPase), calcium dioxide pumps, proton pumps Cotransport of glucose and sodium ions in the small intestine and proximal convoluted tubule The proximal convoluted tube is the first part of the nephron’s tubule. The nephron is the kidneys functional filtration unit. It is located IN the renal cortex, which is the kidney’s outer part. It is responsible for the reabsorption and secretion of water and various solutes. 15. Here you are being tested to know if you know the main basic functions of most eukaryotic organelles. The Smooth ER is to synthesize lipids, such as triacylglycerols. The peroxisome, convert hydrogen peroxide, H2O2, into oxygen and water, detoxifying certain harmful substances. They play an important role in lipid degradation, not synthesis. The lysosomes contain proteolytic enzymes which digest intracellular substances. They DO NOT function in lipid synthesis. The Golgi apparatus is responsible for packaging and modifying proteins for vesicular transport, not lipid synthesis. In the question, being composed predominantly of triacylglycerol, along with chylomicrons being synthesized, is indication of the smooth ER. The partial double-bond character of the peptide bond arises due to the resonance between the nitrogen lone pair and the carbonyl oxygen, which leads to electron delocalization. This delocalization restricts rotation around the carbon-nitrogen axis of the peptide bond. Therefore, option a is the correct explanation for the diminished rotation. 17. which of the following cells is diploid? For the MCAT, you should memorize the sequence of events and cell types that occurs during spermatogenesis. During spermatogenesis, the seminiferous tubules of the testes provide the microenvironment necessary for the development and maturation of sperm cells. The process is regulated by hormones such as follicle-stimulating hormone (FSH) and testosterone. Spermatogenesis is continuous and occurs throughout the reproductive lifespan of males. Stage of Spermatogenesis 1. Spermatogonia proliferation 2. Meiosis I 3. Meiosis II 4. Spermiogenesis 5. Spermiation Cell Type(s) Involved Description Spermatogonia are diploid stem cells located in the seminiferous tubules of the testes. They undergo mitotic divisions to replenish the germ cell population. Primary spermatocytes, derived from spermatogonia, Primary undergo the first division of meiosis, resulting in the spermatocytes formation of secondary spermatocytes. Secondary Secondary spermatocytes undergo the second division spermatocytes of meiosis to produce haploid round spermatids. Round spermatids undergo a series of morphological Round spermatids changes, including elongation and development of tail structures, to form mature spermatozoa (sperm). Mature Mature spermatozoa are released from the epithelium spermatozoa of the seminiferous tubules into the lumen for further (sperm) maturation and storage in the epididymis. Spermatogonia 18. A is incorrect because the acrosome is a region of the sperm cell, which is not relevant to the question here. B is correct because in the question it talks about secreting large quantities of progesterone, which is a steroid, and steroids are lipids, and lipid synthesis occurs in the smooth ER. Therefore, we can infer that progesterone-secreting cells will have extensive smooth ERs. C is incorrect because a corona radiata is NOT a part of a granulosa cell, rather it is a large structure composed of cells that surround the ovulated ovum. Also, the corona Radia is not located IN the ovary, it surrounds the ovum as the ovum travels through the oviduct. Therefore, it cannot be a feature of the ovarian follicle. The corona radiata is the structure found in the female reproductive system, specifically, it is a collection of cells that surround the egg cell (oocyte) within the ovarian follicle. During folliculogenesis, which is the process by which ovarian follicles develop and mature, the egg cells is focused within a structure called the graafian follicle. It serves several important functions: a. Protection, as it provides a protective layer around the egg cell, shielding it from mechanical damage and external influences. b. Nutrient exchange, it facilitates the exchange of nutrients and signals between the surrounding floccular cells and the egg cell, which helps support the eggs development and maturation. c. Ovulation, during ovulation the mature graafian follicle ruptures, which releases the egg cell along with the corona radiata into the fallopian tube. The corona radiata guides the egg cell towards the fallopian tube, where it may be fertilized by sperm. a. After ovulation, if fertilization occurs, the corona radiata also plays a role in protecting the newly formed zygote as it travels through the fallopian tube towards the uterus for implantation. D is incorrect because chromosomes are ONLY visible under the light microscope when they are coiled up during mitosis. 20. Oligosaccharides are short chains of sugar molecules, that contains several simple sugars linked together (monosaccharides). They are SMALLER than polysaccharides but larger than monosaccharides. They can be found on cell surfaces of glycoproteins and glycolipids. Amino acids is the answer because protein hormones such as follicle-stimulating hormone(FSH) and luteinizing hormone(LH), are composed of amino acids. Cholesterol derivatives are compounds that are structurally derived from cholesterol, and they are a type of lipid molecule that are found in cell membranes and are involved in various functions. Although cholesterol is a precursor molecule for the synthesis of FSH and LH, these hormones themselves are NOT cholesterol derivatives. Instead, they are protein hormones that are composed of amino acids, has that is why amino acids is the answer. Lipoproteins are particles that are composed of both proteins and lipids (including cholesterol and triglycerides) that transport lipids through the bloodstream. They play a crucial role in lipid metabolism, serving as carriers for cholesterol and triglycerides. Although they contain both proteins and lipids, they are not the primary constituents of protein hormones like FSH and LH. 22. *23. A residue is nothing more than an amino acid on a polypeptide. Therefore, 4k residue polypeptide has 4k amino acids. When looking at the Pka of an amino acid residue, it refers to the pH at which half of the residues are in their protonated form and half are in their deprotonated form. When looking at aspartic acid, the side chain of carboxylic acid, can either be protonated (COOH) or deprotonated (COO-), depending upon the pH of the solution. When the aspartic acid residue is surrounded by the non-polar(hydrophobic) residues deep inside a protein, it experiences a hydrophobic environment. In such an environment, the non-polar residues tend to repel water molecules and stabilize the hydrophobic core of the protein. Usually, in a hydrophobic environment, the apparent pKa of acidic residues increase. This means that the side chain of aspartic acid becomes less likely to donate a proton and remain in its deprotonated (COO-) form, even at lower pH values. The LOWER the Pka, the higher the acid is, and the more readily it donates a proton. A higher pka value indicates a weaker acid. 25. A monosynaptic reflex arc is a neural pathway that involves only ONE synapse between the sensory neuron and the motor neuron in the spinal cord. It is the simplest form of reflex arc found in the nervous system. Component Description Detects a stimulus, such as a change in muscle length or pressure, and Sensory Receptor initiates an action potential. Sensory Neuron Carries the action potential from the sensory receptor to the spinal cord. Interneuron May be present between the sensory neuron and motor neuron in some (Optional) reflex arcs. Receives the action potential from the sensory neuron and carries it to the Motor Neuron effector organ. Responds to the action potential by executing a rapid and involuntary Effector Organ action, such as muscle contraction. A key characteristic of a monosynaptic reflex arc is the presence of only ONE synapse between the sensory neuron and the motor neuron. This results in a VERY rapid efficient response to a stimulus, as there are fewer synaptic delays compared to the polysynaptic reflex arcs, which involve multiple synapses and interneurons. Characteristic Myelinated Neurons Unmyelinated Neurons Neurons with axons lacking myelin Neurons with axons covered by myelin Definition sheath; they may have small sections sheath, which acts as insulation. of myelin or none at all. Axons are surrounded by myelin sheath, Axons may have no myelin sheath or Myelin Sheath produced by Schwann cells (PNS) or small patches of myelin interspersed oligodendrocytes (CNS). along the axon. High conduction speed due to saltatory Conduction Relatively slower conduction speed conduction, were action potentials Speed compared to myelinated neurons. "jump" between nodes of Ranvier. Less energy-efficient compared to More energy-efficient due to saltatory Energy myelinated neurons due to conduction, requiring less ATP for ion Efficiency continuous conduction along the transport. axon. Found in the peripheral and central Found in various parts of the nervous systems, involved in rapid nervous system, involved in local Function transmission of electrical signals over signaling, autonomic functions, and long distances. reflex arcs. Saltatory conduction is a method of signal propagation along myelinated axons in neurons. The term saltatory means to leap or to jump, and in saltatory conduction, the action potential (nerve impulse) jumps from one node of Ranvier to the next, rather than propagating continuously along the entire length of the axon. Option A thin, myelinated fibers, is not the correct answer because for the monosynaptic reflex arc, you want it to be thick myelinated fibers to increase the velocity of the reflex arc. The monosynaptic has a key characteristic of only having one synapse between the sensory neuron and the motor neuron. This results in a very rapid response to a stimulus; hence you want that increased velocity. Option B, thin, unmyelinated fibers, is also incorrect because again, you want thick myelinated fibers to increase the velocity of the reflex arc. Also, unmyelinated neurons do NOT exhibit saltatory conduction, so their conduction velocity is slower than myelinated neurons, which would make the reflex arc less effective. Section: Psychological and social *2 The ideal design of a study to determine whether people are psychologically capable of foretelling their medical wishes is impossible to conduct in practice. Which of the following best indicates this ideal design and why it is impossible to conduct in practice? Although I got this correct, it would be beneficial to review the types of studies below: a. b. c. d. Case-control study Prospective study Cross-sectional study Retrospective study 6. Eidetic memory: also known as eidetic imagery, is the phenomenon known as “photographic memory.” It is the ability to recall an image from memory with high precision, for at least a brief period of time, after seeing it only once AND without using a mnemonic device. People with eidetic memories have visual sensory memories that last much longer than the average. - This would be the correct answer choice because it states how the student wrote it down exactly how he saw it, which explains the photographic memory part. Context-dependent retrieval: phenomenon where it is easier to recall memories if you are in the same external situation as you were when the memory was encoded. For this passage, this would be the testing room. - Since the student was not in the testing room it would not be context-dependent retrieval. State-dependent retrieval refers to the superior retrieval of memories when the individual is in the same physiological or emotional state as he/she was during the encoding process. - There is not a reason to believe that the student was in the same emotional state that he was when he encoded the memory of the tests. Long-term potentiation: phenomenon that occurs in the brain of all humans when neural pathways are repeatedly fired, leading to increased synaptic efficiency and the establishment of memory. This typically occurs after lots of repetitive practice. - There was no repetitive practice here and also, he was effortlessly recalling them. 7. Pineal body (pineal gland): -Looks like a pinecone - Since it is a gland, it secretes hormones. It secretes melatonin (the sleep hormone). When the sun sets, the amount of light entering our eyes is reduced and your brain detects that. Then, your brain signals the pineal gland to start secreting melatonin, so during the night our melatonin levels start increasing in our body. They will bind to the melatonin receptors in our brain, and the binding allows the body to prepare for sleep. When light gets in our eyes in the morning, it signals our brain that it is morning and it signals the pineal gland to stop secreting melatonin, hence melatonin decreases in the morning. This helps regulate our sleep cycle/circadian rhythm. Lateral ventricles: - The ventricles are the open regions of the brain where cerebrospinal fluid is initially produced. They are NOT involved in emotions OR memories. Limbic system: - Part of the brain involved in our behavioral and emotional response, especially when it comes to behaviors we need for survival: feeding, reproduction, caring for our young, fight or flight responses. Structures of the limbic system are underneath the cerebral cortex, and above the brain stem. Consists of the: o Thalamus o Hypothalamus (production of hormones and regulation of thirst, hunger, mood) o Basal ganglia (reward processing habit formation, movement, and learning) o Hippocampus o Amygdala Medulla oblongata: - - - - Connection between the brainstem and the spinal cord Plays a vital role in controlling autonomic functions that are essential for survival such as: o Heart rate o Blood pressure o Breathing o Reflexes such as swallowing, vomiting, coughing, and sneezing Within the medulla oblongata, there are specialized centers known as the cardiac center and the vasomotor center: o The cardiac center regulates heart rate o Vasomotor controls blood vessel diameter and therefore blood pressure Contains the respiratory centers including: o Dorsal respiratory group o Ventral respiratory group o These centers coordinate the rhythm and depth of breathing by influencing the activity of respiratory muscles. NOT involved in emotional states. *8. Priming: - type of implicit memory because it operates WITHOUT conscious awareness or intentional recollection of prior experiences. Remember, implicit memory refers to the ability to unconsciously and automatically retrieve information or past experiences even with the explicit recall is not possible. Prior activation of certain concepts or associations facilitates subsequent processing without conscious effort or awareness. - phenomenon where exposure to one stimulus influences the response to a subsequent stimulus, which makes the response faster and or more accurate. - occurs when the presentation of one stimulus affects the processing of another stimulus that follows it Example: when we take notice to shape, length, and health of other people’s teeth immediately after we visit the dentist. Our dentist visit has primed our behavior. Regarding this question: the researchers are attempting to determine whether priming the students with rude words will render them more likely to interrupt a conversation. 11. Latent Function: The unintended or unrecognized consequences of a social structure, institution, or behavior. Example: A library at a University functions as serving the students with access to academic resources BUT also functions latently as a scoial space where students can meet, interact, and form study groups. Expressive Function: The function of a social structure or institution that serves to express values, beliefs, or norms within a society, often related to emotional or cultural expression. Example: Tradition of wearing black to a funeral functions as showing respect for the deceased but it serves also as an expressive function by conveying sorrow and mourning to those in attendance. Manifest Function: The intended and recognized consequences of a social structure, institution, or behavior. Example: The manifest function of a hospital is to provide medical care and treatment to patients. Social Construct: A concept or idea that is created and maintained by society, rather than being inherent in nature. Social constructs can include categories, norms, roles, and institutions that are shaped by social interactions and collective understanding. Constructed by people to serve a societal purpose. Example: Race is often considered a social construct, while certain physical characteristics may differentiate individuals, the categorization and meanings attributed to racial categories vary across societies and historical contexts. Race as a social construct influences perceptions, interactions, but its significance and definitions are not fixed or inherent. Structural functionalist point of view: Views society as a complex system composed of various interconnected parts, each with its own function. This perspective emphasizes the ways in which these parts work together to maintain social stability. In terms of the question: It is a latent function because the manifest function of the school is to educate the students, and it actually helps dictate the overall quality of the neighborhood, which becomes the latent function. The neighborhoods being a product of the failing schools is the unrecognized consequence of the school. 13. stigmitization: the assignment of a demeaning label to a group in order to convey an aura of deviance among the groups members. Deindividuation: a phenonenon where a person temperoarily ceases to idnetift with himself/herself in lieu of a group or mob. Example: when everyone in the crowd at a sporting event begins throwing debris at players on the field, even though virtually none of them would act in that manner on their own. Discrimination: actions directed towards a certain group of people rather than labels put on those people. Racism: discrimination towardsa c ertain race. 14. hyperopia: condition that occurs when light focuses behind the retina rather than on the retina. It does not describe the process of changing the curvature. Convergence: eyes turning inward in order to stay focused on an object. It does not involve the shape of the lens. Accommodation: visual accommodation occurs when the ciliary muscles of the eye contract and expand, changing the curvature of the lens and focusing the light rays onto the retina. It is the process by which the eye changes its focus in order to see objects clearly at different distances. - When the eye needs to focus on nearby objects, such as reading a book, ciliary muscles surrounding the lens contract. The contraction causes the lens to become more rounded, increasing its refractive power to bend lights ray more strongly and focus them onto the retina. When the eye needs to focus on distance objects, the ciliary muscles relax. The relaxation causes the lens to flatten, reducing its refractive power so that light rays are focused further back onto the retina. Transduction: refers to the conversion of sensory input into neurological signals (action potentials). It has nothing to do with changing the shape of the lens of the eye. Here we are looking for changing the structure of the eye. This allows for process of elimination. 15. Group think: phenomenon that occurs within a group of people when the desire for harmony or conformity in the group results in an irrational or dysfunctional decision-making outcome. It occurs when the cohesiveness of the group becomes more important than critical thinking or the consideration of alternative viewpoints. *16. Elaboration likelihood model: explains 2 pathways by which people may change their minds about something, one based on the power of the message, and the other based on how the message is delivered. It does not address objective and subjective perceptions of reality. Gestalt principle (relevant to this question): The doctrine described in the question is best attributed to the Gestalt principle. Gestalt psychology emphasizes that human perception is not just a sum of individual sensations but involves organizing sensory information into meaningful wholes. According to Gestalt psychologists, our perception is influenced by our sensory experiences and how our minds organize and interpret sensory input. Therefore, we perceive the world not as it objectively exists but as our senses and perceptual processes allow. In other words, our perception of reality is shaped by the limitations and biases of our sensory modalities. Heuristics: “rule of thumb” or mental shortcuts that are used to speed up a mental process or an experimental investigation. - A common example of heuristic that comes up on the MCAT is trial and error Weber’s law: States that the just noticeable difference of a particular stimulus is proportional to the magnitude of the stimulus. - for example, if there is a tiny bit of ambient noise, but then you notice a whisper, and that is because it is some percent louder than the ambient noise. 17. For the far removed from effectors choice, this really is only applicable to motor neurons. An effector is usually a gland or muscle that a motor neuron synapses with. - You will see answer choices like this on the MCAT that will dissuade you from using your test-taking techniques and make you think that the correct answer has to be one of two opposites, so in this case, close or far from effectors. - When considering proximity to effectors, you consider this when you are dealing with autonomic motor neurons, not somatic sensory neurons. For the in basal nuclei this is incorrect because the basal nuclei are subcortical structures in the cerebrum that have a variety of functions. They are NOT the location of the cell bodies of somatic sensory neurons. For the in close proximity to effectors this answer is incorrect because again, effector is really only applicable to motor neurons. For the answer choice in dorsal root ganglia this answer is correct because they are the location of the cell bodies of somatic sensory neurons. They are located outside the spinal cord. Somatic sensory neurons are a type of sensory neuron that is responsible for transmitting information about the external environment and the body’s internal conditions to the central nervous system, particularly the brain and spinal cord. They are a part of the osmatic nervous system, which is in control of voluntary movements and receives sensory input from the body’s surface and muscoskeletal system. 20. Here, you need to analyze which of the following options have something to do with emotional arousal and performance. The stroop effect: phenomenon in which people take significantly more time to determine the color of a written word if the word is written in a different color than its name. This has nothing to do with emotional arousal and performance. Yerkes-Dodson law: There is a certain level of emotional arousal that leads to optimal performance. Cognitive dissonance is when one’s attitude and behavior about something are in conflict. An example would be if someone claims they are living a healthy lifestyle but smoke cigarettes. 21. two factor theory: this is the Schacter-singer theory, which states that an emotional stimulus will generate general autonomic arousal (the first factor), but then a cognitive component of the mind (the second component) will label the arousal and interpret everything as a specific emotion. - emotions are preceded by physiological arousal, which can result from various stimuli and situations. Physiological arousal can be increased heart rate, sweating, and other bodily responses. - cognitive interpretation, proposed by Schacter and singer, is that the cognitive interpretation of physiological arousal is crucial in determining the specific emotion experienced. They argued that individuals evaluate their current physiological state and then attribute that arousal to an emotion based on the context and cognitive appraisal of the situation. In other words, people label their arousal based on cognitive cues present in their environment. This helps individuals understand the reason for their arousal and determine which emotion is appropriate for the situation. The James-lang theory of emotion: states that after a stimulus, there is first a physiological response (and/or behavior) followed by an emotional experience which is direct result of that physiological response. - Regarding this situation, since all participants were placed in identical situations, this would dictate that their emotional responses should be identical, which is demonstrated in this graph. The opponent process theory of emotion: emotion exists in pairs of opposites: fear/relief, pleasure/pain, etc. The cannon-bard theory describes a complete independence between physiological response to a stimulus and the subjective feeling of emotion that is also a response to that same stimulus. Alternative perspective to James lang theory, suggesting that physiological arousal and emotional experience occur simultaneous, rather than one causing the other. Emotional experiences and physiological responses are independent that occur simultaneously. - If this were supported, you would not expect to see 100% of the uniformed wing to claim they were feeling anger because the emotional experience is not a direct function of the pattern of autonomic arousal. *23. For the mcat, know the following 3 broad categories of research methods: i. descriptive ii. correlational iii. experimental Research Method Description Key Characteristics Descriptive research methods aim to observe, describe, and document characteristics or behaviors of a phenomenon without manipulating variables or - Focuses on describing and Descriptive establishing causal relationships. These methods documenting phenomena. provide a snapshot or overview of the subject of interest. - Does not involve manipulation of variables or establishment of causality. - Common techniques include observational studies, case studies, surveys, and archival research. Correlational research methods examine the relationship between two or more variables to determine if and how they are related. This type of - Focuses on examining Correlational research assesses the degree and direction of relationships between variables. association between variables without implying causality. - Does not involve manipulation of variables. - Provides information about the strength and direction of relationships using correlation coefficients. Experimental research methods involve manipulating one or more independent variables to observe their - Involves manipulation of effects on dependent variables and establish causeExperimental variables to establish cause-andand-effect relationships. This type of research allows effect relationships. for control over variables and random assignment of participants. - Requires control over extraneous variables and random assignment of participants to experimental and control groups. - Allows researchers to infer causality by demonstrating that changes in the independent variable(s) lead to changes in the dependent variable(s). Magoosh FL 2 2/25/24 489 1. A combustion reaction in chemistry is when a substance reacts rapidly with oxygen gas (O2) and typically releases heat and light energy. Usually, the reaction often produces CO2 and H20 as the primary products. #C:9 #H:20 The general chemical equation for a combustion reaction is: CxHy+O 2 →H2 O+CO 2 a. Count the # of carbons and hydrogens b. Set up the general equation for combustion reaction with the C9H20and make sure it is balanced: C9 H 20 +O 2 → H2 O+CO 2 C9 H 20 +14O2 → 10H2 O+9CO 2 Since the question is asking for how many moles of H2O will be released, if 4 moles of 4-ethyl2-methylhexane are combusted, you have to look at the balanced equation. Since 1 mole of C9 H 20 releases 10 H20, then the ratio is 1:10. Therefore, 4x10=40 2. Unknown liquid sample has mass of 195g Placed in furnace with excess O2 Completely combusted according to unbalanced equation: unknown compound+O2 →CO2 +H 2 O All H 2 O completely released is absorbed by Mg(ClO4 )2 filters All of the CO2 released is absorbed by solid NaOH filters All of the oxygen atoms in the H 2 O and CO2 come from the O2 that is pumped into the furnace and NOT from the sample All of the oxygen atoms from the original sample react to form O2 gas and pass through the filter unaffected. From this information, we are supposed to realize that however amount of moles of H20 and CO2 that were released, you are supposed to realize the amount of hydrogen and carbon respectively. For example, if there are 10 moles of H20 then there is 10x2 moles of hydrogen. H 2 O mass:18 CO2 mass:44 190𝑔 18( =some where around 10mols 𝑔 ) 𝑚𝑜𝑙 460𝑔 44( =some where around 10 mols Difference of moles: Mg(ClO4 )2 : 190 moles NaOH:460 moles 𝑔 ) 𝑚𝑜𝑙 If we look at H 2 O , there is roughly 10 moles that were released and absorbed by Mg(ClO4)2. If we look at H specifically, since we want to find the empirical formula for the combustion reaction, there is roughly 20 moles (10x2). If we look at CO2, there is roughly 10 moles that were released and absorbed by the NaOH filter, and since there is only one C in CO2, there is roughly 10 moles of carbon. SO, the mole of carbon to hydrogen ratio is 10:20. This can be simplified 1:2, which is what we can look for in the answer choices. When we look at the following options: C7H14O2, C7H14 has a 1:2 ratio, and we can ignore the O because we are not interested in that. 3. Since we are measuring the masses based on the filters and how much they absorbed, we can say that the absorption masses directly correlate to the amount of hydrogen and carbon in the sample, which are what we used to determine the molecular formula. If something was not absorbed, we would need more information to determine the effect it would have on the calculated mass percent because we are basing it based on what is absorbed. 4. 2 moles of NaOH are required to remove 1 mole of CO2. First, we convert the mass of CO2 to moles: (88g/mol)(1mol/44g)=2 moles of CO2 Removing 2 moles of CO2 would require 4 moles of NaOH. Then, we need to calculate the mass of 4 moles of NaOH: (4moles) (40g/mol) =160g 5. a vacuum is an environment where the pressure is significantly lower than the atmospheric pressure. The frictionless reel indicates that there is no friction pressure between the surface of the reel and the string wound around it. In other words, there is NO resistance to the motion of the string as it winds or unwinds around the reel. This setup is often used to eliminate the effects of air resistance or friction, which allows you to study the motion of objects in idealized conditions or to isolate specific variables in their experiments. This will remove complicated factors such as air resistance or friction. We know the initial velocity is 0 because it is dropped. When we are using a velocity vs time graph, we know that the v(t)=at. The slope of the graph is the acceleration due to gravity, which we know as, 9.9m/s2, which when we convert to cm/ms2 that is 9.8x10^-4. Position-time graphs: a. tell you where an object is located over a period of time. b. The slope will tell you how fast the object is moving. c. The sign of the slope indicates direction of motion. d. The steepness of the line indicates fast an object is moving. e. If an object motions change, the slope of the line changes. how 6. We are looking at the falling object that is accelerating due to gravity. Option A Option B is correct because since an object is being dropped and accelerating due to gravity, we should expect to see the y-axis of the graph to be at 0 because the object is dropped from the initial position. Then, as the object falls, the position changes over time, and this should be reflected within the slope of the graph. The graph of a falling object usually shows a curve, indicating that the objects velocity( and its displacement obviously) is changing with time. The steepness of the curve at any point on the graph represents the objects instantaneous velocity at that moment in time. As the object falls under the influence of the gravity, its velocity increases, which results in a steeper slope on the position-time graph. Since the object is accelerating due to gravity, the slope of the position time graph becomes steeper as time progresses because this indicates the objects velocity is increasing, hence it is accelerating. The direction of the motion is inferred from the slope of the curve, and in the case of the falling object, the slope of the curve would be downward, indicating the object is moving in the negative direction of the y-axis. Option C is incorrect because the linear plot with a slope of 0, represents an object standing still. The mass is clearly not standing still. Option D is incorrect because the straight line shows that the velocity is constant, which is not possible since the mass is accelerating downward under the influence of gravity. 7. Air resistance and reel friction will slow down the mass. Air resistance acts in the OPPOSITE direction to the motion of the falling object. It effectively reduces the net force acting on the object, which in turn decreases the acceleration experienced by the object. The friction from the reel, would slow down the falling mass. The friction would oppose the motion of the mass, reducing its acceleration. Therefore, since we neglected air resistance and reel friction, the value of G would most likely be lower than the actual value of G because these unaccounted forces are acting against it. Linear Resorting forces: 1. These forces are always Directed to equilibrium 2. Magnitude proportional to displacement F1 F2 X1 X2 X1 and X2 are displacement and the force would be in the opp direction because they are restoring and they are proportional to how large the displacements are 𝐹 = −𝑘𝑥 (-) indicates that the force and displacement are in opp directions because they are restoring. K is the proportionality constant, the larger the displacement the larger the force. F=restorative force, the force that the spring needs to restore back to normal X=displacement Hooke’s Law: F=-kx - Force proportional to displacement, but in the opposite direction, or towards equilibrium where equilibrium is 0 displacement. These forces want to take them back to 0 displacement. We know that force is given by: F=ma So we can do the following: -kx=ma Spring Acceleration: a=-kx/m Potential energy of spring: PEs=1/2kx^2 F 4.5cm Solubility is the max amount of solute that can dissolve in a given amount of solvent at a specified temperature and pressure to form a homogenous solution. Here we are converting g/L→mol/L (molarity) Molar mass of PbSO4=303 g/mol