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    what is the effect produced by the prk technique designed to correct nearsightedness?

    James

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    get what is the effect produced by the prk technique designed to correct nearsightedness? from EN Bilgi.

    AAMC MCAT Practice Exam 4 Cp Solutions

    Exam 4 C/P Solutions: Passage 1 1) First thing we want to do to answer…

    MCAT Content / AAMC MCAT Practice Exam 4 Cp Solutions

    AAMC FL4 CP [Web]

    Exam 4 C/P Solutions: Passage 1

    1) First thing we want to do to answer this question is revisit the part of the passage that talks about limestone. The author tells us in Step 1, “The chemist placed a sample of limestone, CaCO3(s), into a sealed chamber and then heated the limestone to 1200 K to generate CO2(g) and CaO(s).” However, there’s a slight change in the question stem. The limestone, CaCO3(s) is only heated to 900 K, but does not decompose and generate CO2(g) and CaO(s). What does that tell us? The reaction is not spontaneous. Let’s jump into some quick background information we’ll use to answer this question:

    ΔG = ΔH – TΔS

    We don’t get the decomposition the author talks about in Step 1, and for nonspontaneous reactions, ΔG is positive. Additionally, entropy increases if we go from a solid reactant to a gas in our products, which would mean a positive ΔS. Given this information, ΔG can only be positive in our above equation when ΔH is positive and is greater than TΔS (which we know is also a positive number in this circumstance).

    positive and less than TΔS. First part of this answer choice is consistent with our breakdown. We mentioned that for nonspontaneous reactions, ΔG is positive. Entropy increases if we go from a solid reactant to a gas in our products, so we have a positive ΔS, and we’re told temperature is 900 K. Given our equation ΔG = ΔH – TΔS, ΔH must be positive and greater than TΔS in order for ΔG to be positive. That means this answer choice is only half right.

    positive and greater than TΔS. This answer choice is entirely consistent with our breakdown of the question. We mentioned some key points in the breakdown that we can reiterate here. For nonspontaneous reactions, ΔG is positive. Entropy increases going from a solid reactant to gas in our products, so we have a positive ΔS. We’re told temperature is 900 K. Given our equation ΔG = ΔH – TΔS, ΔH must be greater than TΔS in order for ΔG to be positive. This is going to be our best answer.

    negative and less than TΔS. This answer choice is the opposite of what we said in our breakdown. We expect ΔG to be positive and greater than TΔS.

    negative and greater than TΔS. This answer choice is only half correct. We expect ΔG to be positive and greater than TΔS. We can stick with answer choice B as our best answer choice.

    2) This question has to do with the same part of the passage we focused on for Question 1. We’re asked about limestone being heated during Step 1.

    We’re told an equilibrium is established and we can write out the equation here:

    CaCO3 (s) ⇌ CO2 (g) + CaO (s)

    The equilibrium constant is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. If we have an equation aA + bB ⇌ cC + dD, we can write out the equilibrium constant expression:

    However, before we write anything for this specific question, what we need to pay attention to is the state of each reactant and product. Pure solids and liquids are not included when writing out this equilibrium constant. That means we are not including CaCO3 (s) and CaO (s) from our initial equation. That means we are only left with the product CO2 (g). Our equilibrium constant is simply [CO2].

    [CaO] We noted in our equation that CaO is a solid. Pure solids and liquids are not included when writing out our equilibrium constant.

    [CaCO3] We noted in our equation that limestone (CaCO3) is a solid. Pure solids and liquids are not included when writing out our equilibrium constant.

    [CO2] This answer choice matches our breakdown. This is going to be our best answer choice.

    [CaO] × [CaCO3] This answer choice incorrectly includes the solid reactant and product instead of excluding these. We can stick with answer choice C as our correct answer.

    3) This is another passage-based question in this set, but once again, the test-maker simplifies our lives a bit by telling us where we will find the necessary information to answer our question. We’re focused on Reaction 2:

    Specifically, we want to hone in on nitrogen. The oxidation number describes explicitly the degree to which an element can be oxidized (lose electrons) or reduced (gain electrons). The number is the effective charge on an atom in a compound. We can look at nitrogen in our reactants and products.

    First our products: The oxidation number of hydrogen is +1 when it is combined with a nonmetal as in CH4, NH3, H2O, and HCl. In uncharged NH3, each hydrogen has an oxidation number of +1 while nitrogen then has an oxidation number of -3.

    In our reactants we have protonation of ammonia so we’re looking at NH4+. Once again, the oxidation number of hydrogen is +1 when it is combined with a nonmetal. In NH4+, each hydrogen has an oxidation number of +1 while nitrogen still has an oxidation number of -3. Together that is where we get the +1 overall on NH4+.

    Yes; it changed from –3 to –4. This answer choice correctly lists the initial oxidation number as -3, but we reasoned out that the oxidation number does not change after the protonation of ammonia.

    Source : jackwestin.com

    AAMC FL4 C/P Q13 *Spoiler* : Mcat

    Can someone explain the correct answer to me? How would it increase if they are flattening it?

    0

    AAMC FL4 C/P Q13 *Spoiler*

    Question 🤔🤔spoiler

    Posted by u/zoomiepug 2 years ago

    AAMC FL4 C/P Q13 *Spoiler*

    Question 🤔🤔spoiler

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    Log In Sign Up level 1 claybrink3 · 2 yr. ago

    516 (130/127/130/129)

    You can also solve this one by knowing what nearsightedness is and know that that image focuses in front of the retina (so the focal length of the lens is too short). To help this, they increased the radius of curvature, which would increase the focal length (because r= f/2) and cause the image to focus on the retina. That was my thought process behind it.

    5 level 2 hotpizzarolls1 · 5 hr. ago

    Just in case anyone comes across this like I did, f = r/2. Not the other way around. Focal length is half of the radius.

    1 level 1 tparty15 · 2 yr. ago

    I had to look this one up too. The flattening causes the arc of the eye to become less extreme. Therefore, the circle that makes the new arc is much larger than the original one, and so the radius of curvature is mush larger.

    https://www.reddit.com/r/Mcat/comments/eoujbl/spoiler_aamc_fl_4_cp_13/

    4 level 2 zoomiepug OP · 2 yr. ago

    That is super helpful, thank you! And I swear I tried looking this up on r/mcat first but maybe I’m not typing it in properly cause nothing came up... but I totally get it now. I remember reading the answer choices thinking like...none of these can happen lolol

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    If you're like me, breathe. You don't need to master every little thing. Don't panic if you don't even understand what the question is asking at first, there are many points to be gained by interpreting the question correctly and reasoning through the answer.

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    fL4 CP Flashcards

    Memorize flashcards and build a practice test to quiz yourself before your exam. Start studying the fL4 CP flashcards containing study terms like solids and liquids (water, etc) are excluded from the, NH3(aq) + CO2(aq) + H2O(l) → NH4+(aq) + HCO3-(aq) During Reaction 2, did the oxidation state of N change? A.Yes; it changed from -3 to -4. B.Yes; it changed from 0 to +1. C.No; it remained at -3. D.No; it remained at +1., The chemist heated enough NaHCO3(s) from Step 5 to produce 1 mol Na2CO3(s), 1 mol H2O vapor, and 1 mol of a gas (Gas X). The chemist collected Gas X for further study. If all of Gas X (from Step 6) is held in a sealed chamber at STP, what will be its approximate volume? and more.

    fL4 CP

    24 studiers in the last day

    solids and liquids (water, etc) are excluded from the

    Click card to see definition 👆

    equilibrium constant

    Click again to see term 👆

    NH3(aq) + CO2(aq) + H2O(l) → NH4+(aq) + HCO3-(aq)

    During Reaction 2, did the oxidation state of N change?

    A.Yes; it changed from -3 to -4.

    B.Yes; it changed from 0 to +1.

    C.No; it remained at -3.

    D.No; it remained at +1.

    Click card to see definition 👆

    C.

    Acid-base reactions.

    While acid-base reactions do not involve oxidation state changes, the starting oxidation state for N is -3

    The part of Reaction 2 that involves nitrogen is the protonation of ammonia (NH3+ H+ → NH4+). Acid-base reactions do not involve oxidation state changes. The oxidation state of N in NH3 is -3. Each H is +1 and is balanced by the -3 of N to make a neutral compound. The oxidation state of N does not change when the N is protonated.

    Click again to see term 👆

    1/55 Created by idknooneknowsme

    Terms in this set (55)

    solids and liquids (water, etc) are excluded from the

    equilibrium constant

    NH3(aq) + CO2(aq) + H2O(l) → NH4+(aq) + HCO3-(aq)

    During Reaction 2, did the oxidation state of N change?

    A.Yes; it changed from -3 to -4.

    B.Yes; it changed from 0 to +1.

    C.No; it remained at -3.

    D.No; it remained at +1.

    C.

    Acid-base reactions.

    While acid-base reactions do not involve oxidation state changes, the starting oxidation state for N is -3

    The part of Reaction 2 that involves nitrogen is the protonation of ammonia (NH3+ H+ → NH4+). Acid-base reactions do not involve oxidation state changes. The oxidation state of N in NH3 is -3. Each H is +1 and is balanced by the -3 of N to make a neutral compound. The oxidation state of N does not change when the N is protonated.

    The chemist heated enough NaHCO3(s) from Step 5 to produce 1 mol Na2CO3(s), 1 mol H2O vapor, and 1 mol of a gas (Gas X). The chemist collected Gas X for further study.

    If all of Gas X (from Step 6) is held in a sealed chamber at STP, what will be its approximate volume?

    The quantity of Gas X was given as 1 mole. One mole of gas occupies 22.4 L at STP.

    Why was it important that the cuvettes containing the glucose oxidase and the blood sample were identical in terms of optical properties?

    A.To enable the comparison of the absorption spectra

    B.To reduce the absorption in the glass walls

    C.To decrease the uncertainty in the wavelength

    D.To increase the absorption in the solutions

    A. The identical optical properties of the cuvettes ensure that the absorbed radiation is due only to the presence of glucose in the blood and not due to the difference in the absorption features of the walls.

    625 ± 5.0 nm

    What is the approximate energy of a photon in the absorbed radiation that yielded the data in Table 1?

    (Note: Use 1 eV = 1.6 × 10-19 J and hc = 19.8 × 10-26 J•m.)

    The solution is B.

    The photon energy is E = hc/λ = 19.8 × 10-26 J•m/(625 × 10-9 m) = 3.1 × 10-19 J ≅ 2 eV.

    According to Table 1, what is the concentration of the glucose in the blood from which the diluted sample was taken?

    A. 60 mg/dL B. 90 mg/dL C. 120 mg/dL D. 150 mg/dL

    From Table 1, the glucose concentration in the diluted sample is (0.20/0.24) × 6.0 mg/dL = 5.0 mg/dL. The blood then has a glucose concentration of 30 × 5.0 mg/dL = 150 mg/dL.

    Suppose a blood sample tested above the range (6.0 mg/dL) of the standards used in the experiment. What modification will provide a more precise reading by data interpolation as opposed to extrapolation using the same standards?

    A.Increase the enzyme concentration.

    B.Increase the oxygen pressure.

    C.Decrease the content of oxygen acceptor.

    D.Dilute the sample with additional solvent.

    The solution is D.

    By adding solvent, the concentration of glucose will be lowered, and the resulting absorbance will fall within the range of the standards. This is easily accomplished, and the resulting calculations that account for the dilution are not difficult.

    A) Increasing the enzyme concentration will increase the quantity of absorber (chromogen) in solution, according to the protocol described in the passage. This will make the resulting absorbance value even further from the values in Table 1 and require additional extrapolation.

    B) Increasing the oxygen pressure will increase the quantity of absorber (chromogen) in solution, according to the protocol described in the passage. This will make the resulting absorbance value even further from the values in Table 1 and require additional extrapolation.

    C) The oxygen acceptor is glucose. Removing glucose from the solution is not feasible and defies the logic of the experiment, which involves quantifying the amount of glucose present. One would need to know exactly how much glucose was removed, and this would require a separate measurement.

    Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1-chlorobutane and 1-butanol by fractional distillation?

    The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol.

    The fact that 1-chlorobutane will have a boiling point that is substantially lower than that of 1-butanol can be rationalized from chemical principles. The molecules have similar molecular weights, but 1-butanol has a hydroxyl functional group that can participate in hydrogen bonding. Hydrogen bonding is a particularly strong force of intermolecular attraction.

    Source : quizlet.com

    Do you want to see answer or more ?
    James 8 month ago
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    Guys, does anyone know the answer?

    Click For Answer