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The equation of thc Wcdi [eression W 6: (Plcu e shrny Voli_Arksots (wd Urciia JceUso tha medal to padict tho icotnd A #Jru: that has Aanndnncu J8,030 paoplo, Runtco...

Question

The equation of thc Wcdi [eression W 6: (Plcu e shrny Voli_Arksots (wd Urciia JceUso tha medal to padict tho icotnd A #Jru: that has Aanndnncu J8,030 paoplo, Runtcon - (Plc 1s4 I6Und Yollt Auswoi a MOd Wllulo mumbtE|

The equation of thc Wcdi [eression W 6: (Plcu e shrny Voli_Arksots (wd Urciia Jce Uso tha medal to padict tho icotnd A #Jru: that has Aanndnncu J8,030 paoplo, Runtcon - (Plc 1s4 I6Und Yollt Auswoi a MOd Wllulo mumbtE|



Answers

Arachidic acid is a $\mathrm{C}_{20}$ fatty acid found in peanut and fish oils. (18.6,18.7)
a. How many cycles of $\beta$ oxidation are needed for the complete oxidation of arachidic acid? b. How many acetyl CoA are produced from the complete oxidation of arachidic acid? c. Calculate the total ATP yield from the complete $\beta$ oxidation of arachidic acid by completing the following:

So once we into separate variables in that interview, both sides we end up with the integral of D. C. Divided by, uh, C s minus C is equal to the integral off k t to the power of negative B DT. So if we solve this, we end up with l n so that it's actually negative. Ln, but we're gonna multiply both sides. Been a good one in the end, So we're going to destroy it. Ln of the absolute value of C s minus C is equal to negative K t e to the T to the power of one, Let's see one minus be divided by one minus speed minus c. So we're gonna let some A is equal to K over one minus b so that we can write this working the an insult for sea so we can say C is equal to CS minus a e to the power of e to the power of negative small a t to the power off one minus B. So therefore we can read our final answer that C is equal to CS times one minus e to the power of negative small 80 to the power of one minus people. All right,

Mm. Hi, welcome to question number 42. In this question, you want to know if we have 25 micrograms of c. 21 age, 30 oh two, We want to know how many moles that is and then how many molecules that represents. So to do this, we are going to need the molar mass of this substance. So we're going to add 21 carbons to the molar mass of 30 hydrogen to the molar mass of two oxygen and I get 314 0.45 So the Mueller mass 3 14 0.45 Of course that would be grams per mole. Right, So I'm going to use this information set up a little bit of dimensional analysis here. First thing I need to do is to get my micrograms two g. So I know that there are one times 10 to the sixth micrograms of anything in one g. Now I can use my molar mass. So there are 314.45 g of C. 21. Age 30 oh two in every mole of the substance. Mhm. Notice I put the grams in the denominator so that they will cancel. So the only unit I have remaining now is moles. So I can go ahead and calculate my answer here. And I'm going to round it to two significant figures since 25 only had two and I get eight point oh times 10 to the negative eighth of the C. 21. Age 30 oh two. Okay, the next thing I want to do then, so let me circle that. That is our first answer. That is the number of moles. Eight times 10 to the negative eight moles. Next thing we want to do is figure out how many molecules this is. So starting with my number of moles eight times 10 to the negative eight moles of C. 21 H 30 02 We need to use of God Rose number because we know there are avocados number of particles in one mole of anything. So six point oh 22 times 10 to the 23rd. That is avocados number. Okay, molecules of the C 21 H 30 02 in every mole of the C 21 H 30 02 What? All right, so mobile cancel and I get a number in number of molecules again rounding the two significant figures. I get 4.8 times 10 to the 16th molecules of C 21 age 30 oh two. So that is the answer to the second part of this question. Thank you so much for joining me.

Running this equation for states of the hydrogen atom for which the orbital quantum number L. Is zero is given in your textbook, Verify that equation 39, which describes the ground state of the hydrogen atom is the solution of this equation. Okay, so the proposed way function, sorry, is equal to went over the square root of pi eight of the 3/2. You did? The negative are over A. So let's substitute this into the right side of the schrodinger's equation and we need to show that it's zero. So first we have that the derivative decide er is negative. One over The Spirit of Pi eight of the 5/2, it is the negative are over A. So in the short anger equation we have one over r squared DDR of R squared do you side? Er So it's one over the square root of pi eight of the five half -2 over our Plus one over a eating the negative are over A. Which comparing it to sigh is one over a -2 over our That's one over a say. So the energy of the ground state is negative and either the fourth over 8 ε not squared eight squared. And we know that the border radius is H squared absalon not over high M. E squared. So that gives us that the energy is equal to negative E squared over eight. Hi. Absolutely not A. And the potential energy is negative E squared over four pi. Absolutely not our. So we have eight Pi squared M over age squared E minus you. Time. Cy just eight Hi squared M over age squared E squared over a pie. Absolutely not. I could have won over a plus two over our. Hi. So this is equal to pi M. E squared for age squared. Absolutely not -1 over a plus two over our sigh is equal to one over a -1 over a plus two over our sigh, which we found before. So the two terms are going to cancel, and so the proposed way function to satisfy the schrodinger's equation.

So here we will start this molecule and I want to treat it with two. And a h is And that will give us this, um, plus two h twos. But those don't necessarily matter right now. Treat. It was ch two b r two and that would give us oh so in siege to combining it. And the last step is to add an age to C C h ch two c l in the a l c 03 And so what? That does it just add? So so 00 on the C h two and then at the very bottom who have ch two ch double wanted to ch two and that is several.


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