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Thc Bnitish neaure body wtght by Ihe stona OCne 20 stone Ovtr {he couri of 2 Tej>Mnard0 687 exh duy t0 RiMnYejr _ 365 djvs L0lb fat 3500 kcallJan krARDo170 LEal...

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Thc Bnitish neaure body wtght by Ihe stona OCne 20 stone Ovtr {he couri of 2 Tej>Mnard0 687 exh duy t0 RiMnYejr _ 365 djvs L0lb fat 3500 kcallJan krARDo170 LEal3LI

Thc Bnitish neaure body wtght by Ihe stona OCne 20 stone Ovtr {he couri of 2 Tej> Mnard 0 687 exh duy t0 RiMn Yejr _ 365 djvs L0lb fat 3500 kcall Jan kr ARDo 170 LEal 3LI



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An adult eats food whose nutritional energy totals approximately $2.2 \times 10^{3}$ Cal per day. The adult burns $2.0 \times 10^{3} \mathrm{Cal}$ per day. How much excess nutritional energy, in kilojoules, does the adult consume per day? If 1 lb of fat is stored by the body for each $14.6 \times 10^{3} \mathrm{kJ}$ of excess nutritional energy consumed, how long will it take this person to gain $1 \mathrm{lb} ?$

So for this problem we have a monk who is completing a a fast to protest some a war it sounds like, so we are given this equation W equals 1 70. So 1 70 is the initial weight times E. To the negative 0.0 H. T. So this is our exponential decay model. If we want to know how much the monk ways after 20 days, we simply plug in 20. And um when we do that we just plug all of this into the calculator at once we get W equals 1 £44.864. Now, how do we figure out the rate at which the weight is changing? So because we're given an exponential model we can turn this into a very basic differential equation. The rate of change is proportional to the amount present. So D. W. T. Equals KW. And since we know what K. Is, it's given to us in our exponential model negative 0.008. And then our W. is the current weight. Because we're trying to figure out at what rate is the weight changing at 20 days. So what is the weight at 20 days? 1:44.864? So I plug this in and I get DWBT is negative 1.159. And that is in pounds per day. So the weight of the monk is changing at negative 1.159 lbs per day.

So the balanced thermo chemical equation for the metabolism of fat in the body is as follows. C. 57 H. 104 oh six at 80 oh two Gives us 57 c. 02 at 52 H 20. Adults H. is negative 3.0-2 times 10 to the four kg jewels. So the molar mass of C. 57 81 oh four oh six is 884.57 g per mole. And so what we need to do is convert the pounds into g. So we have 226 7.95 g from £5. And so we can calculate the amount of heat evolved and we have this amount of grams metabolized by the body. So if you have 3.0 to 2 times 10 to the four kg joules of heat released per 884.57 g. This is equal to X kilo jewels, divided by the new mass 2267.95 g. Where we solve for X, which is equal to 7.748 times 10 to the four killed jules. And so what we can do is convert kilo jewels into calories, and so what we get is 1.852 times turned to the four calories.

Were given that the weight varies directly as the height cubed. So we're given the height and the weight of a person. And we need to use that information to find an equation. And with that equation we need to find the given, we need to find the weight with the given height so varies directly. We have, our weight is equal to K times height cubed. We plug in the values we have so we have 170 is equal to K times 70 cubed. We divide both sides by 70 cubed To 1 70, divided by 70 cubed. Gives us K is .123 five. Story equation is w is equal 2.0005 H cubed. So now that we have a height of 107 we go .0005 times 107 cubed. It was one of us. We get that the weight of this person is 607.16 pounds. Yeah.

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.


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