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11.51 PM Sun Sep88%BackReviewMmmnel Ud MuinmcIuue WsAnswer:Not answered Points out of 1.00Question 15An inhalation aerosol contains 225 mg of metaproterenol sulfate...

Question

11.51 PM Sun Sep88%BackReviewMmmnel Ud MuinmcIuue WsAnswer:Not answered Points out of 1.00Question 15An inhalation aerosol contains 225 mg of metaproterenol sulfate _ which is sufficient for 300 inhalations. How many micrograms of metaproterenol sulfate would be contained in 5 inhalation? Round to the nearest whole number; Do not include units _Answer;Nol answered Points out of 1.,00Question 16olblApproximately 0.02 of a 10O-mg dose of drug has been shown to appear in human breast milk. Calculat

11.51 PM Sun Sep 88% Back Review Mmmnel Ud MuinmcIuue Ws Answer: Not answered Points out of 1.00 Question 15 An inhalation aerosol contains 225 mg of metaproterenol sulfate _ which is sufficient for 300 inhalations. How many micrograms of metaproterenol sulfate would be contained in 5 inhalation? Round to the nearest whole number; Do not include units _ Answer; Nol answered Points out of 1.,00 Question 16 olbl Approximately 0.02 of a 10O-mg dose of drug has been shown to appear in human breast milk. Calculate the quantity of drug detected in milligrams, following single dose; Round t0 the nearest hundredth; Do not include units Answer: Not answered Points out of 1,00 Question 17 A prefilled syringe contains 12 mg of drug in 10 mL of solution: How many micrograms of drug would be administered by an injection of 0.8 mL of the solution? Round answer t0 the nearest whole number. Do not include units_ Answer: Not answered Points out of 1.00 Question 18 How many grams of digoxin ANOXIN) would be required t0 make 90OOO tablets each containing 250 mcg of digoxin? Round answer t0 the nearest



Answers

A single injection of a drug is administered to a patient. The amount $Q$ in the body then decreases at a rate proportional to the amount present. For a particular drug, the rate is $4 \%$ per hour. Thus, $$\frac{d Q}{d t}=-0.04 Q \quad Q(0)=Q_{0}$$ where $t$ is time in hours. (A) If the initial injection is 3 milliliters $[Q(0)=3],$ find $Q=Q(t)$ satisfying both conditions. (B) How many milliliters (to two decimal places) are in the body after 10 hours? (C) How many hours (to two decimal places) will it take for only 1 milliliter of the drug to be left in the body? (D) Graph the solution found in part (A).

Okay, so for part a time, zero with equal zero. Um, so right deep again zero that see on the products just commit the whole expression. Zero. That makes sense. Because, like, right when a drug is administered, like even if it's injected or orally taken in No, we don't know. Um, it has an ad zero time. It's just sitting right where it left the needle of right where the pill is. Um, so at T equals zero. The amount is equal to zero. So just makes sense because it just hasn't had time to get into the bloodstream yet. OK, Part B must know how much is in there after one hour, and so there is going to be 3.42 milligrams per liter of drug concentration. Came part C wants me to sketch a graph of this function so all that really matters is positive time values and positive amounts. So this graph starts at 00 It's gonna go up and then back down as t approach isn't me on the maximum of this one is kind of close to 3 a.m. Every 3.5. Okay, So I d, uh, it levels off to as t approaches infinity, the amount, the concentration. I guess I should market this. No capital C that C is gonna be approaching zero. That makes sense in context, cause your body and your liver start processing it out. So the fact that it never quite gets to zero I mean, at some point, I guess there's always gonna be, like, one little dot of it in your system. But because it's like discrete numbers of Adam's, eventually it will be completely out of your system. And so, uh, party Okay. So funny when it reaches its maximum. So I'm just using the maximum notation. Mr the Time is gonna be actually 3.636 when it reaches its max. Uh, hours. Okay. And then, uh, I'm gonna also plot the line y equals, uh, three milligrams for leader. Just y equals three. And then just find where that intersects the ground. The second time in intersects the graph is gonna be 9.759 hours later will be the second time. It's at three milligrams for leader. Okay, thank you very much.

So between the lithium night treat, the nickel to night tree and the strong team nitrate, the one that we should actually used to precipitate the maximum out of the carbonate ions from the solution should be the strong team nitrate. And the reason for this is has the lowest ks beef value asking here. And so they gave us a case be table with the lithium carbonate, nickel to carbon and the strontium carbonate. And so adding strong human nitrate trying him carbonate is actually what we'd be making. So it would make sense for look at these case p values. And so when a case be value is closer to one, that means that the, um substance dissolves very readily and be squeak. It's only ions and it doesn't like to stay in the forms of for instance, lithium carbonate. The case be was one then what we would have is only the theme ions and only the carbonate ions. We would actually have the lithium carbonate, but here so lithium carbonate has the value closest Thorne and strong from carbonate has value closest zero, which means that it likes the state s surround him carbonate and so once we put the strontium nitrate into the solution and it reacts with the carbon e ions, we get the strong carbonate and have a lot of precipitated. Then for one beer, what we're looking at is for the carbon compound that contains the cattle on chosen in part a, which we chose as strontium nitrate, giving us drunken carbonate determine the concentration of each eye on of that compounded solution equilibrium. So this is the equilibrium, expression or equation. And so what? We have your strong him carbonate solid, solid in equal air be in, which is what these double arrows are with strong team two. Plus, it's the cat eye on and it's a quest. And then the carbon, eh? I unders and own. And that's also Equus. So then, um, what we need now is chaos P is actually just an equation, like it's just a formula that you follow. So they gave us the KSB and Cass be equals the concentration of whatever is a quest on the right side of your equation. So we have the strong tea, um, cat eye on multiplied by the concentration of the carbon A an eye on. So now since they gave us our case, P value, which is just 5.60 times 10 to the negative 10th. We're setting that equal to basically just the concentration of the strong Thiam times, the concentration of the carbonate and what we can do there is that's just x Times X giving us X squared. So now if we just take the square to both sides and we saw for eggs X equals the concentration of the strontium I on which is also equal. So the concentration of the carbonate ion, which overall is equal to 2.37 times 10 to the negative fifth Moeller. And so that's the concentration off each eye on in solution. So now moving on part C one mixing the solution should the student ensure the carbonate solution or the nitrate solution is an access. So for this, what we have to think about it, Which one of these ions would actually help us more? So we are Actual precipitated is strong team carbonated, which is as our CEO three. So we don't need more carbon ian solution because it doesn't matter how much carbonate we have, because that's just gonna stay as an eye on but for the strong tea Um, the way that we got the strong team is from the strong tea of nitrate, which is this So if we don't have enough strontium, then we can't make the strong him carbonate. So the nitrate solution should be an access because in order to create the maximum amount of precipitate, you need enough strong team I owns to react with all of the carbonate ions and having the access strontium ions in the solution doesn't affect the master of the carbonate in the virginal example. So we have a fixed amount of carbonate, but we I don't know, having more having the access won't give us the precipitate but having the excess of strong here Mayan means that we can react all off the carbon e into precipitously and then having the extras drawn to him doesn't affect anything. So then moving on to part dear after titrate and sufficient solution to precipitate out all of the carbonate ions, the student filters the solution before placing in the crucible and heating it to drive off the water after several beatings. The final mass of the precipitate it remains constant is determined be 2.2 grams. So the first thing we wanted to do is determine the number of moles of Pacific it. So what we d'oh is we take our 2.2 grams of our strong team carbonate, and we multiply that by the molar mass, which is basically just a ratio that allows us to go from grand symbols or moral stra graham. So 1 47.63 grams Uh, s or C 03 her one more and overall, that gives us okay after I put it into my calculator one point 37 times tend to the negative seven Negative, second more. And now, um, moving on determining the mass of the carbonate present in the precipitate. What we do is we take these moves that we just got and then we multiply that bye. You have one more of, um, you have one mole of the carbon e ion her one more of the strong him carbonate substance. So then, since it's just a wonder one we end up with 1.37 times tend to the negative second more of the carbonate and then what we need to do is take the Moler Mass off the carbon e ion and multiply that by the moles. And so once the molar mass of carbon e is so carbon is 12 plus the 40 the 48 from the three oxygen. So it's 60. So then once we take 60 times so 0.137 times succeed, which gives us zero point e to two grams of this carbon ee eye on. So then, um, poor E. What we need to do is determine the percent by massive carbonate in the original sample. So the 0.822 grams of the carbonate weakness around that 20.82 grams and just carry the 26 eggs and then you divide that by don't know why I did that. Um, so it's the zero point 82 grams divided by the one point 39 grams from the original sample. And then if you multiply it by 100 that's how you're gonna percent and we end up with 43.4 percent of the carbon E who of the, um carbon e i aunt in the original solution so that moving on too. Oh, you know, I did that. So then moving on to part, as is the original compound most likely lithium carbonate, which is ally to Theo. Three, um, sodium carbonate ing any to Theo three or potassium carbonate, which is you if you go through here. So what we do now is so it's the, um 60.1 for all three, um, on the numerator because what we're looking at for all of them is the carbonate. But then the only difference is the total math. So we're either dividing it by 73.89 because you had the mass of the carbonate plus the mass of dilithium or 16.1 divided by one of 5.99 or 60.1 divided by 1 38.21 Oh, and so you multiply all these by 100 and you enough with anyone 0.2% for that 56.6% for this and 45 0.4% for the pizza, potassium carbonate and


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