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~/3 PDINTSDEVORESTAT9 4.E.034.MY NOTESASK YOUR TEACHERAn article {uggests -hat $uLstut ccncentrzlon (mg'cm (Round "clir Answers decimal Placesmfiventrejct...

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~/3 PDINTSDEVORESTAT9 4.E.034.MY NOTESASK YOUR TEACHERAn article {uggests -hat $uLstut ccncentrzlon (mg'cm (Round "clir Answers decimal Placesmfiventrejctorncmallystributed %IthLSUandU.USWnalIhc probabilily Ital Uie CMIC urbrali:n CdcWnatthe probabllity that the concentrationMcsc#nukl you cha actcriz: thc lalgost 5*4llcnrcertraliou valucTh: Iancat{cucanttaticn *alijc an? abovrTommay n:fMserhe appmpriate tablethe Appendix Tahlasenscer this questio^.EnciJSCCViev' Previcus Question

~/3 PDINTS DEVORESTAT9 4.E.034. MY NOTES ASK YOUR TEACHER An article {uggests -hat $uLstut ccncentrzlon (mg'cm (Round "clir Answers decimal Places mfivent rejctor ncmally stributed %Ith LSUand U.US Wnal Ihc probabilily Ital Uie CMIC urbrali:n Cdc Wnat the probabllity that the concentration Mcsc #nukl you cha actcriz: thc lalgost 5* 4llcnrcertraliou valuc Th: Iancat {cucanttaticn *alijc an? abovr Tommay n:f Mserhe appmpriate table the Appendix Tahlas enscer this questio^. EnciJSCC Viev' Previcus Question Queslinn , 0l 15 View Next Quesrion FlLy AM mctm"



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Indicate the concentration of each ion or molecule present in the following solutions: (a) 0.25$M$ NaNO $_{3}$ , (b) $1.3 \times 10^{-2} M \mathrm{MgSO}_{4},(\mathbf{c}) 0.0150 M \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},(\mathbf{d})$ a mixture of 45.0 $\mathrm{mL}$ of 0.272 $\mathrm{M} \mathrm{NaCl}$ and 65.0 $\mathrm{mL}$ of 0.0247 $\mathrm{M}$ $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .$ Assume that the volumes are additive.

Mm. So in our first example we have some sodium nitrate that's going to break into sodium ion and nitrate ion. and since the sodium nitrate was .25 Moller, the concentration of the sodium ion Is going to be .25 moller. And the concentration of the nitrate is also going to be .25 moller because that compound completely dissociates in our second compound. We've got magnesium sulfate and we're performing mg two plus and S. 04 to minus Again, we're gonna start with .013 more. That is going to completely disassociate. So it's going to give me .013 moller MG two plus 10.013 moller S. 0. 4 to-. Okay and the third example We've got c. six Age 1206. This is a molecular compound so it's not going to disassociate at all. So its concentration is simply what it was given .015 Moller, glucose. Okay now we're gonna take a couple of solutions and combine them so they're gonna get diluted. So we're going to look at this dilution formula. So the new more clarity once it's diluted will be its original more clarity times V one overview too Where V two equals the total volume of two solutions. Okay so we start with point 272 more sodium chloride. So we'll multiply by the volume of that And divided by the total volume which is 1 10 So the concentration is .111 moller and A C. L. So the concentration of your sodium ion is equal to the concentration of your chloride ion And they're both equal 2.111 bowler. Okay, our second solution 0.2 47 moller ammonium carbonate. So we're going to multiply that By its volume, which is 65 million litres over 1 10. So it will be point 0146 Moller of the ammonium carbonate. Okay, so the concentration of the ammonium ion Since there's two NH 4 plus is in there is going to be two times .0146. So that will be 0- 9 to Moller. And the concentration of the carbonate will simply be .0146 Mueller. And those are all your concentrations.

We are going to be combining mixtures. Okay, so we'll start off with two different kinds of solutions of sodium hydroxide. Okay, so feel a stopped their concentrations involved. The first one has 0.17 moles And the second one is slightly higher at .4. Okay Then their volumes. 0.42 L with the other at 0.0 376 leaders. Now we can do the calculation of most is nope concentration times volume to figure out their malls And we will get zero 0071 for moles. And then this one would be 0.015 mi. Okay, so where you know the concentration of sodium will be equal to the concentration of hydroxide of it. And they're combining basically that we're combining the same Calyon and and I on. So well we have to do to get this constitution is trying to the total moles and we can do that by adding these two moles here. And that gives us 0.02-14 moles than to find the volume total concentration is for the total volume used and we'll add this second round here. What we get is 0.796 leaders. Okay. So then if we do that calculation, the final moles concentration of sodium and hydroxide Are equivalent and are 0.0278 multiple later. Yeah, that was just the first example. Next is a slightly bit more involved. Okay, so we'll take sodium sulfate with some salt, potassium. Quite okay. And if we mix come together, what will we do? Yes let's first. Right. Um Are givens. Okay, so here are the concentrations given followed by their volumes. Because up we converted that to leaders because that's actually what's most useful actually sort of this were over a bit to give us some room king. Now we need to calculate the malls simply by multiplying what's above. What we get is 0.0044 moles, sodium sulfate and 0.00375 moles of the potassium chlorate. Okay. Now for going to find concentration of the ions in this solution, we need the proper volume of this solution by adding a second role as we did above. And that gives us 0.069 leaders. Okay, so let's start off by isolating these concentrations. Okay, so we know that the concentration of potassium would be equivalent to that of the chloride. Okay. And we'll just take this these molds here and divided by the volume. Get the total volume like this. And what were ended up with a zero 05 43 moles per liter. Okay then that's going to be concentration for potassium and for the chloride as well. So following for the concentration of the sulfate, we'll take malls of the sulfate invited by The total. Okay, you get above in that will yield of 0.0 0.0637. That's per liter. Then the malls of Sudan will be double that. I think the final problem that we're going through, it's going to be an interesting one. We're going to take potassium chloride and mix it with some calcium chloride. So here we're going to have some additive additive properties as we saw in the very first example. So here, since we were given maths of 3.6 g, we're going to need to remember the molar mass. Okay. Which is that over there? 74 g per mole. Okay. And the malls will simply be mass over molar mass To give the serpent 0483 the walls of this salt. Okay, So then for the calcium chloride We had zero 24 We're sorry to five multiple causing quiet. Okay, so the only volume we had which will call vine total because here we just got a dry mass of the first cop out Will be 0.075 leaders. And the moles then will be fi multiple. What's above your europe 1875. All right. But actually moles of calcium chloride doesn't change. So we can use this to find the malls of the calcium. It will also be the exact same thing here because this one is actually not deluded at all because we're not adding any volume for the rest. We will need to do some work here. All right. So, if we try to calculate concentration of the potassium I am. What we need to do is take this smalls over here here .0483 And divide that by the total volume, which was 0.075 leaders. Okay. And what that gives us is you're 26 44 knows. Okay. So now for the calcium is going to be completely different case when we got in blue. So most of calcium will be the component from the hook passing for it As well as double the component from the calcium cord. Since it's going to associate into two and my arms. Okay. So if we do that addition, What we end up getting is zero 0858 months. Hey, So we would have added it that way and then malls of chloride will be you know that and the concentration Will be most over the total volume to lead 1.1 for four months. The leader. So that was tricky part. There was a need to remember that the chloride ion is coming from both components.

Okay. So for the first part we have any cl which will disassociate as any positive. Um Seeing negative. And then we have calcium chloride. Okay since they're on two chloride irons. Oh we would have still. Thank you too. Are still I used concentration. Okay. Would you want to the next spot? Um So what we are given is KBR MK three P. U. Four. So we have $0.5. Oh can you be off And 3.5 members. Okay. Threepio four. So there is one roommate. I am. But maybe I. So concentration of We are negative is 0.5 more because it's the same in this case concentration of You 4 3 negative is also 2.5 More. Lower. So using. Yeah I am one we one equals two. And we do We didn't have to find five more dollars. B. B. R. Thank you. one militate. Was there divided by one month tvR Plus different five numbers. It's within your four three moved the positive divided by one move. Do you treat the old floor which is a program Number three positive. So this is the answer

Hello. Hope you're doing well. So you've got our function right here for the concert for the drug concentration in the blood stream. Those aircraft with function right here with time in hours on the X axis and our concentration in milligrams per liter on the while says the first were asked to find the highest drug concentration on the graph points. Gonna take place right there. That's our highest concentration. And on an online graphing calculator, you can find the exact value of that point. So that point corresponds to drug concentration of 10.6 milligrams for leader moving on to part B. You can see a CIA purchase infinity on this Part of the graph says T continues getting larger and larger. Our concentration continues. Decrease. We could smaller and smaller, and eventually it approaches the X axis that purchase zero seventies air answer for part B is going to be after an extended period of time. As t approaches, infinity or concentration is going to approach zero. So lastly for lust hurt, see the problem were asked to find out what point the concentration drops below 0.3 milligrams per liter. So again, you can use the online graphing calculator to get exactly that's going to take place right about there. That point corresponds to the 0.0.100 0 point so beyond that after that point passes, the concentration of a drugs in the blood blood stream is blows your 0.3 milligrams for lead. So that means that our that for part C or answers that this takes place at 100 minutes, I said, should answer all the questions for a problem. So it should be about good. I will think so, and I hope that helps.


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