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Question 2 Expressing concentration in different units You have a 0.2 M solution of a compound (molecular weight 200).How many mmoles are there in the following vol...

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Question 2 Expressing concentration in different units You have a 0.2 M solution of a compound (molecular weight 200).How many mmoles are there in the following volumes? 2 mL ii. 5 mL iii. 10 mL iv. 100 mLb) Express this concentration in the following units:mM; uguL; % (wlv); mgdL;ii. iii iv.c) How much of this compound would you have to weigh out if yOu were to make up 100 mL of the solution?If you added 5 uL ofthis solution (0.2 M) to a PCR (polymerase chain reaction) with a final volume of 10

Question 2 Expressing concentration in different units You have a 0.2 M solution of a compound (molecular weight 200). How many mmoles are there in the following volumes? 2 mL ii. 5 mL iii. 10 mL iv. 100 mL b) Express this concentration in the following units: mM; uguL; % (wlv); mgdL; ii. iii iv. c) How much of this compound would you have to weigh out if yOu were to make up 100 mL of the solution? If you added 5 uL ofthis solution (0.2 M) to a PCR (polymerase chain reaction) with a final volume of 100 AL (2 marks) How many moles would the reaction contain? (ii) What is the final concentration of this compound in the PCR assay?



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Calculate the molar concentration for each of the following solutions. (a) $1.00 \mathrm{~g} \mathrm{KCl}$ in $75.0 \mathrm{~mL}$ of solution (b) $1.00 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CrO}_{4}$ in $75.0 \mathrm{~mL}$ of solution (c) $20.0 \mathrm{~g} \mathrm{MgBr}_{2}$ in $250.0 \mathrm{~mL}$ of solution (d) $20.0 \mathrm{~g} \mathrm{Li}_{2} \mathrm{CO}_{3}$ in $250.0 \mathrm{~mL}$ of solution

This question is to calculate the polarity in each of the following solutions. The first solution is 0.195 grams self cholesterol in 0.100 lead is a solution. So you know there's a coconut, the polarity we need to know the mall and also the volley Given the mass weaken determined the mole. Why do you ride in the mass by the molar mass and finally doing everything where the volume to get the malaria. So for this solution import a similarity is determined by taking the mass, which is 0.195 grams. You're waiting by the mill amounts of cholesterol. So you have to compute the mullahs mass off cholesterol here. And when I competed that Thomas I attained the valley off 386 0.661 Graham Spindle Man, You end up dividing by the volume which beginning that's 0.10 point 1200 leaders. This one sold you similarity off life when therefore time sensible minus three Moeller off cholesterol Barbie. We do the same thing. Similar approach. So our polarity is going to be attained by you Read in the move by rolling solution, given the mask or more was going to be the mass read by the molar mass of ammonia. The elements of a morning. 17 points every 31 He's of excluding the units here, but you should include the units one sold in this essentially, for the top part of the units we're going to be in malls and the volume we've been given it considerable five their their leaders. So similarity is going to be you're a point for 999 Moeller Oh, ammonia in part C just like we've been doing Arma levity vehicles will mass on my scariest 1.49 kilograms, just 1004 190 were amps and the mill a mass for isopropyl given by the chemical formula here when I go clear was zero point. Sorry was 60.96 guard the units here as well. The final I said it would have units more per liter and ah, bowling, which has been given this 2.50 leaders. This is going to give us a value. Wolf 9.9 to Mueller off is appropriate. Lastly for Queen 0 to 9 grams off iodine. And when 10 their leaders of solution. Um, polarity. It means to be cool to all mass of flying their planes there to nine grams to edited by the moola. Massive idea, you know, to get the moles minimus abiding. When I sold it, I got 253 When h don't need to write this. Why, William off solution. Just 0.100 Yes, this one so gives the valley of 1.14 terms chance of the bar minus three. Moeller, we're waiting.

Hi everyone. So in this question they ask what is the majority of diluted solution when each of the following solution is diluted to the given final value. So a one leader a point to $15 solution of FN or three prices diluted your final Valium or two later. So a part someone we want you sequel to. I am too we too and therefore Yeah two is equal to I am one V 1 upon. Mhm We too that is equal to and one one in 2 50 to 1 divided by two and therefore um two is equal to blind 125 more love just Ben be part. Yeah. Yeah. M. Two that is equal to and when we won upon me too and that is equal to point for you In 2.1 repeal too divided by 1.250 and that is equal to Mhm. Therefore. Mhm. It is equal to zero blamed 0488. and therefore and do you see equally too entrance Right. And two is equal to 0.04 people ate moller then see part and two and wonderful one. Yeah. Upon we do. That is equal to Why in 3 15 to 2.30 for you upon four. That is equal to zero point 2056- five. And therefore um two is equality zero point 2056. Do five more. Baby the park. Mhm. I am doing is equally and want everyone upon the two. And that is equal to 0.02 for you into 22 point fire divided by I'm ready. That is equal to 0100 56 2 5. And therefore I am too is equality 0.00 56 Do for you Moeller. Mhm.

This question would be nice to find the final concentration when these solutions have been diluted. So important. A nice to find the final concentration when at 1.0 leader. So Lucien of a point, too Clive's era Mueller. I don't nitrate on three nitrate there we did to you the final ruling of two points their leaders. So what is the final concentration? So given that we've been provided on initial volume, an initial concentration in a final volume with this question especially we can use the formula where our initial number of moles should be called to our final number of moles. That is, the moles for the solution should remain questions regardless off the dilution. So with this, we can use the expression or initial concentration, which I'm going to denote by C one times all initial volume children would also by a V one is equal to our final concentration, shall denote us. See too, and I finally why just be too? And in this question will be Knox to find our final concentration, which is C two. So when we solve for C to hear, see, two is going to be equal to see one V one divided by teaching, right. So with this expression here, what we need to do is look for the valleys that denotes the respected variables and solve that expression and that would do to our final concentration. So our initial concentration, according to this question, is 0.250 Moeller. Our initial volume is 1.0 leaders. They would divide by our final ruling, which is two point. Here's your leaders. So for the solution, the final concentration iss Cyr a point 1 to 5 Moeller off. I had three nine tricked. Yeah, right. So in part B, we do the same thing we did for party. We'll be nice to find off the final concentration off a solution of 0.5 centers of our leaders and 0.1 to 2 to Moller that's been deleted to a final. Falling off 1.25 is their leaders. Just like what we've been doing. Our final concentration is equal to our initial concentration times, our initial walling right, and by our final volume financial concentration. Is there a point 1222 Moeller nature of walling 0.5. There's your leaders and this should be divided by a final ruling. Witches 1.25 Their leaders in this one sold gifts the value of 0.0 for 888 Moeller off pro bono in part C. Just like what we've been doing, we have to find our final concentration and our final concentration when the previous examples have done is given by, Like I said, our initial concentration by a final William Divine by Sorry, our initial concentration by our initial ruling divided by our final volume. So this was gonna given by 0.350 Moeller was. That's the initial demonstration by 2.35 leaders was also initial volume. Where to buy. I'll find a blowing on dilution just 4.0 leaders, and this gives the value off. Is there a point to 06 Moeller off close right acid? Mostly for the solution are the less or D ah final concentration ISS. Our initial concentrations air 0 to 5. Well, our time's initial volume. We need to point for I was there milliliters divided by the final volume 100. Well, it's one thing to note that we don't really worry about our units here. So, for instance, all the previous examples we have been given volumes and leaders, and that's the last one forgivable in new leaders. Don't really worry about those because they cancel out anyways and our final results would be in molars. So when solved, this gives us a volley off. 5.6 to 5 extends the former ministry Moeller of this component Sito age 22 or 11.

Where we need to use the following equation. M. One B. One over N. One. It's equal to um to be to over N. Two. So we can plug in some valleys here we have no 20.101 mola multiplied by 51 0 mil, divided by one. That's able to not 10.1 to 0 mola multiplied by B. Two, divided by two. We sold for V. Two. What we got is 83.3 million. Yeah. So next we can calculate the malls of only a wage. This is about five times 10 to the minus three and the miles of each dry so far. This will be half. So it is 2.50 times 10 to minus three miles. If you consider the strike geometric relationship and the balanced chemical equation, so you can calculate the volume of H. Two. So four by taking the moles and dividing it by not 0.125 mola. To get about 20.0 mL. Following this, we have the malls event eight two S 04 to consider which. So we need to mold a mass fast. That's 142.0 g per mole. So the moles here is 752 times 10 to the minus three, divided by 142 g per mole. We get about 5.3 times 10 to the minus three moles. And so then we have the balanced equation for the reaction between B a c L two and n H two S 04 So from the balanced equation we see that we have one more of an A two S 04 That is needed to precipitate one more. Be a two plus. So 5.3 times 10 to the minus three moles of any two are. So four is needed to precipitate 5.3 times 10 to minus three moles of be a two plus. So we can calculate the modularity of B A c L. Two similarity is equal to the malls of the solid, divided by the volume of solution. We can use Melita units and then to multiply by 1000 at the end. So we take 5.3 times 10 to minus three moles, divide that by 55.8 millimeters multiplied by 1000 mL. What we get is not 0.95 lola. So the final part here we can calculate the malls of HcL solution. So hcl moles is equal to not 0.208 mola multiplied by 42.7 mil defied by 1000. We get 8.88 times 10 to the minus three moles balanced. Chemical equation tells us that we have two moles of hcl needed to neutralize one wall of CH two. So the moles of C a oh H two moles is equal to half of that 444 times 10 to minus three moles. Molar mass of CH two is 74 points over nine g per mole. So then we can calculate the malls where we have the malls of CH two is equal to the mass divided by the molar mass rearranged for mass of C a. O H two. That is equal to 4.44 times 10 to minus three moles, multiplied by 74.9 g per mole. That is equal to 9.329 g. Mhm.


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