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Determine the Determine - concentration; Determine I00.0 mL welgnepenccn 2.75 and 8 the mass . 0(4g weighi percent 0.00226 grams, U 8 the lab. Omtin The the sample ...

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Determine the Determine - concentration; Determine I00.0 mL welgnepenccn 2.75 and 8 the mass . 0(4g weighi percent 0.00226 grams, U 8 the lab. Omtin The the sample the original H sample. solution. 0 [ (0 contaln 8 Ppm _Aducmit

Determine the Determine - concentration; Determine I00.0 mL welgnepenccn 2.75 and 8 the mass . 0(4g weighi percent 0.00226 grams, U 8 the lab. Omtin The the sample the original H sample. solution. 0 [ (0 contaln 8 Ppm _ Aducmit



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Water is allowed to evaporate from $100.0 \mathrm{mL}$ of $0.24 M$ $\mathrm{Na}_{2} \mathrm{SO}_{4}$ until the solution volume is $60.0 \mathrm{mL} .$ What is the molar concentration of the evaporated solution?

To find the final concentrations of everything. Cassie my on the Oxlade and I the very my on the bromide ion. We need to figure out how much of the precipitous forms and then knowing how much precipitate form, calculate using equilibrium. Equilibrium concentrations for the IOM's involved in equilibrium involved in the precipitation. Bury him Oxlade, which would be very in my on and the ox late. Those are gonna be two most difficult to determine. And then the potassium bromide can be determined just through a delusion, because they're not part of the a precipitation reaction there. Just a spectator ion. So we'll start with the precipitation reaction, which will write as a K SP reaction, which would be the precipitation reaction occurring in reverse cast T is equal to in the very in concentration multiplied by the oxygen concentration. So to figure out whether or not to precipitate forms and then how much precipitate form, let's assume that all of the Aqsa late reacts and then precipitated does form and then calculate the amount of very amok slate that could be created from all of the Aqsa late. I will do that by recognising we have 100 bill leaders that point to Moller, uh, potassium oxalate this and then we recognize that one Matlock slight produces one mole of the barium Oxlade. So assuming all the ox late does react to form very Moxley, that would be the maximum amount of oxide we could make. 0.2 Moles do the same thing than with Beria. My on you have 150 mil leaders of varying bromide at 0.250 Moeller and then one mulberry in produces one more Beria Moxley. This will be the amount of very mocks late that we could form. Thus we will make the least amount possible. We'll assume then that all of the ox late is consumed. This concentration essentially goes to zero. When the precipitous forms, we will then have to recalculate how much of the Aqsa late is still in solution based on the equilibrium and the Oxlade, then would be the limiting reactant. So the Catholic, then the moles of the burying my on that air left in solution which we need to do to get a concentration of barium left in solution. So we can finally calculate the Oxlade concentration we need to take the moles of barium that we started with 150 milliliters multiplied by the $0.25 and then subtract off the moles of burying that reacted. The mole Siberry in that reacted will be equal to the molds of very mocks late that we're gonna make because of one mole of barium in one mulberry Moxley. This will give us the moles of barium that's still left in solution that will calculate the concentration of barium, which will be the moles of barium left in solution divided by the volume which is 100 milliliters plus 250 milliliters given us 250 milliliters or 2500.25 leaders and then that gives us 0.7 Moeller a very mind. Then we can look up to cast p value for very amok slate, which I did 2.3 times 10 to the negative eight. That then will be equal to the very end concentration which we just calculated, multiplied by the Oxlade concentration Then rearrangement allows us to solve for the Oxlade concentration. Now we know the very in concentration and we know the Oxlade concentration after the precipitate forms in order to solve for potassium concentration, we simply need to take the of all you of potassium obsolete and the concentration of attacks. Potassium ox late. This will give us moles, potassium Oxlade. But there are two potassium and potassium ox lights were moved by that by two. And then divide by the volume, which will be the 0.250 leaders will get 0.16 smaller task in mind. We'll do the same thing with bromide Brookline concentration that is going to be equal to the volume. A very, um, bromide multiplied by the concentration of very old bromide multiplied by two because there are two bromides in Berrien bromide. And then divide that by the new volume in order to get the barium constant. I'm sorry. The Bromine concentration of 20.300

The additional Kato cr toe Full toe Berrien bromide results in tow. Precipitation reaction as fallen the morning off. Both the reactant are added can be calculated us using the formula molar It equals more per unit. Wonder so So malls for okay too cierto or four will be equal to six. Multiply 10 based upon negative four 0.100 later number off molds turn out to be six. Multiply 10 days to plan negative five moves on four billion blue forbade him Ra might It will be equal to 10 based upon negative four multiplied with the low point 150 leader. The number off motor. No Toby 1.5 10 days to put negative five moons The psychometric ratio for burying bromide to baby him too positive. Four billion you might toe Very, um sorry. Toe burial toe. Positive physical one racial one. Similarly, the strike automatically racial for Kato CIA, too. 04 on DDE C two or for the negative is also equal to one racial one. So check whether the precipitation will occur or not. The guy Nick product can be calculated. US concentration off BDM too positive will be equals two more off very um divided by total volume on solution. So moles off baby um is 1.5 10 days to plan negative fight divided by total war Newman Solution that is 0.100 plus 0.150 meters. Richard gives six multiplied 10 trees to plant negative five mullah The concentration off see too, or for two negative will be molds off Kato C 24 divided by total body. More solution. Substituting in the Vandals. We have 6.0 multi pretenders to plant negative five, Divided by 0.100 plus 0.150 meters, which gives us value 2.4. Multiply 10 days to put negative for more. Q For the given iron can be calculated. US. Concentration off very, um multiplied with concentration off Sito for two negative. Substituting these two values into equation. We have Q equals one point for multiply 10 days to one negative eight us. It can be seen the value off cue. The value off cue is quite less as compared to ke SP, so precipitation will not takes. So no precipitation. Now ask first, I cometary baby in Broome might concentration will be beer Negative equals two. Be too positive. So b r negative equals to my private six. 10 days till my negative five equal 1.2 multiplied with 10 days to plant. Negative for Mullah using the strike Romantic creation relation again We have concentration off Cape Positive equal two multiplied with Sito well for two negative substituting the values we have two multiplied with two point for 10 days to plan negative for equal 4.8 multiply 10 days to one negative for for their own The concentration off various I'm solution are be a positive equals 6.0 multi pretending stuff on negative five Be are negative equals 1.2 Multiply 10 days to put negative fool Dasha 4.8 10 days to plan negative for on DDE C 242 Negative turn out to be two point for monte pretenders to plant negative for

When one substance dissolves into another, a solution is formed and a solution is identified as a homogeneous mixture made up of a solid dissolved into a solvent. So here, what we've asked me to do is convert from not point not 10 parts per million into milligrams, polyta, Not quite, not 10 mg/l. And then from this value convert 2g Paletta, which is 1.0 times 10 To the -5 g polyta By multiplying by not point, not not one g per milligram. So therefore the massive lad iron is 1.0 times 10 to the -5 per liter of solution. So next we have the volume of solution that is one liter at the density of water is one g per mil. And so fastly we can calculate the mass of the solution so we have one liter that is equal to 1.2 times 10 to the three g of solution. Where the mass fraction is equal to the mass of the salute, divided by the total mass of solution. What we get is 1.0 times 10 to the -8. That is the mole fraction using the equation here.

Hello. So today we're going to be looking at a situation well. We've taken 2.5 grams of copper sulfates and dissolved it into 900 milliliters of 0.3 mil aren t ammonium. So what exactly is going to happen? Well, copper two plus tends to make complex ions with ammonium, so we're going to see copper two plus ammonium and a complex ion in solution. So first, let's right the equation for the complex ion informations. We have copper two plus in solution and we have ammonium and we're forming copper with Ford Ammonia. Mum's touch to it. Two plus. So I'm gonna have need tohave before here. And so here we have it. Now let's form a nice table So we know the concentration of ammonium already ammonia. We have 0.3 mil arat e. We don't have any of the complex ion yet, so now let's try to find the concentration of the copper. So we have 2.5 grams of copper sulfate, So divide by the molar mass which is 159.609 grams Permal off copper sulfate. So every mole off copper sulfate is one mole of copper two plus I owns and then we divide by the volume we have 900 mil leaders. That's 9000.9 leaders. So we will see that our concentration of copper is 0.174 So 0.174 And so we're gonna have used up one mole of copper. Four moles of ammonium give one mall of the complex ion. So this looks pretty complex right now. So let's try to figure out a way to simplify this. And one way to simplify things is to think of in terms of limiting reagents. So the constant information the include every in constant for this reaction is 5.0 times 10 to the 13th which is really, really big Slam means that this will go almost to completion sort of. But let's take a look at our starting materials. Well, the concentration of ammonia is a lot larger than the concentration of copper, so let's consider that copper is are limiting re agent, so almost all of the copper will react with ammonium to give the complex ion. So then the concentration of our complex ion will be the concentration of copper at the start, which is 0.0 174 polarity. Now let's take a look at Not the Constitution of ammonia, well, ammonia. It takes if a zero point almost 0.174 moles of copper has reacted. That takes four times that amount, so that would be so. It would be 0.3 minus 0.174 times four, which is 0.174 times four is 0.696 Subject that from 3.3 and we get zero point 23 Similarity. So now we have the concentration of ammonium at, um, ammonia and the complex ion. So now let's find the very minuscule concentration of copper two plus that is left. So the constant, the equilibrium constant for this reaction would be the concentration of the complex Ryan over the concentration of ammonium to the fourth. Because remember has a four in front and then the Constitution of the copper two plus so as mentioned before, our great constant is 5.0 times 10 to the 13. The complex concentration will complex ion is 0.174 The concentration off ammonia is 0.23 to the fourth, then is just the construction of covered us. So how about way? Bring the silver to that side? So we'll have 5.0 times 10 to the 13th times 100.23 to the fourth, So 0.23 to the fourth is 0.28 I'm going to multiply that by five times, tend to the 13 so we will get 1.4 times 10 to the 11th. Copper two plus is equal to 0.174 No divide. They will see that we get 1.2 times 10 to the negative 13 polarity, which is very, very small, so you can see that it has. Essentially, this is essentially zero. It's like very, very small. So it's like almost essentially zero, and this has gone essentially to completion almost because this is a very


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