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<Assignment 27: Concentrations &amp;amp;amp; OsmosisProblem 8.30 Enhanced with FeedbackYou may want to reference (Pages 323 327) Section 8.4 while completing

Question:

<Assignment 27: Concentrations &amp;amp;amp; Osmosis Problem 8.30 Enhanced with Feedback You may want to reference (Pages 323 327) Section 8.4 while completing this problem: Part A How many grams of CaClz are present in 30.0 mL of a 0.480 M solution? Express your answer using three significant figures_ AEd Submit Request Answer Provide Feedback


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Assume that a portable reverse-osmosis apparatus such as that shown in Figure $18.17$ operates on seawater, whose concentrations of constituent ions are listed in Table $18.6$, and that the desalinated water output has an effective molarity of about $0.02 M$. What minimum pressure must be applied by hand pumping at $297 \mathrm{~K}$ to cause reverse osmosis to occur? (Hint: Refer to Section 13.5.)

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You guys were doing problems. 46 Chapter 18 in Commissioner's Central Science assume that a portable reverse osmosis separates operates on seawater. His concentration gets a drone, ions are listed in 18.5 and that the D selling and capitalism affective malaria points here. A two Mueller. What minimum pressure must be applied by hand pumping at 287 Kevin to cause reverse osmosis to occur.

They were going to use this formula. Pi is equal to ay and Marty, and it's going to be our effective concentration. Are is going to be a 0.8 to 6. Next is going to attempt a return to 90. Assuming Kelvin, I is going to be 11.

Because if you look at table 18.5, you have 11 as you have a 11 11 even, uh, ions that air ions in the water and you can approximate the eye for each iron by the S. So basically, each electrolyte there Jevon water produces, and I i of one. So you would add those 11 together and that gives you absolutely leather. This is a way to approximate it. So we're gonna have pie.

It's equal to 11 that is 0.2 players here. 0.0 age, 206 times to 90 seven. Yeah, tow the pressure that we're going to be applying. That's going to be five 0.36 Um.

First step is to determine whether solid, edgy I owe three forms. Don't do this. We need to calculate the concentration off silver. Ah, nde are your three negative mix. Using these formula? We can determine the concentration substituting the values we have.

Emily Michael played with 0.200 divided by 100 move, which gives us the value off little 0.1 mullah for the 2nd 1 substituting the value stable points you 100 divided by 100. The value to know Toby 0.0 five Moola. Now we need to determine the value off Q, which will be equal to the concentration off product substituting the randy okay, multiplied by 0.0 fight. It gives us the Vandy off for you multiply 10 days to problem minus six. In this case, the value off cue is larger than ks being, so precipitation will occur.

No. The next step is to run the precipitation equilibrium reaction as represented over here, along with before and after concentrations for dough. We need to note that excess off Ayotte really means after precipitation reaction goes to completion, so concentration off I want three negative excess frequents. 0.400 Divided by 100 equals 0.0 for Mullah. Although we have assumed that Eiji Positive is completely consumed, we know that Eiji positive concentration will not be zero at equilibrium.

Let me represent it. Reaction Eiji, are your tree gifts e g border? Do yes. Are you three negative and gai sp? He quote concentration off 80 positive on DDE are you? Three negative equals three points. It'll 10 days toe or negative eight Now we'll be right Initial and equally broom concentration I see will be initial concentration on DDE e C will be equally grim concentration. So for a G for the two equals zero Molo on for this would we have X concentration off? I owe three negative 0.4 And for the right side equilibrium the 0.4 plus x as x more poor leader A g r u tree dissolves So make the reaction reached equilibrium No! We calculate the KSB as 3.0 10 days.

Tobar. Negative eight equals the concentration off Eiji positive on DDE. I owe three negative substituting in the value we have X multiplied with 0.4 plus x x that is concentration off a G positive at equally broom. Don't know Toby seven point for you multiply 10 days to burn minus six Mullah..

So let's start by calculating the moles that we need. So moles, osmolarity Times Leaders. So our polarity this .15 moller and We've got 175 mm. So that's 175 L. So that's going to be Went 0- 6- five moles.

I can't measure out moles. So let's measure that. Let's change that to grams Mueller Masses 1 69 88. So that's going to give us 4.46 g. So we're going to dissolve 446 g of silver nitrate in enough water To form 175 ml of solution.

Be sure that you don't say that you're adding 175 million liters of water because then you'll end up with two too much water. And your solution will be to dilute In our second example, we're trying to dilute more concentrated solution. So let's use M one V one equals M two V two, Which is a good solution for dilution. Good equation for dilution. So .5 Moller Times 100 ml is 3.6 smaller B two.

So we grew up with 14 mm. So what we need to do Is at 14 mm of the 3.6 Mueller solution again to enough water to form the volume of solution that we want. Okay, so to form 100 ml of solution Again, don't say you're going to add 100 of water. You just add it to have a total volume of 100 ml of solution.

This question has three parts that are all unit conversion calculations where polarity will be used as a conversion factor One mil leader of solution can be converted into leaders by dividing by 1000. Well then convert the leaders of solution into moles of the solute by multiplying by the polarity of the solution. Then when we have mold salute, we simply multiply by the molar mass in order to get the mass of the salute. So there are 0.48 g sodium carbonate in one mil leader a 10.450 molar sodium carbonate solution. If we have 10.5 million liters, we can convert it into leaders by dividing by 1000 then multiplied by the polarity of the silver nitrate solution to get mold silver nitrate and then multiply by the molar mass to convert the mold silver nitrate into grams silver nitrate.

So there are 4.46 g silver nitrate In 10.5 million liters of a 2.5 molars, silver nitrate solution. And for the last one We have 28.4 ml will divide by 1000 to get leaders multiplied by Mila Mueller to get the million moles of the hcl. Then we'll convert the milling moles into moles. By dividing by 1000. There's 1000 million moles per mole.

Then, knowing the moles of HcL, we can multiply by the molar mass ticket grams hcl and we get 6.21 times 10 to the negative three g hcl In 28.4 million liters of a $6 million dollar hcl solution..

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