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Table 2: Sample data table for the results and observations trial Trial 1 Trial 2 Room Pressure 1 atm 1 atm Room Temperature 23 Celcius | 23 Celcius] Mass NaHCOz (g...

Question

Table 2: Sample data table for the results and observations trial Trial 1 Trial 2 Room Pressure 1 atm 1 atm Room Temperature 23 Celcius | 23 Celcius] Mass NaHCOz (g) 3.561 7.656 Volume of 3.0 M Acetic Acid (mL) 12 mL 20 mL Total Mass of Bag _ and Reactants g) 17.6 g 31.4 g Height of Bag after Reaction (cm) 8 cm 7 cm Time needed for maximal inflation (s) Moles NaHCO3 Added Moles HCzH3Oz Added Limiting Reactant Theoretical mols COz Produced Theoretical Volume COz Produced (mL) Observations

Table 2: Sample data table for the results and observations trial Trial 1 Trial 2 Room Pressure 1 atm 1 atm Room Temperature 23 Celcius | 23 Celcius] Mass NaHCOz (g) 3.561 7.656 Volume of 3.0 M Acetic Acid (mL) 12 mL 20 mL Total Mass of Bag _ and Reactants g) 17.6 g 31.4 g Height of Bag after Reaction (cm) 8 cm 7 cm Time needed for maximal inflation (s) Moles NaHCO3 Added Moles HCzH3Oz Added Limiting Reactant Theoretical mols COz Produced Theoretical Volume COz Produced (mL) Observations



Answers

A student set up an experiment, like the one described in Chemistry You Can Do (p. 143), for six different trials between acetic acid, $\mathrm{CH}_{3} \mathrm{COOH}$, and sodium hydrogen carbonate, $\mathrm{NaHCO}_{3} .$ $\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{NaHCO}_{3}(\mathrm{~s})$ $$ \mathrm{NaCH}_{3} \mathrm{CO}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ The volume of acetic acid is kept constant, but the mass of sodium bicarbonate increased with each trial. The results of the tests are shown in the figure. (a) In which trial(s) is the acetic acid the limiting reactant? (b) In which trial(s) is sodium bicarbonate the limiting reactant?

Okay, so it was considered a reaction Ch three secret hedge plus and a let's see your three leading to formation of N A CH three, c. 0 two Plus, C. or two along with water. As the product increases, the volume of carbon dioxide gas also increases. So for option A there is no change in volume as the mass of sodium bicarbonate increases hens. In trials for 12, 5 and 12, the limiting reactant is ascetic acid. We want to part B as the mass of sodium bicarbonate increases the volume of balloon also increases in trial three Hands. for trial one, two and three, the limiting reactant will be sodium bicarbonate in all the trial. In all these six clients, the volume of ascetic asset, the volume of ascetic asset is constant. That means the mass of gastric acid is same throughout all the violence, the mass of sodium bicarbonate increases. With each trial Has the mass of sodium bicarbonate increases the volume of balloon also increases up to trial three. Entrails four, five and six. There is no change in the volume as the mass of sodium bicarbonate increases. This means that no more sodium bicarbonate is there to combine with ascetic acid to lead out the liberation of carbon dioxide. Yeah, hence sodium bicarbonate is limiting reactant in Trial 1, 2 and three.

In this question. We already been given the balanced equation and many of the variables. We just need to mess with them a little bit to get them in the correct format. So it's important for us to know that them ocular mass sodium bicarbonate is 83.9 grams per mole. I just calculated this using a periodic table. So you just need to add up on the different elements from the different numbers of them in order to get them looking at mass for the total compound. No, for N A H CEO three. We were given that amount in crimes, but we really wanted in moles. I'm in order to use the ideal gas law, so we just need to convert from any H 03 grams to moles. So that's why we need that Markkula Manus armor gonna programs in the bottom. So that means, um um, like those milky masks gonna go on the bottom, the 83.9, and we're just gonna put one of the top two convertibles. I'm just gonna give us your 0.33 moles of n a h c O. C. Three. And since the equation is full of balanced. We know that the coefficients are all one. So if we have 0.33 moles of NHC or three, we're gonna produce Israel 30.33 moles of Seo Ji. Since there's no. Two and Brennan CEO do you don't have to multiply or anything. It's just a straight 121 conversion. We have a value and Mel's little scent. Then we just need to change our temperature from 40 degrees Celsius to the covens. We had 273. Then we just plug everything in. We've got to your 0.33 moles of co two. Um, we're gonna multiply that by leaders. So we're gonna up most of the pilots they canceled. Multiply it by the STP 22.4 leaders. And then we're gonna multiply by the change in temperature. You've got 293 divided by 273. Both Calvin and we're gonna get our leaders for Seo to actually equation. The new volume is 7.9 litres of Seo Ji

This question asked us to complete this table using Boyle's law and assuming that temperature and moles are constant. So the first thing that I want to do is you'll notice that I've drawn to same chart here, and the first thing I want to do is unit conversion. So in for Boyle's law, we need our pressure to be in atmospheres and are volume to be leaders. So I'm going to convert everything that is not in those units to those units on. And I'm gonna write those down in this second table. So for this first one, this is the millimeters Mercury. What we needed to be in atmospheres. So to convert, that will divide that by 7. 60. And that gives us 3.993 atmospheres. Then, um, coming down here 1 92 Bellamy's, um, rectory about it by 76 to you gives us 0.253 atmospheres and this one is already in atmospheres. 2.11 This is good. Already in leaders and with this one mil leaders, two leaders, we just divide by thousands of sweaty 20.382 Oops, 32 liters. Um, then we have 85 millimeters of mercury, so we'll divide that by 7 60. And that gives us 1.16 atmospheres and then or 0.32 atmospheres and 3.82 atmosphere is already good. And then, in this column, 2.88 leaders is already good. And then again, for mill leaders, leaders justified by it wasn't so point for eight to leaders and 0.125 liters. All right, now that we have everything in the correct units now, we'll go. How to use Boyle's law So Boyle's law is p one times e one equals P two times V two. So for this, actually put that down here, you want to be one equals p to be to. So for this first row here, we're looking for me to so we have P one. That is 10.993 a. T. M. Times 2.85 leaders RV, one people to pee to, which is 1.16 It's, um, and we're solving for B two. If you go ahead and do that math, you will get a B to hear of 2.43 liters. So I needed to get. That was multiply 0.993 times 2.5 and then divided by 1.16 and they gave me 2.43 So that is my volume are might be, too for that row going on to the next row we're solving for P one so P one times 1.33 leaders That's R V one is equal to R P. Two of 4.32 atmospheres. Times are view two of 2.88 leaders, and we do that Math the same way we did above will get a P one equal to 9.35 atmospheres for the next row we're solving for P two So R P one is 10.253 atmospheres Times are we one of 0.382 leaders is equal to P two times point or 82 leaders. And if we solve for that P two we get Let's see 0.53 I was falling 382 divided by Don't worry, too. That gives us point to atmospheres and then for the last one. So that one in blue So we're solving for B one here, so we have P one of two point 2.11 atmospheres. Times be one people to PTO 3.82 atmospheres times the Jew of 0.1 to 5 leaders and hopefully solve their We end up with 0.226 leaders for B one. So now I want to do is go back and put all these in the same units that there Rose were in. So for the top row, we were in millimeters of mercury and leaders and our answer was already in leaders select and stay the same. 2.43 leaders for the second role were in atmospheres and leaders again. That one won't have to change that still 9.35 atmospheres for the throat. Third row We were in a millimeters of mercury and, um, mil leaders. So I want to convert this, um, 0.2 atmospheres. Two millimeters of mercury said to do that, I'm just gonna multiply it by 76 seek. That's our conversion factor here. And that will give me 1 52 millimeters of mercury and or the last row. We're in atmospheres and mill leaders. So to convert my leaders to mill leaders. I'm just gonna multiply by 1000 so I get to 26 mill leaders there.

Hi everyone. So in this question they ask you, is the molecular volume provided with each compound To calculate its been sitting in grandpa um, L A N H C 03 sodium bicarbonate or sodium hydrogen carbonate, also called baking soda 0.3 89 liter per mold. Then be eye to eye ring 0.5 went for it later caramel and mercury 0.1476 liter per mole. And uh and a seal common table salt that is 0.0269 Men Little Parmalee. So they have provided molecular volume of everything. So you have to calculate the rain city. So a part They have provided uh, you know masa and 803 loss of molecular mass up N H C 03 morning Q lad. Mhm months of Amen H 03 19 is equal to 84 grand. Then then city is equal to 84 upon 38.9. And that is equivalent to two points. Well for you. 94 grand. Bottom more grandpa mold. So Then city of an h. c. 0. 3 auto making sort of therefore the city of and the edge see what we is two point one for you. 94 crime Baltimore. Then the part. Sure. Yes. B but the density of Irene. You know my ass off. I do. That is equal to 126 points 9042 19 is equal to 253 808 g. Yeah in belfour you know what in the hell given is 0.05148 L for um all And that is equal to 51.48. And then so therefore the city is equal to mass upon volume. That is equal to 253 eight 08 A 1 51.48. That is equal to four points 90 three zero two for you grandpa mold and therefore the city of Irene in the city of I owe him is equal to four 930-5 graham. Our model then see part Yeah. Most of them are recruiting moss. Both my goody That is equal to 200.159 than one limb. You sequel to 0.014 76 liter. The basic 12 14.76 ML. Now the city is equal to mass upon welding and that in sequence to 200 point 15 9 divided five 14.76 And that is equal to 13559 crime body man and therefore the city of marrakech, very therefore the city of Mercury That is equal to 13.159 Graham Bird Mall then depart. Okay, most of finished cl myself initial. That is equal to 58.5 Grand then volume. That is equal to and so you don't play and see it all 2699 leader that is equal to 26.99 ML. So then set up Initial. Then the city of a nation is equal to mass upon volume and that is equal to You'll be 8.5 upon 26.99 and that is equal to two 1675 Grandpa and mel. And therefore the density of financial uh huh. The city of an A C. L is is two point 1675 Graham's argument. Thanks.


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