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Pmccdure Producc Disinfectant Spray. (20%) You will usc isopropyl alcohol and distilled water produce 70.070 # disinfectant spray. The distilled "ulcr docs nOl...

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Pmccdure Producc Disinfectant Spray. (20%) You will usc isopropyl alcohol and distilled water produce 70.070 # disinfectant spray. The distilled "ulcr docs nOl nced = bc sterile #utoclaved water for this purpose . SinC € will not he "pplicd c ditectly to human skin and since an alcohol content of above 64%& Will kill essentially all pathogens What volumc of isopropyl aleohol (from your supplies listed above) to You Inecd t0 make [0 Lof70.0% disinfectant spray? What volume of dist

Pmccdure Producc Disinfectant Spray. (20%) You will usc isopropyl alcohol and distilled water produce 70.070 # disinfectant spray. The distilled "ulcr docs nOl nced = bc sterile #utoclaved water for this purpose . SinC € will not he "pplicd c ditectly to human skin and since an alcohol content of above 64%& Will kill essentially all pathogens What volumc of isopropyl aleohol (from your supplies listed above) to You Inecd t0 make [0 Lof70.0% disinfectant spray? What volume of distilled water d you necd? Show work for full credit Procedure Produce Hand Sanilizer. (40*1 You will usc thc supplics lisicd above proxlucc hand sanitizer. Since glyecrine an incredibly sticky liquid, Wc have decided 10 wcigh the elycetine instead of measuring by volurne_ Whui volumc of isopropyl ulcohol (from the supplies listcd above} do You Iccd t make [0 L of hand annilize What volumc of hydrogen pcruxide do you nced? What volume AND what mass 0f glycering do You nced? ( Hint: what I8 tne density - glcerine?) What volume of sterile Wuler do You need? Show wotk for full credit;



Answers

You want to clean a 1.00 -L bottle that has been used to store a $0.500 M$ solution. Each time the bottle is emptied, $1.00 \mathrm{~mL}$ of solution adheres to the walls, and thus remains in the bottle. One method (method 1) of cleaning the bottle is to empty it, fill to $1.00 \mathrm{~L}$ with solvent, and then empty it again. An alternate method (method 2) is to empty the bottle, pour $9.00 \mathrm{~mL}$ of solvent into the bottle (to make $10.00 \mathrm{~mL}$ total), swirl to mix uniformly, and then empty it. This process is repeated twice, for a total of three rinses. (a) What is the concentration of the solution that remains in the flask after the single rinse in method $1 ?$ (b) What is the concentration of the solution that remains in the bottle after the triple rinse in method $2 ?$ (c) Compare these two methods of rinsing, in terms of the amount of solvent used. (d) Discuss the implications of this comparison for an industrial process in which manv such bottles must be rinsed.

Okay, Um, for this particular one, we need to determine the order of the reaction with respect to each reactant before we can write the rate law To do that, we can take the information given for one of the experiments. Rate is equal. Decay multiplied by one reactant concentration raised two x multiplied by the other reacting concentration raised toe. Why? What we can do then, is for a second experiment where one of the reactant concentrations is staying constant and the other one is changing. We can then rewrite the rate law rate for Experiment two is equal to rate constant multiplied by the concentration of one reactant raced to the X, the other concentration raised to the why you'll see then that the cave values cancel in the concentration associated with the second reactant. Now also cancels and all we have left is this ratio and this ratio raised to the X. So we get three equals three rates to the X, so x must be one. We do the same thing now where the reactant concentration of a different one is changing while the other one's days constant. So we have. The rate is equal to que multiplied by a concentration of first species raised to the X concentration of second reactant race to the y and then another experiment. We choose the experiment appropriately so that the concentration for the reactant that we just determine the value of the exponents stays constant so we can determine the concentration. We can determine the exponents for the reactant that has a changing concentration case. We're going to cancel the concentrations raised to the X or going to cancel. And we get 0.335 equals 0.333 Raised to the why so wise one for the rate law is going to be equal to K equals concentration of Cielo, raised to the first power multiplied by the concentration of O to minus raised to the first power. Then all we have to do is take any experiment anyone that we want. I'm gonna look at the first one and write this rate law where we have all the information except K. And then we can solve for K. So the rate of the first experiment is equal to K multiplied by the concentration of one reactant 10.6 raise to the Now we know first power well played by the concentration of the other reactant. Now we know. Raised to the first power, we can rearrange the equation and solve for K and K ends up becoming 0.24 divided by the 0.6 multiplied by 0.3 and we get 13.81 over Moeller. Second, we make sure that we have the appropriate units knowing that second order overall and second order overall has units of one over polarity time.

So we know that isopropanol can decompose into acetone and hydrogen. So let's start with the information they have given us about the isopropyl alcohol. Okay, We know that. We've got 20 ml. Uh we've been given a density. So let's change milliliters two g using that density. So 0.785 g in one millimeter. And then let's go ahead and change grams two malls. And we use the molar mass of our isopropyl alcohol, which is 60.11 And we'll get .261 moles. And then since we're going to want to work with gases and a partial pressure, so we'll use the ideal gas law PV equals NRT right? Or P is an R. T over B. So we'll go ahead and plug in our moles there .261 ours point oh 8 to 1. Our temperature was 180. Mhm. So we'll add to 73 to that to get it from C to Kelvin. So that gives us 453 Kelvin. So go ahead and plug that in here for 53 and then we'll divide this by our volume, Which is five L. So we should end up with 1.94 a. t. m. So that's the initial partial pressure of our isopropyl alcohol. So I'm gonna go ahead and set up an ice box here to keep track of what's happening. So I as initial C is change is equilibrium. And initially with her isopropyl in Isopropyl alcohol, we have 1.94. Okay. And we know that we don't have any acetone and we don't have any hydrogen. So as these react, it's all 1 to 1 to one. So this would be minus X plus X. Cossacks. So we end up with 1.94 minus X. And an X. And an X. So, if we go ahead and write our expression for K, That's the partial pressure of our acetone. Right? Times a partial pressure of our hydrogen divided by the partial pressure of our isopropyl alcohol. Okay, so we've been given K. So we'll go ahead and plug that in. 2.45 is X squared over 1.94 minus X. So if we multiply it kind of cross multiply that We'll get .87 -15 x equals X squared. So I'm just going to rewrite this. So it's set equal to zero. Okay, so X squared Plus .045 x minus 87 Equals zero. So now we can go ahead and use that quadratic equation or you can graph it and your values will come out to be either negative 1.18 or 0.73 Obviously the negative one isn't going to work. So we know that X is .73. So we can use that now to find the partial pressure of our ice appropriate isopropyl alcohol. Right? So that we said was 1.94 -1 -13. So that's going to give us 1.21 ATM. So now we can figure out percentage of that that has associated is the amount that I get Divided by the original 1.94 times 100. And we'll get that. It is 62.4 un dissociated. Okay, so This is the part that still remains isopropanol that's not dissociated into our products. So that's the 62.4% there.

So we know that isopropanol can decompose into acetone and hydrogen. So let's start with the information they have given us about the isopropyl alcohol. Okay, We know that. We've got 20 ml. Uh we've been given a density. So let's change milliliters two g using that density. So 0.785 g in one millimeter. And then let's go ahead and change grams two malls. And we use the molar mass of our isopropyl alcohol, which is 60.11 And we'll get .261 moles. And then since we're going to want to work with gases and a partial pressure, so we'll use the ideal gas law PV equals NRT right? Or P is an R. T over B. So we'll go ahead and plug in our moles there .261 ours point oh 8 to 1. Our temperature was 180. Mhm. So we'll add to 73 to that to get it from C to Kelvin. So that gives us 453 Kelvin. So go ahead and plug that in here for 53 and then we'll divide this by our volume, Which is five L. So we should end up with 1.94 a. t. m. So that's the initial partial pressure of our isopropyl alcohol. So I'm gonna go ahead and set up an ice box here to keep track of what's happening. So I as initial C is change is equilibrium. And initially with her isopropyl in Isopropyl alcohol, we have 1.94. Okay. And we know that we don't have any acetone and we don't have any hydrogen. So as these react, it's all 1 to 1 to one. So this would be minus X plus X. Cossacks. So we end up with 1.94 minus X. And an X. And an X. So, if we go ahead and write our expression for K, That's the partial pressure of our acetone. Right? Times a partial pressure of our hydrogen divided by the partial pressure of our isopropyl alcohol. Okay, so we've been given K. So we'll go ahead and plug that in. 2.45 is X squared over 1.94 minus X. So if we multiply it kind of cross multiply that We'll get .87 -15 x equals X squared. So I'm just going to rewrite this. So it's set equal to zero. Okay, so X squared Plus .045 x minus 87 Equals zero. So now we can go ahead and use that quadratic equation or you can graph it and your values will come out to be either negative 1.18 or 0.73. Obviously the negative one isn't going to work. So we know that X is .73. So we can use that now to find the partial pressure of our ice appropriate isopropyl alcohol. Right? So that we said was 1.94 -1 -13. So that's going to give us 1.21 ATM. So now we can figure out percentage of that that has associated is the amount that I get Divided by the original 1.94 times 100. And we'll get that. It is 62.4 un dissociated. Okay, so This is the part that still remains isopropanol that's not dissociated into our products. So that's the 62.4% there.

In this question that mass percentage of isopropyl alcohol in given solutions would be calculated. We know must percentage is equal to myself and a light divided by Markov some but market by by 100 given mass off isopropyl alcohol is equal school 2.67 g and myself solution is 6.01. Yeah. Therefore mass percentage of of isopropyl alcohol. It's supposed to massive and right that is massive. Isopropyl alcohol divided by mass of simple. That is massive solution. This is my private conduct. Therefore, mass of ice. Super quiet is given that is 3.67 Grand Divided by mass of solutions. 6.01 run modified 500 Equals to 61 1% that's mass percentage of isopropyl alcohol and given solution is 61.1%.


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