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89.56 MHz0.04 ml : 0.5 ml CDCLHe = 1.5 +Imi Hd?nHb 1 Ha = 110...

Question

89.56 MHz0.04 ml : 0.5 ml CDCLHe = 1.5 +Imi Hd?nHb 1 Ha = 110

89.56 MHz 0.04 ml : 0.5 ml CDCL He = 1.5 + Imi Hd?n Hb 1 Ha = 1 10



Answers

The volume of ' 10 vol.' of $\mathrm{H}_{2} \mathrm{O}_{2}$ required to liberate $500 \mathrm{~mL}$ $\mathrm{O}_{2}$ at NTP is: (a) $50 \mathrm{~mL}$ (b) $25 \mathrm{~mL}$ \begin{tabular}{l} \hline \end{tabular} (c) $100 \mathrm{~mL}$ (d) $125 \mathrm{~mL}$

Let's try to figure out the number of milliliters of nitric acid that we would need to use. Um if we wanted to react it with 1.30 g of barium hydroxide, They will do that. We need to know the polarity of the nitric acid that we would need. And we're going to use a .125 molar solution. So you have to do some solutions to it geometry here to figure this out. So we'll go ahead and we'll use the value were given our 1.30 g of our barium hydroxide and we'll convert that to moles by using molar mass Mueller massive barium hydroxide is 171 34g. And we want to know about nitric acid. So we're going to use the mole ratio. Now we need to moles of nitric acid For every one mole of barium hydroxide that we use coming from the coefficients of our balanced chemical equation. But now we need to go from moles of nitric acid, two ml and so we can use the polarity to help us get there. And so the polarity .125 moles per liter Can also be written as one leader is .125 malls. But we don't want to know about leaders. We want to know about millet leaders, So we have to convert to say that there's 1000 ml in a leader double check that our units work grams, cancel out moles, cancel out moles, cancel out leaders, cancel out for left foot milliliters. And we wind up with needing 121 millilitres of the nitric acid solution.

So in this problem, we're trying to figure out the mass of magnesium iodide. We need to add to a solution of 0.876 Moller potassium iodide in 250 mill leaders in order to create a solution that has a iodide ion concentration of 0.100 Moeller. And the way we do this is first by figuring out, um, how many moles we have in solution already on the way we do this is by multiplying are concentration year by our volume solution and then by our mole ratio. So we have our to your 0.876 Moller potassium iodide, both blood by our volume, which is 0.250 leaders. And then we can multiply by our mole ratio, which is 1 to 1. And this gives us, um, we have a point 0 to 19 rules of I died. I on in solution already and now let's figure out how many moles we want total. So to do this, we take our desired concentration 0.100 Moeller in iodide and multiplied by our volume which again is 0.250 leaders and This shows us that we want a total of 0.0 to 50 moles of iodide in our solution. Now we can subtract thes two to figure out how many moles of I died. We need to add so we can take 0.250 moles of iodide minus 0.21 0.219 those of iodide and this tells us that you need to add zero point 0031 most of iodide to get to our desired concentration. Now, using this value how many moles of idea and we need We can perform some dimensional announces to figure out how many milligrams of magnesium iodide is will give us this many moles of iodide ions. So we start with our 0.31 Most of iodide and weaken multiplied by are molar ratio of magnesium iodide to iodide ions. Two goals I died and it's 1 to 2 Because of our subscript Here we have one or one more of magnesium. I died for every two moles of iodide irons. Then after that, we can multiply by our molecular weight 278.1 grams Permal and then we can convert two milligrams by multiplying by 1000 those in milligrams per one gram. And this gives us 431 0.1 milligrams of museum I had died needed.

Yeah. Here in this question we have to find the volume in millilitres of 0.125 moller 90 Cassidy which is required to react completely with 1.30 grandma. Barium hydroxide. According to the following equation. Now we'll find the molecular weight up very um hydroxide. The atomic mass of Idiom is 137 1 37.32 plus two oxygen attempts O H hold twice. So let's do this one, oxygen is 15.99 plus hydrogen 1.1 times two. So we'll get 171.33 g bar more. There's the molecular weight. Now Mass of barium hydroxide given is 1.30 Brown. So 1.30 g barium hydroxide is 1.30 divided by 1 71.33. This is the ground and here ground far more which is 7.58 times 10 to the power negative three. Yeah. According to the reaction one mole of very um hydroxide. Yeah. With two months off 18 03. There are four Mall saw eight You know, three required. Yeah. These two times 7.5. 8 times 10 to the 4 -3. Morning. Let the volume Okay or 0.125 Mueller nitric acid is x millimeter number one Number of moles of 1803. Yeah is equal to the volume of 1803 in Liga times concentration Mueller concentration. Number of months of age. You know to just calculated two times 7.58 times 10 to the power negative three and volume of uh, paktika cities X millimeter means it's by 1000 later. And Mueller concentration is 0.125 mall park later. Therefore, X is two times 7.58 times 10 to the poor negative three times 1000 divided by 0.125. There are four x is 121.2 millimeters. There are four actually volume of Nitric acid 1803 required. Yeah, is 1:21.2 millie linda.

So the p H equals p k add log NH three over NH four plus equals 9.26 by the concentration of our hydroxide ions is 1.8 times 10 to the minus five. We can then determine our KF, which is our products concentration divided by our starting material concentration. All held this document trick coefficients, which equals 1.1 times tense past 13 then are covered supers catatonic concentration is 1.4 times tends to minus 11. Our USP is C U two plus multiplied by O h minus which he goes 4.5 times 10 to minus 21 which is less than r. K s P value. Therefore, we do not have precipitation occur.


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