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Benecneedua -Pukc 6 087Wala shecl Part I: Dcterinfning (he Mass Percent = NMNH Aa; ol sour product lor Jnubee Uruli Volume of eruplc 07a9 Iriul > -Vitaatlon Vlunic uu HCI uelded Jiu" A HCI added Eleile" Aolnt alu tralion Concentration Handar MOu Actual mole, purct HCused LLAC "cutralize NH; [o Mus (mulc' Mc7 Tialat ~ A 40042 & ol 4= Nhou 'nsumed 6nu litration o # of molc of HC 0 ckmd neutralize NIL 000j0 oaenal DucrreflrdlA Actjze mult OHC ruttilli @Iils Calculat

Benecneedua -Pukc 6 087 Wala shecl Part I: Dcterinfning (he Mass Percent = NMNH Aa; ol sour product lor Jnubee Uruli Volume of eruplc 07a9 Iriul > -Vitaatlon Vlunic uu HCI uelded Jiu" A HCI added Eleile" Aolnt alu tralion Concentration Handar MOu Actual mole, purct HCused LLAC "cutralize NH; [o Mus (mulc' Mc7 Tialat ~ A 40042 & ol 4= Nhou 'nsumed 6nu litration o # of molc of HC 0 ckmd neutralize NIL 000j0 oaenal DucrreflrdlA Actjze mult OHC ruttilli @Iils Calculat molc uf WH LALLis Calculate the rass of NH; sunnit Wacd Losal Hfqou rcpeated lttatl Iwice,take mnaverape IASS Caculau TUD utc Culculuta Parf II: Determining the Mass Percent N7?" Ton [ AI(NILJCL) Auclchall Ma Tor Nicke ahanina Valunnieon HCI addcd: Abenrhincr sample druun culibrliu M Concenmatamak nlrenn alarantachlni DILM Kolnndl Her Vle ealnla Recall (hat WtTRW " #mole, Ni-S87 Chculate Ii: ol Nickel in Ile beaktt = Oii



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Moles within Moles and Molar Mass Part 1: How many hydrogen and oxygen atoms are present in 1 molecule of $\mathrm{H}_{2} \mathrm{O}$ ? How many moles of hydrogen and oxygen atoms are present in $1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}$ ? What are the masses of hydrogen and oxygen in $1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} ?$ What is the mass of $1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}$ ? Part 2: Two hypothetical ionic compounds are discovered with the chemical formulas $\mathrm{XCl}_{2}$ and $\mathrm{YCl}_{2}$, where $\mathrm{X}$ and Y represent symbols of the imaginary elements. Chemical analysis of the two compounds reveals that $0.25 \mathrm{~mol} \mathrm{XCl}_{2}$ has a mass of $100.0 \mathrm{~g}$ and $0.50 \mathrm{~mol} \mathrm{YCl}_{2}$ has a mass of $125.0 \mathrm{~g}$ What are the molar masses of $\mathrm{XCl}_{2}$ and $\mathrm{YCl}_{2}$ ? If you had 1.0 -mol samples of $\mathrm{XCl}_{2}$ and $\mathrm{YCl}_{2}$, how would the number of chloride ions compare? If you had 1.0 -mol samples of $\mathrm{XCl}_{2}$ and $\mathrm{YCl}_{2}$, how would the masses of elements $\mathrm{X}$ and $\mathrm{Y}$ compare? What is the mass of chloride ions present in 1.0 $\mathrm{mol} \mathrm{XCl}_{2}$ and $1.0 \mathrm{~mol} \mathrm{YCl}_{2} ?$ What are the molar masses of elements $\mathrm{X}$ and $\mathrm{Y} ?$ How many moles of $\mathrm{X}$ ions and chloride ions would be present in a 200.0 -g sample of $\mathrm{XCl}_{2}$ ? How many grams of $Y$ ions would be present in a $250.0-\mathrm{g}$ sample of $\mathrm{YCl}_{2} ?$ What would be the molar mass of the compound $\mathrm{YBr}_{3} ?$ $\operatorname{Part} 3:$ A minute sample of $\mathrm{AlCl}_{3}$ is analyzed for chlorine. The analysis reveals that there are 12 chloride ions present in the sample. How many aluminum ions must be present in the sample? What is the total mass of $\mathrm{AlCl}_{3}$ in this sample? How many moles of $\mathrm{AlCl}_{3}$ are in this sample?

The cancer drug Sis Platon can be made by reacting um NH 42 ptcl four with the most ammonia to produce this Platon and the by product of ammonium chloride. Let's write a balanced equation for this starting from any or two PTCL four plus two and h three p t and h 32 c l two And this is our sis Platon and our product of ammonium chloride to an h four. See also here's our balanced equation for part B. We have two questions were asked first of all, uh, to obtain 12.50 g of cisplatin What mass of NH 42 PTCL for is required. So we're gonna start with the 12.5 programs. It's just Platon Moeller mouths 300.5 g of these six Platon in one more and one more 21 mole and H 42 ptcl for on one more in a 42 PTCL four has a mass had 372 point 97 g and h 42 ptcl four. This would yield 15 point 15.54 g of N h 42 ptcl four this'll would be required to produce 12.50 g of CIS platinum. The second part to this question asks what volume of 0.1 to 5 Mueller NH three is required. So we're going to start with It's again 12.50 g assist Platon convert to Mel's 300 05 g So spartan in one more one more losses Klatten to two moles of any three when were given the morality of age three 0.1 to 5 moles of NH three per leader and this'll give us Yeah, 0.666 six Leaders, um, of 0.1 to 5 Mueller and age three required to produce 12.50 g of thesis platinum for part C were asked to um, cisplatin can react with period in to form a new compound rust to form. Find the formula of the unknown compound. So first of all, for part C, we're going to find the moles of Period Dean and were given the three Thai tradition data with HCL. So let's find the moles spiriting that reacts. So for part C, um, gonna start with 37. Yeah, let's Correcto leaders 37 mL is 370.37 Leaders Times and Liberty of HCL 475 Moles of HCL per leader on Do we know that there's 1 to 1 more ratio? One mole of HCL toe one more of C five h five n, and this would be equal to 1.76 times 10 to the negative two moles of C five h five n Let's calculate the um no, this would be the period e that's left over. Let's calculate the original amount of period Dean using its density and Mohler Mouse. So we have 150 mL times uh, 0.979 g c five h five n per millimeter times Moeller Mass. 79.10 g In one mole of C five h five n, this would work out to 1.86 times 10 to the negative two moles of C five h five n, and this is theater mount. That is the initial amounts. So to solve for the amount consumed uh, C five h five n consumed, too reacted. It would be the initial amounts minus the leftover amounts, and this would be equal to you 1.0 times 10 to the negative three moles of theme the C five h five n that actually gets reacted. So let's calculate the moles of cisplatin actually used. So we have, um sorry. Okay. And we have a 0.150 g of cisplatin. And let's convert this two moles 300.5 g of cisplatin in one more. This would work out to 5.0 times 10 to the negative four moles of this cisplatin. So let's determine the historic geometry here for the unknown compound. Um, by dividing the amount of period in by the amount of assists Platon so x here 5.0 r three 1.0 times 10 to the negative three moles divided by 5.0 times 10 to the negative four moles. And we get this to work out to be exactly two. So the formula for the unknown compounds looking at the ratio of the purity into the cisplatin. We see that, uh, for every cisplatin, there would be two period means so PT n h 32 c l to c five h five n and now ex we figured out is going to be to

Here again is one of those ridiculously long questions with three long parts, which again very easily could be separated into three or more questions. So let's get started. First part letter A. How many hydrogen and oxygen atoms are present in one molecule of H 20. This can be determined by the looking at the sub scripts. The two on the hydrogen means that there are two atoms of hydrogen And no subscript on oxygen means that there's one atom of oxygen in one molecule of H 20. These sub scripts can also be used for moles. If we have one mole of the compound, then we would have two moles of hydrogen and one mole of oxygen, Part C. What are the masses of hydrogen And oxygen in one mole of water? Well, if we have one mole of water, we have two moles of hydrogen we can multiply the two moles of hydrogen by the molar mass of hydrogen. In order to get the grams of hydrogen in one mole of water, 2.016 g. One mole of water will contain one mole of oxygen, which we can then multiplied by the molar mass of oxygen. To get the grams of oxygen in one mole of water, 16.00 g oxygen. So in part d, what is the mass of one mole of water? Well, it'll be the summed mass of the g of the elements in one mole of water, Or 18.016 g per mole of water. Now, for part two we have two imaginary compounds, X cl two and Y C L two. And they give us some information associated with the moles of each compound and the corresponding masses. To calculate the molar mass which has units of grams of the compound, per moles of the compound. We simply need to take the grams and the moles that were provided in this problem and divide them. So for XC two There's .25 moles for every 100 g. So for every 100 g there's .25 moles which gives us 400 g per mole of the compound. Now for YC. L two There's .5 moles for every 125 g or for every 125 g there's five moles giving us 250 g per mole has the molar mass then. And let her be. It asks if you had one mole samples of each of these compounds, how would the number of chloride ions compare? Well, this goes back to their chemical formulas. There are two C. Ls for every chemical formula. So if we have one mole of each of these compounds, we will have two moles of chloride ions. So they'll be the same. They'll both be two moles of chlorine. Let her see, asks if you had one mole of each of the compounds, how would the masses of elements X and Y compare? Well this is where we're going to have to go back to the molar masses. The amount of chlorine is the same in each case, But XC two has a greater Mueller mass. That greater Mueller mass must be coming from the greater Mueller massive X. And why Cl two has a smaller Mueller mass? Because the amount of Cl two is the same, then that smaller Mueller mass must be coming from a smaller Mueller mass of Why? So, because X cl two has a greater Mueller mass than why Cl two Than one mole of each of them would have a greater massive ex than a mass of why letter D. What is the mass of chloride ions present in one mole of each of the compounds? Well, as I mentioned in one mole of each of the compounds, there will be two moles of chlorine. So all we need to do is convert the two moles of chlorine into mass chlorine, recognizing that one mole of chlorine is 35.453g. This value coming from the periodic table Given us 79 1 g chlorine letter E asks, what are the molar masses of elements X and Y. Well, this is where we're going to go back to the molar masses for each of the compounds. The Molar Mass for XC two is 400. So we'll take the total molar mass minus the mass that corresponds to the chlorine in the compound which we just calculated and the rest will be the molar mass of X 3 29.1 g from all. We'll do the same thing with Y C L. Twice the L two. The molar mass of Y. C. L two is 2 50. The mass of to see L's is 70.91. Subtracting these two will give us the molar mass of why 1 79.1 g per mole for letter F. How many moles of X. Science and chloride ions would be present in 200 g of sample X. L. Two. Well Remember Excel two has a molar mass of 400. So if we have 400 g we have one mole. So if we have 200 g we're going to have half of them all. If we have half of them all of X. L. Two, then that means we have half of them all of X and we will have half of two or one mole of Y. So how many moles of X and C. L. Will we have will have a half a mile of X and one mole of cl letter G. Asks how many grams of why ions would be present in 250 g of the sample. Well, remember the molar mass of Y. C. L two is 2 50. So if we have 250 g we will have one mole of cl And we will have to, I'm sorry, one mole of Y and two moles of cl. If we have one mole of why we will have the molar mass of why? 179.1 g. So that will be the mass of why that we have letter H. What would be the molar mass of the compound? Why? b. r. 3? Well, we already know the molar mass of why it's this right here. So we would take the molar mass of Y plus three times the Mueller massive BR because we would have three moles of BR for every one mole of Y. B. R. Three Giving us 418 8 g per mole. Now for part three minute sample of A. L. C. L. Three is analyzed with chlorine. The analysis rebuilds 12 chloride ions. So how many aluminum ions would we have? Well, from the chemical formula we know that there are three chloride ions for everyone, aluminum ion. So 12 chloride ions would correspond to four aluminum ions. Thus we would have four Formula Units of L. c. l. three. Part A. They snuck in an extra question here, part A what is the total mass of L. C. L three in the sample? Well, if we have four aluminum ions and there's one aluminum ion for every aluminum chloride formula unit, then we'll have four aluminum chloride formula units. We can cover it the number of formula units into molds by dividing by avocados number. Then when we have the moles of aluminum chloride, we can multiply it by the molar mass of aluminum chloride in order to get the total mass of the sample 8.86 times 10 to the -2 g. And for part B, they probably should have asked part B before Part A how many moles of aluminum chloride or in the sample? Well, it's the exact same calculation. We're just going to stop right here with the moles. We won't multiply it by the molar mass and we get 664 times 10 to the negative 24 moles of aluminum chloride.

For this problem. We're gonna try to find the math percent of 10 it Good. Now I can compare the mass of this to the lands of the sample and get a present. So is Jerry Mouse over my was given turns 100 Mascarenhas Your 1000.27442 The last game was your avoid 5 300% of 51.

We are provided with Delia Dogs die, Trey, Shin, hear oxidation and reduction half can be presented out. Oxidation will be essence toe s and too positive. Plus two electron reduction will be n 03 Negative. Bless four height origin. Last three electron gives me and no, let's to edge too. This reaction for the Thai tradition can be summed up as three s n plus toe. You know, three negative less e th for the duke gives me three ascend to positive plus two and no plus four water. From the above chemical reaction we have three more off sn will be equal to two more off n 03 negative. The equal villains point is open that point 0344 litre. We calculate the number off moles off sm present. Here is the simple calculation. We have the mullahs Marceau fasten That is 118.71 from which we can determine the mask off sn using unitary method. So the marks off a scent turned out to be two point 74 months. Pretenders toe negative one ground which is equally villain to 51.3% off. December


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