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.