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DATA TABLEMass of cnicible (g)15.772Mass of crucible and hydrated sumple (E)16.872Mass of hydrated smple (A)Mass of crucible and dehydrated samnple (E)16.642Mass of...

Question

DATA TABLEMass of cnicible (g)15.772Mass of crucible and hydrated sumple (E)16.872Mass of hydrated smple (A)Mass of crucible and dehydrated samnple (E)16.642Mass of dehydrated sample (g)Mass of water evolved (g)Mass of empty watch glass (g)8.234Mass of watch glass and copper (g)8.643Mass of copper (g)DATA ANALYSIS How many moles of water were in your sample of copper chloride hydrate?How many moles of copper were in your sample of copper chloride?How many moles of chlorine were in your sample of

DATA TABLE Mass of cnicible (g) 15.772 Mass of crucible and hydrated sumple (E) 16.872 Mass of hydrated smple (A) Mass of crucible and dehydrated samnple (E) 16.642 Mass of dehydrated sample (g) Mass of water evolved (g) Mass of empty watch glass (g) 8.234 Mass of watch glass and copper (g) 8.643 Mass of copper (g) DATA ANALYSIS How many moles of water were in your sample of copper chloride hydrate? How many moles of copper were in your sample of copper chloride? How many moles of chlorine were in your sample of copper chloride? Write the proper chemical formula for the compound that you tested:



Answers

A hydrate of copper(II) chloride has the following formula: $\mathrm{CuCl}_{2} \cdot x \mathrm{H}_{2} \mathrm{O}$. The water in a $3.41 \mathrm{~g}$ sample of the hydrate was driven off by heating. The remaining sample had a mass of $2.69 \mathrm{~g}$. Find the number of waters of hydration $(x)$ in the hydrate.

Is there a problem? 1 34 of chapter three in chemistry Milica approach. Ah, hydrate. You know Hydra of copper tube chloride with formula. See, You see, l two of that X H 20 It has the water, the water and 343.41 gram sample Example was of hydrates given off by a sample Has a the many samples of mass of 2.69 grams. So that means that C c you seal to that has 2.69 grams Because the water's gone, that's the only thing is love. Find number of waters of the high tree borders of hydration in the hydrate. So this pretty sure forward to find number. I kind of mastered the water We take 3 40 3.41 minus 2.69 And that gives us That's the mass of our water. Nine. This is 0.72 grams names of water and then we're going to find the empirical form us. Now let's take a 2.69 We're gonna Cooper convert them both into moles. So one more over 18 grams and then one more. There There's one copper into chlorine. So copper has a molecular weight of 63.5 by post two time 35.5 that's 134.5 grams. That's gonna equal 0.0 for moans, and this one's gonna equal 0.2 moles. So now we divide both by 0.2 there's gonna equal to and there's gonna equal one. So that means that we have to water molecules forever. You see, You see, L too. So that means that X is equal to two. And our formula is See you seal to to H 20 and we're done.

Good day. We know that a hydrate is a compound that has salt containing a certain amount of water in it. So in this case are hydrate is given to have a formula. So that's just write it first here that this is it is a salt with water embedded in it. So in this case our hydrate is given with the formula of CCL to a copper to current with X here being the amount of water that is embedded in it. Okay, for that is embedded per formula unit of the compound. So we are given that the mass of the hydrate is equal to 3.41. So this is the mass of this corporate to chloride hydrate. And uh it's given here that after heating the mass of the anhydrous high and hydro's or this is the mass of the hydrate after water molecule is driven off. Yeah. And it's given to be 2.69. So note that when hydrate is heated, its water molecules Uh gets evaporated. And so what is left is only the the ionic compound or assault. So in this case these 2.69 g Also is equal to g of CUCL two. So it follows that we can solve for the mass of water in the hybrid, which is just going to be equal to the mass of the hydrate. Less the mass of the it's closed. Yeah. And this gives us zero coins 72. We can convert this into most of water. It's going to be our step to convert this to most of water by dividing the mass By the molar mass of water, which is equal to 18. And that gives us 0.72 over 18. Where that's going to be equal to zero point 04. We can also uh let us also solve for the most of the Salt Sea UCL two which is g of CUCL two over its own molar mass. So we have to see her two points 69 right over 134 has its own water mess. So you will get 0.02. And at the state down the ratio of the most of um water And the most of CUCL two. So please. So this ignoring this. So we have most of uh H 20. Again divided by the most of cl cl 2/0 200.2. And this is give, this gives us to over one. So what does this ratio mean? This means that in the in the hydrate for every one mole of copper to chloride, we have two months of water. Therefore, the formula for the hydrate will be ceo cl 22 H 20. So that means the X. Value that we have presented above is going to be equal to two. So I hope that this is clear

Let's forget the problem from the chapter chemical composition. This problem is based on finding the masses of the substance using their moles. Here we are given some samples along with their moles. We will find out the Martin Graham using their moles as well as the molar masses. Martin Graham, from a substitute for a substance from their mold, can be calculated by multiplying the moles of substance with Mueller Mass of substance. So let's first of all right, the more masses of all the substances given for carbon dioxide it is 44 grams per mole for night wasn trickle right the masses 120.5 g per mole for ammonium nitrate, the molar masses, 80 g per mole for water. The more masses 18 g per mole and for copper sulfite that is copper two sulphate. The more mass is 159.5 g per mole. Now let's calculate each one by one first of all, 1.27 million more of carbon dioxide that is 1.27 times 10 days to power minus three MOL multiply 44 g per mole give the result as 0.56 g carbon dioxide, then four point 12 times tenders power three. Mol multiply 120.5 g per mole. Give the value of Mars as 4.96 times tends to power. 5 g My chosen Try chloride, that is N C. L three. Then we have a money. My trade 0.451 mould Multiply 80 g per mole. Gives the value as 0.3 6 g ammonium nitrate. Then we have 18 grandma, 18 more of water, so 18 mole Multiply 18 g per mole. Give the mass of water as 324 g, then copper two sulphate 62.7 mol multiply 159.5 g per mole Gives the mass of copper sulfate as 10 days to power. 4 g. These are the masses calculated for the substances from the Earth given number of molds using their molar masses

Let's solve the problem from the chapter chemical composition. This problem is based on finding the masses of the substances using their molds and the molar masses. Here we are given molds of some substances using the value of molds and the moral masses of substance. We will calculate the mass in grams mask, and Graham can be concluded by multiplying the moles of substance with molar mass of the substance. For this, first of all, we will right there molar masses of all the substances for deportation nitrate. The Marama says 101 g per mole for ethanol. It is 46 grand per bowl for calcium oxide. The molar masses 56 g per mole for gold. Three bro might. The molar mass is 437 g per mole for water. The more masses 18 g per mole and for silver nitrate, the Bahamas is 170 gram per moon. Now we will calculate massing Graham Mass deportation nitrate and Graham will be equal to 2.41 multiplied industry power minus three MOL multiply 101 g per mole, and the mass value comes as point 24 3 g for ethanol, we can write 8.91 mole times 46 g per mole and on solving the value comes out as 400 9.86 g. Then we have for calcium oxide at this 0.141 mall, multiplied by 56 g per mole on calculating. The value comes as 0.7896 g calcium oxide, then for gold three and provide. It has 1.91 mole times 4 37 g per mole and the value in solving comes as 834.67 g gold, three barometer. Then we have water at this 0.117 mall multiply 18 grand per mole on solving the value comes as 2.1 nod six times 10 days to power minus four grand water, Then the lasted silver nitrate. Here it goes like 2.68 moles times 1 70 g per mole, which on solving gives the value is 455.6 grams. And when I train So these are the masses of the substances calculated using their mobile masses and a given number of moles in the problem


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