Problem

A sample that weighs 107.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the sample?

Relevant Solution
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Welcome back everyone in this example, we're given Constantine which is a copper nickel alloy containing 55% copper and 45% nickel were given the density of the alloy as the seller here. And we're told that the sphere made from Constantine has a volume of 58.5 cubic centimeters. And to calculate the number of our copper and nickel atoms in our sphere. So beginning with our number of copper atoms, Because we recall that density is equal to mass divided by volume. We can recall that we can take the density times the volume to get our mass of our alloy and from the prompt were given the volume of the alloy as 58. cc. Where we're going to utilize our density given the prompt as a conversion factor to cancel out our cubic units cubic centimeter units of our alloy. So according to the prompt, we have a density of 8.885 g of our alloy for one cc of our alloy. And so this allows us to cancel out cubic centimeters of our alloy. And because we want our final unit to be Adam's of copper, we need to get rid of. Now this gram unit of our alloy and go into grams of copper. So we should recall that from our prompt. We are given percent by masses of each of our elements here. And for copper we have a percent of 55%. So we would assume This is out of 100 g of our alloy. And so we can use that as a conversion factor by saying in our denominator, we have 100 g of our alloy which is equal to 55 g of our copper. So this allows us to cancel out our units of our grams of alloy. And we want to get rid of now grams of copper to get into molds of copper where we can then end up with atoms of copper. So getting rid of grams of copper first, we want to recall from the periodic table are molar mass of copper, which gives us a mass of 63.5 g of copper for one mole of copper. And now we can go ahead and focus on converting from moles of copper. Two atoms of copper. So actually let's go ahead and take this conversion down below. So we'll continue on and say that we have in our denominator moles of copper where we can now move into atoms of copper. So canceling out our gram units and canceling out moles. Next, we would recall avocados number which tells us we have 6.22 times 10 to the 23rd power atoms of copper for one mole of copper. And so now we're left with just atoms of copper as our final unit. And what we're going to get for our value of our quotient above is a value of 2.71 times 10 to the 24th power copper atoms in our sphere. So this is going to be our first answer here and now. We can go ahead and figure out our number of our nickel atoms. And so we'll use a different color for this part of our solution. So our number of nickel Adams is going to be here. So following the same processes above, we're going to start out with that volume of our cube given from the prompt or sorry, our volume of our sphere Given from the prompt as 58.5 cc of our alloy. Where we would go ahead and utilize our density given in the prompt as 8.885g of our alloy per one cc of our alloy. And this allows us to again cancel out units of cubic centimeters of our alloy. Now moving from grams of our alloy, two g of our next adam nickel here we are going to recall from the prompt and sorry, we should be using purple. So from the prompt We're told that we have a percent by mass of 45% for nickel. So this is going to be assumed out of 100g of our alloy where we have 45 g of our nickel. And so this allows us to cancel out grams of our alloy. And now moving from grams of nickel to our next unit, which is going to be going from grams of nickel two moles of nickel. We're going to recall our molar mass from the periodic table for nickel, which is going to be 58.6934 g of nickel for one mole of nickel. And this allows us to now cancel out our units of grams of nickel. And now we're left with moving from moles of nickel to now Adams of nickel as our final unit. So we should recall avocados number which tells us we have 6.22 times 10 to the 23rd power nickel atoms for one mole of nickel, canceling out our moles of nickel. We are left with atoms of nickel and we're going to get a result here equal to. And this is going to give us a result of 2.40 times 10 to the 24th power nickels. And so this is going to be our number of nickel atoms in our sphere here. And so this would be the second part of our answer. So what's highlighted in yellow represents our final answers to complete this example. And this corresponds to choice be in our multiple choice. So I hope that everything that I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video