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A cube made of platinum (Pt) has an edge length of 1.0 $\mathrm{cm}$. (a) Calculate the number of $\mathrm{Pt}$ atoms in the cube. (b) Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all of the available space. If only 74 percent of the space inside the cube is taken up by $\mathrm{Pt}$ atoms, calculate the radius in picometers of a Pt atom. The density of Pt is $21.45 \mathrm{g} / \mathrm{cm}^{3}$ and the mass of a single $\mathrm{Pt}$ atom is $3.240 \times 10^{-22} \mathrm{g}$. [The volume of a sphere of radius $\left.r \text { is }(4 / 3) \pi r^{3} .\right]$

(a) $6.624 \cdot 10^{22}$(b) $1.387 \cdot 10^{2} \mathrm{pm}$

Chemistry 101

Chapter 2

Atoms, Molecules, and Ions

Atoms, Molecules and Ions

Carleton College

University of Maryland - University College

Brown University

Lectures

03:18

A cube made of platinum (P…

01:01

08:50

Suppose you are given a cu…

05:52

07:22

(a) You are given a cube o…

Hi there. In this question, we have a cube, and our cube has the dimensions one centimeter by one centimeter by one centimeter. Since it's a cube, the length, the width and the height they're all the same. And we want to know first of all, how maney platinum atoms we can put into that cube. After that, we then went to calculate the radius of a single platinum atom, Um, or trying to figure out how many atoms Aaron, this platinum cube or how maney Platinum Adams can fit this cube. We have been given some information about platinum. In this problem, we know its density, and we know the mass of a single atom. We can use that information to calculate how many atoms air in there. Let's see what I mean by that. First of all, I need to know the volume of this box we know volume, miss length, times with times height. Since this is a Q, all three of those dimensions are the same. It's one centimeter by one centimeter by one centimeter. Or, in other words, the box is 1.0, cubic centimeters. All right, I'm going to use that volume of the box and the information given in the problem. So we have 1.0 cubic centimeters. That's the volume that we have for the space we are given the density of platinum. The density of platinum is given is 21.45 g of platinum. Can you in every cubic centimeter? Look what that will do for us that will get us to grams of platinum because Centimeters Cube will cancel. We are also given the mass oven Adam of Platinum. That information is 3.240 times 10 to the negative, 22nd grams, so grams will cancel. So if I saw this now, I will have number of platinum Adams. That is gonna be 6.6 times 10 to the 22nd. I rounded to just two significant figures because our initial measurement of the box was only given into significant figures. All right, so this is our answer for part A. This is how many atoms are in that box in part B. We want to know the Radius and Adams are, of course fears, so we can use the equation that tells us volume of a sphere. But before that, we need to know the volume occupied by a single atom. In other words, I am trying to find the volume of one Adam. Well, I just calculated in part a. I knew I found out in part A That one cubic centimeter of space in that box is occupied by a total of 6.6 times 10 to the 22nd Adams. So if I divide by that, that will tell me how many atoms for space. But hold on. We know that Adams Air spherical. So there's empty space in there. In fact, we know that for every cubic centimeter of space on Lee, 0.74 cubic centimeters are taken up by the platinum atoms because they told us on Lee 74% So 0.74 of one is 74% tonight. So let's look at cancels here. Centimeters of space cancels Adams cancel and I am left with the volume per platinum atom of 1.12 times 10 to the negative 23rd centimeters cube. For every platinum atom, it's okay. I will round to the correct number significant figures when I get to the end of his calculation, because I want to keep those same two significant figures, but for now we will leave it as is he and we are ready to calculate the radius of a single platinum atom because volume is equal to four thirds. Hi are cute. That's the formula first fear. And we just calculated the volume of a single platinum atom so I can solve for radius. So let's rearrange this equation radius cubed. He's going to be equal to three times the volume divided by four times pi. It's a little more writing space here. All right, so let's plug in some values here three times that volume that I just calculated 1.12 times 10 to the negative. 23rd centimeters Cube divided by four times pi. This gives me a radius cube of 2.67 times 10 to the negative. 24th centimeters cube, right? But I just want our I don't want our cute, So I'm gonna take the cube root of both sides using my calculator. Gonna take Cuba route and I will get R is equal to 1.4 times 10 to the negative. Eight centimeters. Right, So that's the radius and centimeters but taking this one step further because the problem asked us for the answer in PICO meters. So in every centimeter there is one times 10 to the 10th p commuters. That means our answer is going to be 1.4 times 10 to the second Pekka meters as the radius of a single platinum atom. And that makes sense. It's 140 p commuters, which seems to be right in range with the size average size of atoms. So that is our answer for part B. 1.4 times 10 to the second Pekka meters would be the radius off a platinum atom. Thanks for watching.

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