The diagrams different below size and represent three shape: Arrange samples of the the letters same substance cach hving 4 from largest of the samples to show the ...

Use the graph lines in Figure P 10.8 to determine the densities of the three liquids in SI units. If you place them in one container, how will they position themselves? How does the density of each liquid change as its volume increases? As its mass decreases? Compare the masses of the three liquids when they occupy the same volume. Compare the volumes of the three liquids when they have the same mass.

Solving problems. 68. I'm chapter toe. So we have four diagrams. Um, 123 and four. And we need to match, um, each of them with one off the following sentences. So let's start reading the first sentence A and let's try to understand to which diagram can be matched. So sentence a, say the cube US A. Greater density than water so means cube, then city greater than water density. So if the cube that is emerging water as an iron density, for sure, it has to be under the level of water. So the only diagram where the cube is completely submerged in water and actually is at the bottom is diagram tree. So we match sentence A we diagram tree, so sentence be on the cube as a density. The tease between 0.60 and 0.80 gram over l. So cube cube density is between 0.16 and 0.18 drum over ml. So in order to solve this, we need toe remind the water density equal one over Emmel. So it means that the cube a za lower density than the water, but it's pretty close to it, so if we take a look at the diagram that are left that are want toe and four, um, it cannot be one because it looks like that the desert of the cube is exactly off. Um, as the density of the water, in fact, exactly. Off off the Q b submerge and the other after is is out of the water is floating on the water. Um, you cannot be too, because basically, the cube is behaving like if he was water. So it means in this case, the two days it is the same. We read the sentry. So, um, the matching for sentence to is for In fact, the cube is not at the bottom because because the density is lower compared to the water. And ah, but, um, is dense enough that more than half off the cube is emerging water. So sentence be you match with diagram for see, the cube has a density that these one off the density of the water. So, as we said in this case, we can see that off exactly. Off of the cube is on top of the water is floating. The other off is, um, some merging water So in this case, the Cuba PSA density that is off the density of the water. So we manage diagram. See? So it is right now. Ncube, then city. He's one off cold water density. And we mentioned with one. So now only the is left were cube density, equal water than city. And this to you. Fact, if you if you see the cube is no completely submerge but is no even floating, Um, about whether in any fraction. And this is because the cube and the water in this case, they have the same thing.

In question 89. We have four diagrams showing a cube placed in water. Remember water has a density of 1.0 grams per centimeter cubed or grams per mil leader if something is less dense than water, then it will float. If it is more dense than water, it will sink. If it is the same density as water, then it will be completely submerged but remains suspended and not sink to the bottom. So part a ask which one has a greater density than water? That would be the one that sinks number three. Which one has a density? That is about 10.8 g per mil leader or 80% that of water? Well, if it's 80% that of water, then it's going to be about 80% submerged, which would be number four. If the density of the object is half that of water, then it's going to be about half submerged, which would be number one. And then if the density is the same as that of water, it'll be completely submerged but not sink, and that will be number two.

Question 80 shows a few pictures and then ask some questions associated with the properties of gas properties such as density, pressure, temperature so forth. So let's first reproduce these figures. A Has seven molecules that air die Atomic B has four molecules that air die. Atomic C has four molecules that are have three atoms Tri Atomic and then D has seven or seven atoms that air present. The first question asks, Are these gas is at the same temperature and pressure? Well, there's really no information that says anything about the temperature or the pressure. However, if they are all at the same temperature, we could say that A and D, which have the same number of species seven, would be at the same pressure and C and B, which have the same number of species. Four would be at the same pressure, and we could also state that A and D that has higher moles of gas would have a higher pressure than C and D. The next question asks if the Mueller Mass of B is 38 and the gas and that gas of see it is 46 than which gas is more dense well, this one's fairly straightforward. If you remember the density equation for a gas that was derived from the ideal gas law, density is in direct relationship with the Mueller mass of the gas. So assuming this equation works thus as Mueller Mass increases, density increases. So gas see that has a larger Mueller Mass would have a greater density for see it says to make the densities of gas is B and C equal. Now that we know that they're different, what should Which one should expand well, We need to expand the more dense gas, expand the volume of the more dense gas in order to have its density decreased to equal that of the less dense gas thistles. Because density is equal to mass over volume and as volume increases, density decreases for the last one, it says. If the densities of gasses and A and C or equal what is the relationship between their Mueller masses? While the relationship between their Mueller masses would be found with the relationship between the number of species that air present, there's four and one and seven in another, so gas a so in that ratio is 0.571 So gas a then is going to have a Mueller Mass. That is 0.571 times that of C in order for the density of both of these gasses to be equal.

Okay, so this question wants you to calculate areas and densities using these given values. So you know, that area is equal to a link Times wit. So you're given a rectangle that has sides measuring three times 10 to the power once and meters and the width of three times 10 to the power of negative juice. Centimeters. Sorry. There should be a carrot here. So because you know the area's equal length times with, we have to multiply these two values together. So you have three times 10 to the first power of one centimeters times three times 10 to the minus two centimeters. So remember, when you're multiplying together numbers that are in scientific notation, you want to multiply, you're coefficients and then you're going thio, add together your ex opponents, which is here, in here. So if you multiply three times three so three times three is equal to nine. So this is your coefficient. And for your exponents, you have one plus negative too, which then equals negative one. So when you combine these together, your answer is equal to nine times 10 to the negative one centimeters clean for be again. You know the area is you goto length times with. So now Because again, both of our units are the same. So now we can just multiply these two numbers. So you have one times 10 to the third centimeters times five times 10 to the negative one centimeters. And so this for your coefficient, you multiply them together. So you multiply one times five, which equals five. And for your exponents, you add them together. So I have three What negative one, which equals two. And so now again, you just combine everything So your final answer with your coefficient is five. So it would be five times 10 to the power of two centimeters, actually centimeter square, because now you're multiplying centimeters and centimeters to get centimeters squared. And so again, over here with your unit's centimeters and centimeters, Final answer should be in centimeters squared. So it's really important to pay attention to you're, um, the value of your units. So part C wants you to now calculate didn't see. So if you remember, density is going to be equal to mass over volume. So you're giving a mass of nine times 10 to the power of five grams and volume of three times 10 to the negative one power of centimeters cubed. So in order to calculate density, you're going to have to take your mass and divided by your volume. So nine times 10 to the power of my grams divided by three times 10 the negative one centimeters. Cute. So unlike multiplication, when you're dividing, you're going to divide your coefficients. So in this case, you do nine divided by three, which would be equal to three. And you're going to subtract your exponents so you have five minus negative. One would be equal to six. And when you combine everything together, you get a final answer of three times 10 to the six and we're dividing our units so grams per centimeters cubed. And for the final problem again, you want to calculate the density. So you know density. You know that density is equal to your mass over your volume, so your mass. In this case, the density is equal to four times 10 to the power of negative three grams, and your volume is two times 10 to the negative, too. Sorry, that should be. And negative too. Centimeters cute. And so for your coefficients you're dividing so you would do for divided by two is equal to And you would subtract your exponents So you do. Negative three minus negative too would be equal to negative one. So your density is equal Thio two times 10 to the power of negative one grams per centimeters deep and that would be your final answer.