CHE 107 Lab Homework 2: Understanding Density and Proportional Reasoning - Prof. Martin Br | Assignments Chemistry

CHE 107 Chemistry Lab homework 2

CHE 107

LAB HOMEWORK 2

DENSITY

Developing quantitative reasoning skills

Most of the material below is deriv ed from McDermott, et al., Physics by Inquiry, vol. I (199 6), Wiley

This activity is an introduction to quantitative reasoning. Every arithmetic step taken with numbers in

solving a problem must have a justification. The reasoning that goes into a calculation is as important a

part of science as its laws. Becaus e mass and volume are proportional, the reasoning used in this section is

called proportiona l reasoning. Read through the materials below, and then answer th e 8 questions on the

final page. This will be due next w eek at the third lab.

Sample problem

Suppose we have a stone with a mas s or 15 g and a volume of 5 cm3. What is the mass of 1 cm3 of

this stone?

Sample solution

At right is a diagram of the stone. It is not a picture of the stone. It is not supposed to look like the

stone; it is just something to help us think about the problem. The actual stone might be any shape.

In the diagram, the stone is cut into five equal pieces, each having a volume of 1 cm3. If all these 1

cm3 pieces share the mass equally, then we can find the mass of 1 cm3 of stone by splitting up the total

mass into 5 equal shares. The way to split up the total mass (15 g) into 5 equal shares using

arithmetic is to divide (15 g)/5 = 3 g . Since each 1 cm3 has just one of these shares, th e mass of

each cubic centime ter of stone is 3 g .

Comment

This problem illustrates one of the reasons division is used in calcu lations. We know that the mass of 5

cm3 is 15 g. We want to find the mass of just 1 cm3. To do this, we divid e 15/5. When we want to find ou t

how much of one quantity goes with just one unit of a second quantity, we divide. Th e result of division in

this kind of problem tells us the amount of the quantity in the numerator (the numb er on top) for each unit

of the quantity in the denominator (the number on the bottom).

The mass of one unit of volume of a substance is an important quantity. Calculations we do with

mass and volume often can be reasoned through by thinking about breaking objects into 1-cm3 packag es.

For the stone above , the package has a volume of 1 cm3 and a mass of 3 g. The following problems show

how to reason with these packages.

Sample problem

Suppose we have a larger piece of the same kind of stone with a volu me of 120 cm3. What would

its mass be?

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CHE 107

LAB HOMEWORK 2

DENSITY

Developing quantitative reasoning skills Most of the material below is derived from McDermott, et al., Physics by Inquiry, vol. I (1996), Wiley This activity is an introduction to quantitative reasoning. Every arithmetic step taken with numbers in solving a problem must have a justification. The reasoning that goes into a calculation is as important a part of science as its laws. Because mass and volume are proportional, the reasoning used in this section is called proportional reasoning. Read through the materials below, and then answer the 8 questions on the final page. This will be due next week at the third lab. Sample problem Suppose we have a stone with a mass or 15 g and a volume of 5 cm^3. What is the mass of 1 cm^3 of this stone? Sample solution At right is a diagram of the stone. It is not a picture of the stone. It is not supposed to look like the stone; it is just something to help us think about the problem. The actual stone might be any shape. In the diagram, the stone is cut into five equal pieces, each having a volume of 1 cm^3. If all these 1 cm^3 pieces share the mass equally, then we can find the mass of 1 cm^3 of stone by splitting up the total mass into 5 equal shares. The way to split up the total mass (15 g) into 5 equal shares using arithmetic is to divide (15 g)/5 = 3 g. Since each 1 cm^3 has just one of these shares, the mass of each cubic centimeter of stone is 3 g. Comment This problem illustrates one of the reasons division is used in calculations. We know that the mass of 5 cm^3 is 15 g. We want to find the mass of just 1 cm^3. To do this, we divide 15/5. When we want to find out how much of one quantity goes with just one unit of a second quantity, we divide. The result of division in this kind of problem tells us the amount of the quantity in the numerator (the number on top) for each unit of the quantity in the denominator (the number on the bottom). The mass of one unit of volume of a substance is an important quantity. Calculations we do with mass and volume often can be reasoned through by thinking about breaking objects into 1-cm^3 packages. For the stone above, the package has a volume of 1 cm^3 and a mass of 3 g. The following problems show how to reason with these packages. Sample problem Suppose we have a larger piece of the same kind of stone with a volume of 120 cm^3. What would its mass be?

Sample solution The problem says “with a volume of 120 cm^3 .” Using the operational definition of volume, we can translate this phrase to mean: “there are 120 one cubic centimeter packages of stone.” Now, we know that 1 cm^3 of this stone has a mass of 3 g, so all we have to do to find the mass of this large stone is add up 3 g 120 times – once for each 1 cm^3 package. A fast way to do this addition is to multiply 120 x 3 g, which gives 360 g. Comment This solution illustrates the reasoning that leads to the use of multiplication in a problem. We use multiplication here because we want to add the same number repeatedly. Sample problem What is the volume of 180 g of the same kind of stone? Sample solution In this problem we are asked to find the volume. First, we translate this request into: "Find the number of 1 cm^3 packages of stone it takes to have a mass of 180 g." We again use our knowledge that the mass of each 1 cm^3 package of stone is 3 g. It can help to draw a diagram of the 180 g piece of stone, shown cut up into 1 cm^3 , or 3 g packages. The problem is to find out how many 1 cm^3 , 3 g packages there are in 180 g. We can calculate how many 3 g packages there are in 180 g by dividing: 180/3 = 60. Thus, there are 60 of these 3 g packages in 180 g, and each one of them has a volume of 1 cm^3. Therefore, the volume (number of 1 cm^3 packages) of the 180 g of stone is 60 cm^3. Comment Here we see another, different justification for dividing. We are not trying to split up 180 into three equal shares. Instead, we want to find out how many threes there are in 180. These two quite different justifications for division are both used frequently. When explaining the reason for dividing, we must choose between these two interpretations of division (and possibly others). Usually, only one will make sense. In this course, detailed explanations of reasoning are often required. It is most important to explain why to divide or multiply. The key parts of such explanations are given below. Multiplication An explanation of why to multiply should contain something equivalent to the following: We need to add up 120 of these 3 g packages. A quick way to do this is to multiply 120 x 3. Division At this stage, we have encountered two explanations of why to divide. One is: 5 cm^3 has a mass of 15 g. We can find the mass of just 1 cm^3 by dividing 15/

Comments This is not a difficult problem. The only trouble might arise from a misunderstanding of what counts as an interpretation. For example, the following two responses are not acceptable interpretations of the number 18/25. Inadequate response 1: “18 gm is the mass of the block and 25 cm^3 is the volume.” Although this is certainly correct, it is an interpretation of two numbers , 18 and 25. The problem asks for the interpretation of 18/25. 18/25 is one number : 0.72. Inadequate response 2: “18/25 is the density of the wood.” Again this is true, but it only tells the name of the number. It is not an interpretation of what the number means. Not all calculated numbers have clear meaning. For example, aluminum has a density of 2. g/cm^3. Suppose we have a 5.4 cm^3 piece of aluminum. Consider the number obtained by dividing the second quantity by the first, 5.4/2.7, in other words, 2. It is difficult to imagine any reason for doing this calculation, and indeed the number has no simple physical interpretation in this instance.

Name Date

These homework problems are due at the time of the third lab assignment. Separate this homework page from the instructional materials above, and turn it alone as a single sheet.

1. Explain every step in your reasoning for the following problem. A piece of clay has a mass of 14.1 g and a volume of 12.2 cm^3. A. What is the mass of 1 cm^3 of this clay? Draw a diagram that illustrates the thinking involved in the calculation. B. What is the mass of 15.2 cm^3 of the same type of clay? C. Suppose we have a 68.8 g piece of the same clay. By how much would the mass increase if we added a lump of clay with a volume of 3 cm^3? 2. In this problem, the calculating has been done for you. The questions ask about the reasoning involved at each step. Problem : 15.0 cm^3 of iron has a mass of 118 g. What is the volume of a piece of iron that has a mass of 413 g? 118/15 = 7.87 (step 1) 413/7.87 = 52.5 (step 2) A. How do you interpret the number 7.87? B. Use a diagram to explain why the division is done in step 1. C. How do you interpret the number 52.5? D. Use a diagram to explain why the division is done in step 2. In the following problems, you are not expected to do the math, but to provide an interpretation for the kinds of math suggested. 3. A piece of plastic has a mass of 650 g and a volume of 720 cm^3. Give an interpretation, if there is one, of each of the following numbers. A. 720/650 B. 650/720 C. 720 x 650 D. 720 + 650 4. Suppose we have a piece of iron with density 7.87 g/cm^3 and volume 6.25 cm^3. Give an interpretation, if there is one, of each of the following numbers. A. 7.87/6.25 B. 6.25/7.87 C. 7.87 x 6.25 D. 6.25 x 7. 5. Suppose we have a piece of tin, with a density of 5.75 g/cm^3 and mass 8.31 g. Give an interpretation, if there is one, of each of the following numbers. A. 8.31/5.75 B. 5.75/8.31 C. 8.31 x 5. 6. Each pound of bananas costs 30 cents. We bought 4.3 pounds of bananas. Give an interpretation, if there is one, of each of the following numbers. A. 4.3/30 B. 30/4.3 C. 4.3 x 30 D. 4.3 + 30

CHE 107 Lab Homework 2: Understanding Density and Proportional Reasoning - Prof. Martin Br, Assignments of Chemistry

Partial preview of the text

Download CHE 107 Lab Homework 2: Understanding Density and Proportional Reasoning - Prof. Martin Br and more Assignments Chemistry in PDF only on Docsity!

CHE 107

LAB HOMEWORK 2

DENSITY

Name Date