CH 121 Worksheet 1 Key PDF

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Chemistry 121 Oregon State University Worksheet 1 – Notes Dr. Barth 1.

In a reaction, a total mass of 78.1 g of reactants are consumed. If there are two products, and 48.3 g of one is produced, how much of the other product is produced? The Law of Conservation of Matter states that in a chemical reaction, matter is neither created nor destroyed. This means that the total mass of the products has to be equal to the total mass of the reactants. For this reaction, that means that we have: 78.1 g = 48.3 g + mass of reactant 2 Solving for reactant 2 gives us: mass of reactant 2 = 78.1 g - 48.3 g = 29.8 g

2.

Identify an element, and identify a compound that contains that element. There are many possible answers here - a few examples: - Carbon (C) is an element that is in the compound carbon dioxide: CO2 - Oxygen (O) is also an element in the compound carbon dioxide - Phosphorus (P) is an element in the compound phosphorous pentachloride: PCl5 Please note that diatomic elements (such as O2 and Cl2) are molecules but are not compounds, as they only contain one type of atom.

3.

Identify each of the following as a physical or chemical change: a. ice melting - physical change. This involves a change in how water molecules interact with each other, but there are no bonds between atoms broken or formed; it’s still water molecules b. alcohol evaporating - like before, this is a physical change. Any time we have a phase change (like melting or evaporation), it’s a physical change. c. iron rusting - chemical change. This involves breaking and forming of bonds between atoms because solid iron and rust (iron oxide) are chemically different species. d. glass breaking - physical change

e. wood burning - chemical change - the compounds in the wood are combusting in oxygen to yield other compounds 4.

Express the following in scientific notation a. 357,462 3.57462 x 105 b. 0.035

3.5 x 10-2

c. 0.00000972

9.72 x 10-6

d. 12.8 1.28 x 101 - an important thing to note here is that we can express any of these numbers in scientific notation. A number like 12.8, while you should know how to put it into scientific notation, is more practically written as is, so for the purposes of these class we will mostly be writing numbers however it’s more practical to do so (whether it is decimal or scientific notation form will depend on the magnitude of the number) 7.33 x 102

e. 733

5.

Convert the following: a. 0.459 m to km For this, we’ll use the relationship 1 km = 1000 m: 0.459 m •

1 km 1000 m

= 4.59 x 10-4 km

b. 8.32 x 10-6 g to ng The relationship we need here is 1 g = 1 x 109 ng: 9

8.32 x 10-6 g •

1 x 10 ng 1g

= 8.32 x 103 ng

Please note that this will work the same way if you use 1 ng = 1 x 10-9 g for your ratio c. 642 cL to L For this, we’ll use the relationship 1 L = 100 cL: 642 cL •

1L 100 cL

= 6.42 L

d. 4.813 x 104 Mg to mg For questions where we’re not converting from or two the base unit, I find it best to convert by way of the base unit. The relationships we need are: 1 g = 1 x 10-6 Mg and 1g = 1 x 103 mg 4.813 x 104 Mg •

3

1g 1 x 10

-6

Mg

•

1x10 mg 1g

= 4.81 x 1013 mg

It is also important to recognize here that in this context “Mg” is mega grams, and not the symbol for magnesium. e. 8.2 µ m to mm The relationships we need are: 1 m = 1 x 106 µm and 1 m = 1 x 103 mm 8.2 µm •

1m 6

1 x 10 µm

3

•

1x10 mm 1m

= 8.2 x 10-3 mm

Again, it’s valid here to go directly between µm and mm, but I find that I make fewer errors when I go through the base unit. 6.

a. 74.7 mL of a compound has a mass of 837.0 g. What is the density of this compound in g/mL? Density is defined as:

mass volume

in this case, the units of mass will be g and the units of volume will be mL (as given in the question. So we can plug in the g and mL given: 837.0 g 74.7 mL

= 11.2 g/mL

We’ll cover significant figures in a few questions, but for this one, we follow the multiplication rules, and the 74.7 mL term has the least number of significant figures with 3, so we report our answer with three significant figures. 7.

In each of the following, identify which of the digits are significant: For each, the digits that are significant are underlined. Explanations are added as necessary.

8.

a.

18.6

All non-zero digits are significant, so in this case we have three significant figures

b.

18600

We know that the three non-zero digits are significant. The two trailing zeroes are ambiguous - they’re before the decimal, but there’s no indication that they’re significant. With our book, we will only treat the three non-zero digits as significant. A great way to clarify which of the digits are meant to be significant would be to express this in scientific notation: as 1.86 x 104, 1.860 x 104, or 1.8600 x 104, depending on which (if any) of the zeroes are significant.

c.

18600.

In this case, the decimal after the trailing zeroes clarifies that they’re significant, so all five digits are significant.

d.

4.9203 x 10-6 For this number, all of the non-zero digits are significant. Additionally, the only zero is internal - it’s between two nonzero digits - so it is significant as well.

e.

0.0004020

Leading zeroes are not significant. Trailing zeroes that are after the decimal are considered significant, however, as is the internal zero. We have a total of four significant digits in this number.

f.

350.0

We have two non-zero digits here (which we know are significant. We have a zero before the decimal; the presence of the decimal tells us that it’s significant. Also, we have a zero after the decimal; trailing zeroes after the decimal are considered significant. This means all four digits shown are significant.

For each of the following calculations, express the result with the correct number of significant figures: a. 99.1 + 4.38483 = The rules for addition/subtraction state that our answer can’t have more precision than our least precise term. We determine that by what decimal place the term goes to. 99.1 goes to one digit past the decimal, so our final answer needs to go to one digit past the decimal. Sometimes this means we gain a digit. In this case: 99.1 + 4.38483 = 103.5

That answer has four significant figures, even though 99.1 only has three, but it’s correct, as it goes one digit past the decimal. b.

84.07 - 82.627 =

We apply the addition/subtraction rules here. 84.07 is the least precise term and goes two digits past the decimal, meaning that our answer will go two places past the decimal. With subtraction, this means that we could lose digits. In this case: 84.07 - 82.627 = 1.44 Our answer goes two digits past the decimal but only has three digits (where the least precise term has four) because we no longer have a digit in the tens place. c.

4.27 x 10-4 * 5.6 =

The multiplication rules state that our final answer will have the same number of digits as the term with the fewest number of significant digits. In this case, 5.6 has two significant digits. That gives: 4.27 x 10-4 * 5.6 = 2.4 x 10-3 d.

0.963 / 1.8976 =

With our multiplication/division rules, we’ll use the term with the fewest number of significant digits. 0.963 has three significant digits, so our answer will as well. 0.963 / 1.8976 = 0.507 e.

(7.88 + 0.8) * 3.1554 =

When we have multiple steps in a calculation, we follow the rules at each step, and track the digits that are significant in each intermediate answer. We don’t, however, round before our final answer, as that can lead to calculation errors. Our first step is an addition, and since 0.8 is the least precise, our intermediate answer will be significant to the tenths place (one digit past the decimal). We don’t round after the first step, though, so we’ll underline the last significant digit in that answer. 7.88 + 0.8 = 8.68

Our next step is a multiplication, so we use the number of digits from the term with the fewest number of significant digits. That’s the 8.68 which has two significant digits (remember that the underline tells us what the last significant digit is!) 8.68 * 3.1554 = 27 Note that here, we’d end up with the same result whether we use 8.68 or round it to 8.7 before the second calculation. In some cases it will make a difference though, so it’s important not to round too early. f.

(29.7 - 29.2) / .0455 =

Our first step in this calculation follows the addition/subtraction rules: 29.7 - 29.2 = 0.5 This only has one significant digit. In the second step, we follow the multiplication/division rules: 0.5 / 0.0455 = 1 x 101 9.

a. A compound has a density of 4.92 g/mL. What is the volume (in mL) if 26.2 g of the compound is present? We’re starting out with g, and we want mL for our final units. We have the density (in g/mL) that can serve as a conversion factor. For the units to cancel properly, we need to invert the density: 26.2 g •

1 mL 4.92 g

= 5.33 mL

Note that the ‘1 mL’ in the ratio doesn’t limit our significant figures, because the density of 4.92 g/mL is 4.92 g in exactly 1 mL. b. Another compound has a density of 0.549 g/mL. What is the mass (in g) if 9.06 mL of the compound is present? For this, we’re starting with mL and we want final units of g. Since the density is in units of g/mL, we can use it as is to cancel out units the way we want: 9.06 mL •

0.549 g 1 mL

= 4.97 g

10.

A sprinter is running at 22.5 miles/hr a. what is this speed in m/s? With questions like this, it’s good to come up with a conceptual plan before we start combining numbers. We have two units in our starting figure (miles and hours), and two in the final answer (meters and seconds). We will need to convert both of those units. Starting with converting miles to meters, we can do the following: miles ➝ km ➝ meters In the denominator, we can do the following: hours ➝ minutes ➝ seconds Overall, as a step-by step, that will look like this: miles hour

➝

km hour

➝

m

➝

hour

m

➝

minute

m s

When we set this up with numbers, we just need to be careful that we use the correct conversion factors and set them up in a way that the units cancel properly. For the first conversion: 22.5 miles 1 hour

•

1.61 km 1 mile

=

Note that since we have units of miles in the numerator of the first term and in the denominator of the conversion factor that miles cancels out in the answer. Also notice that we bring one extra digit through in our intermediate answer than what is significant. This helps us avoid rounding errors. We will round to the correct number of digits at the end. 22.5 miles 1 hour

•

1.61 km

=

1 mile

36.23 km 1 hour

Next we can set up the conversion from km to m: 36.23 km 1 hour

•

1000 m 1 km

4

=

3.623 x 10 m 1 hour

When we convert the denominator from hours to seconds, we can set up the conversions all at once, as long as we’re careful to be sure the units cancel: 4

3.623 x 10 m 1 hour

•

1 hour 60 min

•

1 min 60 sec

4

=

3.623 x 10 m 3600 sec

=

10.1 m sec

Note that we round to three significant figures in the final answer. b. if the sprinter ran at speed for 45.0 s, how far (in m) would they run? We already have the speed converted to m/s, so now we can use that ratio to convert from seconds to meters: 45.0 s •

10.1 m 1s

= 455 m

c. how long would it take the sprinter to run 200.0 m at this speed? Similarly, we can use the m/s conversion in this problem, we just need to invert it for units to cancel properly. 200.0 m •

1s 10.1 m

= 19.8 s

It’s always a good idea to check how reasonable our numbers are to make sure we’ve done things correctly. The world record in the 200.0 m for men is 19.19 seconds. That means that our sprinter is very fast, but still in a reasonable range for an answer. 11.

A cube measures 0.598 inches on one side a. what is the volume (in cm3) of this cube? (use 1.00 in = 2.54 cm) When we do this conversion, it’s important that we convert from inches to cm before we cube the number. Otherwise our conversion becomes a bit more complex (if we cube inches first, then we’d need to apply the inch to cm conversion in all three dimensions). 0.598 inches •

2.54 cm 1 inch

= 1.519 cm

The edge lengths in cubes are all identical, so we can get the volume by cubing the length we’re given. (1.519 cm)3 = 3.50 cm3 b. what is the volume (in mL) of this cube? This is a fairly straightforward conversion, as 1 mL = 1 cm3: 3.50 cm3 •

1 mL 1 cm3

= 3.50 mL

c. if the cube has a mass of 0.0065 pounds, what is the density of the cube in g/mL? (use 1.0 pounds = 454 g) We have the volume in mL already, so we need to convert the mass from pounds to g and then we can get the density. 0.0065 pounds •

454 1 pound

= 2.95 g

Dividing the mass by volume to get density: 2.95 g 3.50 mL

12.

= 0.84 g/mL

An object has a total volume of 632.4 mL. What is this volume in ft3? To do this conversion, we need to first convert from mL to cm3, and then from cm3 to ft3. 1 cm3

632.4 mL •

1 mL

= 632.4 cm3

Next, we need to convert from cm3 to ft3. When we do this, we need to remember that these are three dimensional units, and convert all three units or we won’t have the correct conversion. To illustrate what I mean by this, let’s apply the cm to ft conversion (1 ft = 30.48 cm) once: 632.4 cm3 •

1 ft 30.48 cm

= 20.748 ft•cm2

The reason for this is that units of cm3 are the same as cm•cm•cm . When we apply the cm➝ft conversion one time, we only convert one of the dimensions, so we end up with ft•cm•cm . We need to apply the conversion once for each dimension that we’re working in to do the conversion fully, so let’s apply it two more times: 20.748 ft•cm2 •

1 ft 30.48 cm

•

1 ft 30.48 cm

= 2.233 x 10-2 cm3

Whenever your units have an exponent, be sure you’re tracking that to make sure you’re converting in all directions!

13.

Convert the following temperatures: a. 56 ° C to K To convert from °C to K, we add 273.15: 56 + 273.15 = 329 K b. 56 ° C to ° F To convert from °C to °F, we multiply by 9/5 and add 32: 56 (9/5) + 32 = 133 °F c. 48.9 ° F to ° C To convert from °F to °C, we subtract 32, and then multiply by 5/9: (48.9 - 32) 5/9 = 9.39 °C d. 262 K to ° F To convert between K and °F, it’s most useful to go through °C. First, we’ll subtract 273.15: 262 - 273.15 = -11.15 °C Then we’ll convert from °C to °F as before: -11.15(9/5) + 32 = 11.9 °F...