Pooled Exam 1 Worksheets - Keys

CHEM 107 Problem Solving (Ch.1 – Ch. 4)

1 Chemistry 107 – Problem Solving Solution Key 1. Express 625,000 as 6.25 x 10 n . 625000 = 6.25 x 100000 = 6.25 x 10 5 (for numbers greater than 10, move decimal to the left ➔ (+) exponent) 2. Express 0.000707 as 7.07 x 10 n . 0.000707 = 7.07 x 0.0001 = 7.07 x 10 -4 (for numbers less than 1, move the decimal to the right ➔ (-) exponent) 3. Carry out the operations indicated on the following exponential numbers and write the answers in exponential notation. Please use your calculator to do these problems as your instructor is more interested in whether you know how to use your calculator. a. 3.24 x 10 3 + 2.40 x 10 3 (3.24 + 2.40) x 10 3 = 5.64 x 10 3 b. 4.76 x 10 2 + 7.7 x 10 1 4.76 x 10 2 + 0.77 x 10 2 = (4.76 + 0.77) x 10 2 = 5.53 x 10 2 c. 6.75 x 10 3 – 1.74 x 10 3 (6.75 – 1.74) x 10 3 = 5.01 x 10 3 d. 3.74 x 10 5 – 2 x 10 3 3.74 x 10 5 – 0.02 x 10 5 = (3.74 – 0.02) x 10 5 = 3.72 x 10 5 4. Determine the value of each of the following numbers in exponential form: a. 2 5 ) 10 x (2.0 (2.0) 2 x (10 5 ) 2 = 4.0 x 10 10 b. 6 10 x 9.00 3 6 10 x 3.00 10 x 9.00 = c. ) 10 x (2.5 x ) 10 x (1.0 4 6 (1.0 x 2.5) x (10 6 x 10 4 ) = 2.5 x 10 10

3 d) Solve q = mCΔT for C . e) Solve q = mC(T f – T i ) for T f . f) Solve q = mC(T f – T i ) for T i . g) Challenge Question: Solve for n i . ➔ ➔ ➔ ➔ ➔ ➔ At this point, n i is isolated and you would be finished. However, one could clean up the equation even more if desired (see below). ➔ ➔ ➔ T m q C  = ( ) mC q T - T i f  = mC q T T f i − = i f T mC q T + = ( ) T mC q T - mC q T - T f i i f − =  = T mC q - T f i  + =         − =  − 2 2 18 1 1 10 18 . 2 f i atom n n J x E 2 2 18 1 1 10 18 . 2 f i atom n n J x E − =  − 2 2 18 1 1 10 18 . 2 i f atom n n J x E = +  −         +  = − 2 18 2 1 10 18 . 2 1 f atom i n J x E n         +  = − 2 18 2 1 10 18 . 2 1 f atom i n J x E n         +  = − 2 18 1 10 18 . 2 1 f atom i n J x E n ( ) ( ) ( )( ) ( ) ( ) ( )( )         +  = − − − J x n J x J x n E n n f f atom f i 18 2 18 18 2 2 10 18 . 2 10 18 . 2 1 10 18 . 2 1 ( ) ( ) ( ) ( )( )         +  = − − J x n J x E n n f atom f i 18 2 18 2 10 18 . 2 10 18 . 2 1 ( )( ) ( ) ( ) ( ) J x E n J x n n atom f f i 18 2 18 2 10 18 . 2 10 18 . 2 − − +  = 1 1 10 18 . 2 2 18 2 =         +  − f atom i n J x E n

4 7. Write the correct units for the answers to each problem below in the blank boxes provided. Note that the numerical value has already been calculated for you. a) 2.0083 cm 3 b) 15 _kg 2 _ m 2 • s c) x = 0.00374 g sodium nitrite / slice d) (562 g) x (1 cm 3 / 0.832 g) = 675 cm 3 e) (56 g) x (1.000 cal / g °C) x (-45 °C) = -2500 cal f) 22 L g) E = 4.02 x 10 -22 kJ =             = − nm 10 m 1 nm 495 1000 ) / 10 00 . 3 )( 10 (6.626 9 8 34 J kJ s m s J hc  atm 1.54 K) K)(346 mol / atm L 06 mol)(0.082 (1.19 V = =         meat g 1,000,000 nitrite sodium g 200       slices 3 meat g 56 mL 1 cm 1 x L 10 mL 1 x L 2008.3 3 3 =                   =                     s kg 1 x g 1000 kg 1 x m g 26 x m 565 3

6 Measurement, Scientific Notation, and Significant Figures Name(s)_____________________________________ 1. Read each measurement to the correct number of significant figures. _____________________ _____________________ ___________________ 2. How many significant figures are there in the following numbers? a. 2 1 sig. fig. b. 200 1 sig. fig. c. 2000. 4 sig. fig. d. 200.0002 7 sig. fig. e. 0.0002 1 sig. fig. f. 0.000200 3 sig. fig. g. 2000 1 sig. fig. h. 20.00 4 sig. fig. 3. Round each of the following off to the indicated number of significant figures. a. 0.08205783 (5 sig. figs.) 0.082058 b. 14159 (3 sig. figs.) 14200 c. 6.0221367 x 10 23 (4 sig. figs.) 6.022 x 10 23 d. 6.62607 x 10 -34 (2 sig. figs.) 6.6 x 10 -34 e. 0.000333000 (4 sig. figs.) 0.0003330 f. 10.0000111 (5 sig. figs.) 10.000 KEY 73.5 mL 88.2 °C 645 mL

7 4. Express each of the following in proper scientific notation a. 123,000 1.23 x 10 5 b. 101,001,000,000 1.01001 x 10 11 c. 0.00000000012 1.2 x 10 -10 d. 349,987 3.49987 x 10 5 e. 0.002000 2.000 x 10 -3 f. 0.00350 x 10 -1 3.50 x 10 -4 (note: 0.00350 x 10 -1 = 3.50 x 10 -3 x 10 -1 = 3.50 x 10 -4 ) g. 0.170 x 10 2 1.70 x 10 1 (note: 0.170 x 10 2 = 1.70 x 10 -1 x 10 2 = 1.70 x 10 1 ) Strategy: If a number is already in scientific notation, but is not in PROPER scientific notation, first write the standard number in proper scientific notation and then multiply through the given power of 10 to get to the correct PROPER result. SEE THE LAST TWO EXAMPLES ABOVE!!!! 5. Perform the following calculations. Be sure to round the result to the correct number of significant figures. a. (0.102)(0.0821)(273) / 1.01 2.28616 / 1.01 = 2.2635 ===> 2.26 b. (0.14)(6.022 x 10 23 ) 8.4308 x 10 22 ===> 8.4 x 10 22 c. (6.626 x 10 -34 )(2.998 x 10 8 )(7.34993 x 10 -7 ) 1.460045 x 10 -31 ===> 1.460 x 10 -31 d. (2.00 x 10 6 ) / (3.000 x 10 -7 ) 6.6667 x 10 12 ===> 6.67 x 10 12 6. Combination problems. Round the result to the correct number of significant figures. a. (4.184)(100.62)(25.27 – 24.16) = (4.184)(100.62)(1.11) = 467.3 ===> 467 Note: 25.27 – 24.16 = 1.11 (answer has 3 sig. figs. but has two numbers right of the decimal according to the sig. fig. rule for subtraction). The 3 sig. figs. in 1.11 then carries through for the multiplication rule for sig. figs. to get to the final answer of 467. b. (8.27)(4.987 – 4.962) = (8.27)(0.025) = 0.2068 ===> 0.21 Note: Same idea applies here as in the previous problem. 4.987 – 4.962 = 0.025 (answer has 2 sig. figs., but three digits right of the decimal due to the sig. fig. rule for subtraction). The 2 sig. figs. in 0.025 then carries through for the multiplication rule for sig. figs. to get to the final answer of 0.21. c. [(.120)(0.08206)(273.15 + 23)] / (230 / 760.0) = [(.120)(0.08206) (296.15)] / (230 / 760.0) = [2.9162] / (0.3026) = 9.636 ===> 9.6 d. (2.3232 + 0.2034 – 0.16)(4.0 x 10 3 ) = (2.3666) (4.0 x 10 3 ) = 9466.4 ===> 9500 or 9.5 x 10 3

9 Review of the Metric System Name(s)______________ KEY _______________________ The metric system is so important in science that not understanding it becomes a huge disadvantage in science coursework. Therefore, a review of the metric system is warranted. A table of metric conversions can take on many forms; one example is shown below. It is very important to understand the difference between a base unit and metric prefix. When combined, these make a metric unit. Examples include: Base Unit Metric Prefix Metric Unit Name *Metric Symbol grams (g) milli- (m) = milligrams mg liters (L) micro- (μ) = microliters μL meters (m) nano- (n) = nanometers nm bytes (B) tera- (T) = terabytes TB calories (cal) kilo- (k) = kilocalories kcal Joules (J) kilo- (k) = kilojoules kJ *In calculations, metric symbols ARE the metric units. Now you try it! 1. Write metric units for the following metric unit names, OR, write the name for the following metric unit. a) gigabyte ➔ GB b) centigram ➔ cg c) millijoule ➔ mJ d) teragram ➔ Tg e) micrometer ➔ μm Largest Unit of Measure Smallest Unit of Measure a f p n  m c d k M G T P E atto femto pico nano micro milli centi deci kilo mega giga tera peta exa 10 -18 10 -15 10 -12 10 -9 10 -6 10 -3 10 -2 10 -1 10 3 10 6 10 9 10 12 10 15 10 18 SI Unit Prefixes base unit

10 f) femtosecond ➔ fs g) deciliter ➔ dL h) attometer ➔ am i) second ➔ s (This is simply a base unit; no metric prefix is required) 2. What is wrong with the following statement? “The doctor prescribed me a dose of 500 Mg of amoxicillin per day to be taken before bed time.” The problem is in the metric prefix used. The metric symbol (Mg) refers to megagrams when it should be milligrams (mg). One Mg is a billion times larger than a mg. Thus, if the pharmacist actually filled the prescription according to the wrong metric unit (Mg), one tablet would kill someone due to overdose. This is a prime example of how knowing your metric units can really pay off (and avoid a wrongful death lawsuit). The key to understanding the metric system is to be able to write unit conversions (a.k.a. equalities) from the table while ascribing metric symbols to each number. Once you know what the correct conversion is, then a metric conversion problem can be easily solved. The simplest way of using metric conversion tables is to take the power of ten from each prefix and then swap the powers of ten while keeping the units in place . Remember, the base unit always has 10 0 as its power of ten in the table (even though many metric tables do not show this). Also remember, 10 0 = 1. Take the following equalities as examples: Incorrect Equality (directly from the table) Swap Numbers Correct Equality 10 -3 mg = 10 0 g ====> 10 0 mg = 10 -3 g 10 -9 nm = 10 -6 μm ====> 10 -6 nm = 10 -9 μm 10 -9 nm = 10 3 km ====> 10 3 nm = 10 -9 km 10 -15 fg = 10 6 Mg ====> 10 6 fg = 10 -15 Mg

12 c) 6890000 s to Ms d) 0.00000000542 L to nL e) 1 x 10 -10 g to ng f) The distance between 2 atoms was determined to be 11.4 nm. What is this distance in centimeters? g) 6975.0 fs to ns 6975.0 fs x 10 -15 ns 10 -9 fs = 0.0069750 ns or 6.9750 x 10 -3 ns h) 553 cm 2 to m 2 Note: Since the units are squared, the conversion must be done twice! i) 11.4 nm 3 to dm 3 Note: Since the units are cubed, the conversion must be done three times! Ms 6.89 s 10 Ms 10 x s 6890000 6 0 = nL 5.42 L 10 nL 10 x L 542 0.00000000 9 - 0 = ng 0.1 g 10 ng 10 x g 10 x 1 9 - 0 10 - = cm 10 x 1.14 nm 10 cm 10 x nm 11.4 6 - 2 - -9 = 2 0 -2 0 -2 2 m 0.0553 cm 10 m 10 x cm 10 m 10 x cm 553 = 3 23 - 1 - -9 1 - -9 1 - -9 3 dm 10 x 1.14 nm 10 dm 10 x nm 10 dm 10 x nm 10 dm 10 x nm 11.4 =

13 4. Is it possible to convert 5.0 μm 2 to units of cm 3 ? Explain why or why not. If not, what further information is needed? No, you cannot directly convert square units to cubic units. HOWEVER, you could do it in a situation where another measurement of length was included in the square units. 5. Is it possible to convert 5.0 L to units of cg? Explain why or why not. If not, what further information is needed? No, you cannot directly convert volume units to mass units. HOWEVER, you could do it in a situation where the density of the substance in question is given. 6. For the following, circle which of the two units compared is a larger quantity. a) 12 cm or 12 μm Note: See metric table on which units are smallest or biggest. b) 59 dL or 59 kL Note: See metric table on which units are smallest or biggest. c) 96050 fs or 78 μs Since 9.605 x 10 -5 μs << 78 μs, then 78 μs is the larger quantity! d) 238 Mg or 8.6 x 10 13 μg Since 86 Mg < 238 Mg, then 238 Mg is the larger quantity! s   10 x 9.605 fs 10 s 10 x fs 96050 5 - 6 - -15 = Mg 86 g 10 Mg 10 x g 10 x 8.6 6 -6 13 =  

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