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

2 d. 5 - -2 10 x 2.0 10 x 4.0 3 5 - -2 10 x 2.0 10 10 x 2.0 4.0 = e. ) 10 x 3 . 9 )( 10 x (2.0 10 x 4.0 6 5 - -2 -4 10 x 2.2 = Note for part e, one possible sequence of keys you would need to punch on your scientific calculator, depending on the calculator model, would be (boxed in bold): 4.0 => EE or EXP => (-) or +/- => 2 => ÷ => ( => 2.0 => EE or EXP => (-) or +/- => 5 => x => 9.3 => EE or EXP => 6 => ) => = or ENTER 5. Solve the following linear equations for the unknown (x). a. 4x = 7 + 1 2 x 8 4x =  = b. 2(x+2) = 14 5 x 10 2x 14 4 2x =  =  = + c. 3x 2 = 27 3 x 9 x 2  =  = d. 5.5 2 - 3x 33 = 6 2. x 16.5x 44 11 - 16.5x 33 2) - 5.5(3x 33 =  =  =  = 6. Solve the following equations for the stated variable by using basic algebraic principles. a) Solve for m. b) Solve for V. c) Solve PV = nRT for R. V m d = V m d =   = V d m V d m  = nT PV R = d m V =

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