ANSWER SOLUTION for PROBLEM SET #1

Graded out of 21points.  Problems 1.14, 1.20, 1.24, 2.24, and 2.58 were worth 1 point each and only graded for completion (a check mark on these problems indicates 1 point credit, but not that the answer is necessarily correct).  Problems 1.36, 2.30 (a-d), 2.64, 2.78 (c-f) were all graded more carefully for 4 pts each.  See Heidi if you have questions about grading.  

Some general comments:

show work - answer as if you were writing a solution key for someone else to read.

significant figures and scientific notation - people had particular trouble with this, make sure you understand.

always include units!

see homework guidelines at bottom of vital info section - we will take off points next time

box answers

see us in office hours if you need help!

 (Specified page numbers are from the text Introduction to General, Organic, and Biological Chemistry. Matta, Wilbraham, and Stanley.)

1.14    To solve, use the equation that is “solved” for the unit of temperature desired (p. 9), then substitute in the given temperature.

(a)   

Given: 103°F                      

(b)    Find: K

Given: 39.4 °C (from part a)  K = °C + 273

Solution:   K = 39.4 + 273 = 312.4 K

1.20    To answer, understand the definitions of a solid, liquid and gas (p 11; especially look at Figure 1.5 for a good visual representation of these three states of matter).

Solution:

(a)    3

(b)   1, 2, 3

(c)    1, 2

1.24    Homogeneous mixtures are completely uniform.  A heterogeneous mixture is not uniform (see definitions p.12)

Solution:

   homogeneous:  b, c, d, e, f, h

   heterogeneous:  a, g

The components of soil and ocean water (not just salt water, but everything floating around in ocean water) are not uniformly mixed while the other given substances are uniformly mixed.

1.36    All are physical properties except for (e).  Choices a, b, c, d, f all describe an observable condition of silicon without having to change its composition.  Choice e  describes how silicon will chemically react with fluorine.

2.24    To solve, keep the first three significant numbers in each measurement, (be careful not to count the place holding zeros as a significant number).  Then round the last (third) significant figure, using the number to its right (underlined below) to determine whether to round up or down.  Use scientific notation when appropriate. 

Solution:

(a)    98.473 L = 98.5 L

(b)   0.00076321 cg = 0.000763 cg = 7.63 x 10^-4 cg

(c)    57.048 m = 57.0 m

(d)   9500 s = 9.50 x 10³ s

(Note that 9500 really only has two significant figures.  It does not really make sense to add another significant figure.  That would be like measuring something to within 1 cm and then latter saying you measured it to within 1mm.  The answer shown above assumes that all four significant figures in 9500 are significant.   A better way to write that would be 9.500 x 10³, which emphasizes that the last two zeroes are significant, not just place holders.)

(e)    12.17 °C = 12.2 °C

(f)     0.0074983 x 10⁴ mm = 0.00750 x 10⁴ mm = 7.50 x 10¹ mm = 75.0 mm 

2.30    Rule of Thumb for Significant Figures (sig figs.):

When multiplying and dividing, use the least number of significant figures.

Ex.  5.0 x 4.34 = 22  (since 5.0 contains the least sig figs, which is two)

When adding and subtracting, use the least number of decimal places.

Ex.  5.0 + 13 = 18 (since the ones place in 13 is the least decimal place)

Solution:

(a)    37 mg  (since the ones place in 21 is the least decimal place)

(b)   2.0 x 10¹ m²  (since 8.4 has the least sig figs; 8.4 has two sig figs,  Note that 20 is not correct because it technically only has one sig fig in that the zero could simply be a place holder)

(c)    51.6 mL  (since the tenths place in 87.2 is the least decimal place)

(d)   28.7 g  (since 0.225 has the least sig figs; 0.225 has three sig figs)

2.58    Since the specific gravity is 0.42, the density of the wood is 0.42 g/cm³ (see page 52).  If we know the volume of the block, we can calculate how tall it is from volume = (length)(width)(height).

  We know the mass and to interconvert between mass and volume we use density:

  First: find the volume:

Although I did not round the volume, keep in mind that we only know it to two significant figures, limited by the fact that we only know the density to two significant figures.  I carry the digits along to prevent the accumulation of rounding errors.  If the question asked for the volume, the proper answer would be 5.9 x 10² cm³.

Then, find the missing dimension

volume = (length)(width)(height)

Rearranging for height yields:

height = volume / (length * width) = 585.71 cm³ / (4.2 cm * 8.5 cm) = 16.41 cm = 16 cm

As we know the length, width and volume to two significant figures and because we are multiplying and dividing we know the height to two significant figures as shown.

2.64.  The specific heat relates temperature change to heat input (or output).  

heat = (mass)(DT)(specific heat)

Note that this equation should make sense:  The specific heat (if in units cal g^{-1 o}C^-1) tells you how many calories it takes to raise 1g of material by 1^oC.  Two grams takes twice as much so we multiply by the mass.  Raising by 2^oC takes twice as much so we multiply by the temperature change as well.

Solving this for the change in temperature DT, we get.  

 DT = heat / [(mass)(specific heat)]  (Note: both mass and specific heat are in the denominator-on the bottom)

We know the heat and mass to two significant figures, so we know DT to two significant figures (multiplication and division take least number of sig figs)  resulting in DT = 19^oC.

Hence, the final temperature is T_final = 15^oC + 19^oC = 34^oC.

2.78   

a. Directly:

However, it is usually easier to think of doing this in two stages by going through meters. 

b.

c.

d.

e.

f.