Chapter 1
Chemistry: Matter and Measurement

1.1 (a) Cu (b) Pt (c) Pu

1.2 (a) silver (b) rhodium (c) rhenium (d) cesium (e) argon (f) arsenic

1.3 (a) Ti, metal (b) Te, semimetal (c) Se, nonmetal (d) Sc, metal (e) At, semimetal (f) Ar, nonmetal

1.4 The three "coinage metals" are copper (Cu), silver (Ag), and gold (Au).

1.5 (a) The decimal point must be shifted ten places to the right so the exponent is -10. The result is 3.72 x 10-10 m.

(b) The decimal point must be shifted eleven places to the left so the exponent is 11. The result is 1.5 x 1011 m.

1.6 (a) microgram (b) decimeter (c) picosecond (d) kiloampere (e) millimole

1.7

1.8 (a) K = oC + 273.15 = -78 + 273.15 = 195.15 K = 195 K
(b)

(c) oC = K - 273.15 = 375 - 273.15 = 101.85oC = 102oC


1.9

1.10

1.11 The actual mass of the bottle and the acetone = 38.0015 g + 0.7791 g = 38.7806 g. The measured values are 38.7798 g, 38.7795 g, and 38.7801 g. These values are both close to each other and close to the actual mass. Therefore the results are both precise and accurate.

1.12 (a) 76.600 kg has 5 significant figures because zeros at the end of a number and after the decimal point are always significant.

(b) 4.502 00 x 103 g has 6 significant figures because zeros in the middle of a number are significant and zeros at the end of a number and after the decimal point are always significant.

(c) 3000 nm has 1, 2, 3, or 4 significant figures because zeros at the end of a number and before the decimal point may or may not be significant.

(d) 0.003 00 mL has 3 significant figures because zeros at the beginning of a number are not significant and zeros at the end of a number and after the decimal point are always significant.

(e) 18 students has an infinite number of significant figures since this is an exact number.

1.13 (a) Since the digit to be dropped (the second 4) is less than 5, round down. The result is 3.774 L.

(b) Since the digit to be dropped (0) is less than 5, round down. The result is 255 K.

(c) Since the digit to be dropped is equal to 5 with nothing following, round down. The result is 55.26 kg.

1.14 (a)
This result should be expressed with 3 decimal places. Since the digit to be dropped (7) is greater than 5, round up. The result is 24.612 g (5 significant figures).

(b) 4.6742 g / 0.003 71 L = 1259.89 g/L

0.003 71 has only 3 significant figures so the result of the division should have only 3 significant figures. Since the digit to be dropped (first 9) is greater than 5, round up. The result is 1260 g/L (3 significant figures), or 1.26 x 103 g/L.

(c)
This result should be expressed with 1 decimal place. Since the digit to be dropped (9) is greater than 5, round up. The result is 41.1 mL (3 significant figures).

1.15 The level of the liquid in the thermometer is just past halfway between the 32oC and 33oC marks on the thermometer. The temperature is 32.6oC (3 significant figures).

1.16 (a) Estimate: oF 2 x oC if oC is large. The melting point of gold 2000oF.

Calculation:

(b) r = d/2 = 3 x 10-6 m = 3 x 10-4 cm; h = 2 x 10-6 m = 2 x 10-4 cm

Estimate: volume = r2h 3r2h 3(3 x 10-4 cm)2(2 x 10-4 cm) 5 x 10-11 cm3

Calculation: volume = r2h = (3.1416)(3 x 10-4 cm)2(2 x 10-4 cm) = 6 x 10-11 cm3

1.17 1 carat = 200 mg = 200 x 10-3 g = 0.200 g

Mass of Hope Diamond in grams =
1 ounce = 28.35 g
Mass of Hope Diamond in ounces =

1.18 An LD50 value is the amount of a substance per kilogram of body weight that is a lethal dose for 50% of the test animals.

1.19 mass of salt = = 281.2 g or 300 g

1.29 The rows are called periods, and the columns are called groups.

1.31 Elements within a group have similar chemical properties.

1.38 F, Cl, Br, and I

1.59 (a) deciliter (10-1 L) (b) decimeter (10-1 m) (c) micrometer (10-6 m) (d) nanoliter (10-9 L)

1.62 (a) 5 pm = 5 x 10-12 m

5 x 10-12 m x = 5 x 10-10 cm

5 x 10-12 m x = 5 x 10-3 nm

(b) 8.5 cm3 x = 8.5 x 10-6 m3

8.5 cm3 x = 8.5 x 103 mm3

(c) 65.2 mg x = 0.0652 g

65.2 mg x x = 6.52 x 1010 pg

1.68 (a) To convert 453.32 mg to scientific notation, move the decimal point 2 places to the left and include an exponent of 102. The result is 4.5332 x 102 mg.
(b) To convert 0.000 042 1 mL to scientific notation, move the decimal point 5 places to the right and include an exponent of 10-5. The result is 4.21 x 10-5 mL.
(c) To convert 667,000 g to scientific notation, move the decimal point 5 places to the left and include an exponent of 105. The result is 6.67 x 105 g.

1.72 (a) 4.884 x 2.05 = 10.012
The result should contain only 3 significant figures because 2.05 contains 3 significant figures (the smaller number of significant figures of the two). Since the digit to be dropped (1) is less than 5, round down. The result is 10.0.

(b) 94.61 / 3.7 = 25.57
The result should contain only 2 significant figures because 3.7 contains 2 significant figures (the smaller number of significant figures of the two). Since the digit to be dropped (second 5) is 5 with nonzero digits following, round up. The result is 26.

(c) 3.7 / 94.61 = 0.0391
The result should contain only 2 significant figures because 3.7 contains 2 significant figures (the smaller number of significant figures of the two). Since the digit to be dropped (1) is less than 5, round down. The result is 0.039.

(d)
This result should be expressed with no decimal places. Since the digit to be dropped (3) is less than 5, round down. The result is 5526.

(e)
This result should be expressed with only 1 decimal place. Since the digit to be dropped (3) is less than 5, round down. The result is 87.6.

(f) 5.7 x 2.31 = 13.167
The result should contain only 2 significant figures because 5.7 contains 2 significant figures (the smaller number of significant figures of the two). Since the digit to be dropped (second 1) is less than 5, round down. The result is 13.

1.77 (a)

(b) (6 x 2.5 x15) hands3 x

1.81

= 1.452 mg = 1.5 mg

1.84
K = oC + 273.15 = 3422 + 273.15 = 3695.15 K or 3695 K

1.91 d = 2.40 mm = 0.240 cm
r = d/2 = 0.120 cm and V = r2h

1.95 (a) Element 117 is a halogen because it would be found directly below At in group 7A.
(b) Element 119
(c) Element 115 would be found directly below Bi and would be a metal. Element 117 might have the properties of a semimetal.
(d) Element 119, at the bottom of group 1A, would likely be a soft, shiny, very reactive metal forming a +1 cation.

1.107 35 sv = 35 x 109
(a) gulf stream flow = = 2.1 x 1018 mL/min

(b) mass of H2O = = 3.0 x 1021 g = 3.0 x 1018 kg

(c) time = = 0.48 min