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Practice problems Ch1 Key pt1

CHEM115 – Chapter 1
1.Although naturally occurring, smallpox was eradicated by a superbly coordinated effort including
the World Health Organization and health care providers worldwide, the classification of smallpox as
a Category A bioterrorism agent has renewed interest in its treatment and prevention. Moreover,
although vaccination against the disease is considered relatively safe for most individuals, it is not
entirely without risk.
The CDC estimates that 14 to 52 out of every million people who are vaccinated for smallpox will
suffer serious, potentially life-threatening reactions to the vaccine. In these cases, immediate medical
attention is required. The first course of treatment is with vaccinia immune globulin (VIG). If a patient
does not respond to treatment with VIG, a second option is cidofovir, a drug that currently is
approved by the Food and Drug Administration (FDA) to treat specific viral infections of the eye in
individuals with compromised immune systems. Both drugs are available for use in the treatment of a
serious reaction to the smallpox vaccine only through the FDA's Investigational New Drug (IND)
Cidofovir, marketed under the name Vistide, is distributed in vials containing 375 mg of the drug
dissolved in 5 mL of water. The manufacturer specifies that the drug should be kept at room
temperature (18°C–20°C). The vial contents are first diluted with saline and then administered
intravenously with a recommended dosage of 5 mg cidofovir per kilogram of body weight.
(a) Convert cidofovir's recommended storage-temperature range to the Kelvin Scale
18 + 273.15 = 291.15 K
20 + 273.15 = 293.15 K
(291.15 K – 293.15 K )
(b) If the fluid in a single vial of cidofovir has a volume of 5.00 mL and a mass of 5.89 g, what is the
density of the fluid? (Report the density to the appropriate number of significant figures)
Density = mass/ volume
d = 5.89 g / 5ml
d = 1.178 g/ml
or d = 1178 g/L
(c) What mass of cidofovir should be administered to a 177-lb man (1 lb = 0.4536 kg)
177 lb to kg
177lb * 0.4536 kg/lb = 80.29 kg
“a recommended dosage of 5 mg cidofovir per kilogram of body weight”
5 mg = 5 * 10-6 kg
So 5 * 10-6 kg per 1 kg
5 * 10-6 kg = ????????kg
1 kg
80.29 kg
mass = 4.014 * 10-4 kg
(d) Convert the density in part (b) to g/L and to kg/m3
1178 g * 1 ml
1 ml
10-3 L
= 1178 g/L
1.178 g * 1 kg * 1 ml * 100 cm3
1 ml
1000 g
1 cm3
= 0.1178 kg / m3
2. Classify each of the following substances as an element, compound, homogenous mixture or
heterogeneous mixture:
(a) hydrogen element
(b) water compound
(c) gold element
(d) sugar compound
(e) seawater it can be both homogenous and heterogeneous, homo because of the salts and hetero
because of the dust …
(f) helium gas element
(g) sodium chloride compound
(h) a bottle of Pepsi homogeneous
(i) a milkshake heterogeneous
(j) air in a bottle homogeneous
(k) concrete heterogeneous
3. Determine whether the following statements describe chemical or physical properties:
(a) Oxygen gas supports combustion. Chemical
(b) Fertilizers help to increase agricultural production. Chemical
(c) Water boils below 100°C on top of a mountain. Physical
(d) Lead is denser than aluminum. Physical
(e) Uranium is a radioactive element. Chemical
3. The density of ethanol is 0.798 g/mL. Calculate the mass of 205 mL of the liquid.
d=m/V So
0.798 g/mL * 205 mL = 163.59 g
4. The density of platinum (Pt) is 21.5 g/cm3 at 25°C. What is the volume of 87.6 g of Pt at this
d=m/V So
87.6 g ÷ 21.5 g/cm3 = 4.074 cm3
5. Express the answers to the following calculations in scientific notation and report to correct
significant figures:
(a)145.75 + (2.3 × 10−1)
146 = 1.4598 * 102
(b)79,500 ÷ (2.5 × 102)
320 = 3.2 * 102
(c)(7.0 × 10−3) − (8.0 × 10−4)
(7.0 × 10−3) − (0.8 × 10−3 ) = 6.2× 10−3
(d)(1.0 × 104) × (9.9 × 106)
1.1 × 105
6. In water conservation, chemists spread a thin film of a certain inert material over the surface of
water to cut down on the rate of evaporation of water in reservoirs. Scientists found that 0.10 mL of
oil could spread over the surface of water about 40 m2 in area. Assuming that the oil forms a
monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule in
Volume = length * width * height
Area = length * width
So, volume = area * height
Height * 40 = 0.10
Height = 2.5*10-3 m
Nanometers = 2500000*10-9 nm