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Problem Set 2 with Solution Key - Elementary Inorganic Chemistry | HCHE 111G, Assignments of Inorganic Chemistry

Material Type: Assignment; Professor: Lawrence; Class: General Inorganic Chemistry; Subject: Chemistry; University: Morehouse College; Term: Fall 2007;

Typology: Assignments

Pre 2010

Uploaded on 08/04/2009

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Chemistry 111 – Elementary Inorganic Chemistry
Morehouse College – Problem Set #2
ANSWER KEY
1. Density – To determine the density of acetone, a 55.0 gal drum is weighed twice.
The drum weighs 75.0 lb when empty and 437.5 lb when filled with acetone. What is
the density of acetone, expressed in grams per milliliter?
To determine the density of acetone, we require both its mass and the volume it
occupies.
lbs 362.5 lbs 75.0 - lbs 437.5
empty
drum
Mass
-
acetone)
drum
(Mass
acetone
Mass
=
=
The volume of acetone in the drum can be taken as the volume of the drum itself
Volume acetone = 55.0 gal
The density is then the ratio of the mass to the volume of acetone
gal 55.0
lbs 362.5
acetone Density
acetone Mass
acetone Density
=
=
Unfortunately, this result does not have the correct units so we require a few unit
equivalences to make a conversion.
mL
g
0.791
mL 1000
L 1
L 3.78
gal 1
kg 1
g 1000
lbs 2.204
kg 1
gal 55.0
lbs 362.5
acetone Density
=
=xxxx
pf3
pf4

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Download Problem Set 2 with Solution Key - Elementary Inorganic Chemistry | HCHE 111G and more Assignments Inorganic Chemistry in PDF only on Docsity!

Chemistry 111 – Elementary Inorganic Chemistry

Morehouse College – Problem Set

ANSWER KEY

  1. Density – To determine the density of acetone, a 55.0 gal drum is weighed twice. The drum weighs 75.0 lb when empty and 437.5 lb when filled with acetone. What is the density of acetone, expressed in grams per milliliter?

To determine the density of acetone, we require both its mass and the volume it occupies.

362.5 lbs

437.5lbs-75.0lbs

Massacetone (Massdrum acetone)-Massdrum empty

The volume of acetone in the drum can be taken as the volume of the drum itself

Volume acetone = 55.0 gal

The density is then the ratio of the mass to the volume of acetone

55.0 gal

362.5lbs

Densityacetone

Massacetone Densityacetone

Unfortunately, this result does not have the correct units so we require a few unit equivalences to make a conversion.

mL

g

1000 mL

1 L

3.78L

1 gal 1 kg

1000 g 2.204lbs

1 kg 55.0gal

362.5lbs Densityacetone

= x x x x

  1. Measurement – A culinary institute student is given an assignment in candy making that requires a sugar mixture to be brought to a “soft ball” stage (234 – 240 °F). If the student borrows a thermometer having a range from -10 °C to 110 °C from the chemistry laboratory, will this help them?

Using the relationship between temperature scales, one can find the maximum measurable temperature in °F.

(110) 32 198 32 230 F

T(C) 32

T( °F)= ° + = + = + = °

Based on the above calculation, the thermometer will not work for its intended purpose.

  1. Scientific Notation – Express each value in scientific notation / exponential form. Where appropriate, include units in your answer.

a. The speed of light in a vacuum, 299 792 458 m s-

2.99792458 x 10^8 m s-

b. The height of Mount Everest, 29,035 ft, in meters

8.8499 x 10^3 m

c. The average diameter of a human hair, 90 μm, in kilometers

9.0 x 10-8^ km

d. The average household income in 2006, $48,201.

4.8201 x 10^4 dollars

e. The population of India, 1,129,866,

1.129866154 x 10^9 people

f. The circumference of the Earth, 40,075.16 km, in miles

2.490155 x 10^4 mi

  1. Many acceptable answers. After all, an opinion is just an opinion.
  1. Avogadro’s Law – “Equal volumes of different gases under the same conditions of state have equal numbers of particles.” This is Avogadro’s hypothesis. Explain the experiment that led him to conclude this and show how it also suggests that equal numbers of particles of different gases occupy the same volume.

To conclude that “equal volumes of different gases under the same conditions of state have equal numbers of particles,” Avogadro used the relative densities of common gases. That is, he collected samples of hydrogen, oxygen, and nitrogen in a glass bulb under conditions in which the pressure and temperature of the gases were equal. The ratio of masses was not the same, indicating that particles of different gases had different masses. More importantly, however, he recognized that the proportions of molecules present could be confirmed by chemical proportions. For example, he observed the combination of two volumes of hydrogen and one volume of oxygen to form two volumes of water could be used to easily explain by his assertion.

Thus, the number of molecules in a specific volume of gas is independent of the size or mass of the gas molecules when related to an ideal gas.