## Chapter 1 Practice Problems

The practice problems are mainly to solve dimensional analysis
and unit conversion
problems. In all calculataions, we assume that you give correct
significant figures *at the end of the
calculations*. Please carry all digits until the
end, then round.

None of trivial examples are included in this.

** Example:**1.85 The density of silver is 10.5
g/cm

^{3}. What is the mass in kg of a cube of silver that measures 0.62 m on each side?

The volume of the cube is (0.62 m)^{3}. The question is
how many kg are there in (0.62 m)^{3}?
$$?kg={\left(0.62m\right)}^{3}{\left(\frac{{10}^{2}cm}{1m}\right)}^{3}\left(\frac{10.5g}{1c{m}^{3}}\right)\left(\frac{1kg}{{10}^{3}g}\right)=2.502444\times {10}^{3}kg=2.5\times {10}^{3}kg$$

** Example:**1.52 Carry out the following conversions:

a. 5 pm = __ cm = __ nm

b. 8.5 cm

^{3}= __ m

^{3}= __ mm

^{3}

a. Ask, how many cm are there in 5 pm? $$?cm=5pm\left(\frac{1m}{{10}^{12}pm}\right)\left(\frac{{10}^{2}cm}{1m}\right)=5\times {10}^{-10}cm$$ Similarly, $$?nm=5pm\left(\frac{1m}{{10}^{12}pm}\right)\left(\frac{{10}^{9}nm}{1m}\right)=5\times {10}^{-3}nm$$

b. Little tricky. But, the similar question, how many
m^{3} are there in 8.5 cm^{3}?
$$?{m}^{3}=8.5c{m}^{3}{\left(\frac{1m}{{10}^{2}cm}\right)}^{3}=8.5\times {10}^{-6}{m}^{3}$$
Notice we're cubing the *whole parenthesis*!!
Then, we can do the second part, as well.
$$?m{m}^{3}=8.5c{m}^{3}{\left(\frac{1m}{{10}^{2}cm}\right)}^{3}{\left(\frac{{10}^{3}mm}{1m}\right)}^{3}=8.5\times {10}^{3}m{m}^{3}$$

** Example:**1.96 When an irregularly shaped chunk of
silicon weighing 8.763 g was placed in a graduated cylinder containing
25.00 mL of water, the water level in the cylinder rose to
28.76 mL. What is the density of silicon in g/cm

^{3}?

Since
$$d=\frac{m}{V}$$
The volume of silicon is given by the difference of the water levels
before and after placing the silicon in, which is 3.76 mL.
$$d=\frac{8.763g}{3.76mL}=2.330585g/mL=2.33g/c{m}^{3}$$
The conversion factor 1 mL = 1 cm^{3} is used in the
last equality.

** Example:**1.105 Vinaigrette salad dressing consists
mainly of oil and vinegar. The density of olive oil is 0.918
g/cm

^{3}, the density of vinegar is 1.006 g/cm

^{3}, and the two do not mix. If a certain mixture of olive oil and vinegar has a total mass of 397.8 g and a total volume of 422.8 cm

^{3}, what is the volume of oil and what is the volume of vninegar in the mixture?

Therefore, the volume of oil then is $${V}_{oil}={V}_{total}-{V}_{vin}=442.8c{m}^{3}-110c{m}^{3}=313c{m}^{3}$$

** Example:**1.112 Brass is a copper-zinc alloy. What is the mass in grams
of a brass cylinder having a length of 1.62 in. and a diameter of 0.514 in. if the
composition of the brass is 67.0 % copper and 33.0 % zinc by mass? The density
of copper is 8.92 g/cm

^{3}, and the density of zinc is 7.14 g/cm

^{3}. Assume that the density of the brass varies linearly with composition.

*m*= 39.7956210 g = 39.8 g.

_{alloy}

** Example:** A calibrated pycnometer with its volume 25.000 mL was filled
with ethyl alcohol. By weighing the flask before and after adding the alcohol, it was
determined that the pycnometer contains 19.7325 g of alcohol. In a second experiment,
a 25.0920 g of metal beads were added to the pycnometer and the pycnometer was filled
with alcohol to 25.000 mL mark. The total mass of the metal beads and alcohol in the
pycnometer was deternined to be 38.4704 g. What is the density of the metal beads
in g/mL?

Then, the density of metal beads is then $${d}_{metal}=\frac{25.0920g}{8.050mL}=3.11702g/mL=3.117g/mL$$

** Example:** What is the length of a 10-lb spool of 12 gauge aluminum wire?
A 12 gauge has a diameter of 0.0808 in. and the density of aluminum is 2.70 g/cm

^{3}.

We can visualize the question by unfolding the spool and straighten the wire to a cylindrical form as follows.

The volume of cylinder is given by
$$V=\pi {r}^{2}l$$
where *r* is the radius of the circular face, and *l* is the length in question.
Solving for *l*, we have
$$l=\frac{V}{\pi {r}^{2}}$$
Let's calculate the volume first. Since the length we want is in *m*,
the volume we calculate should be in *m ^{3}*; so this is what
we're looking for. What is our given? The given should be 10 lb. 12 gauge
and 0.0808 in. are measures of diameter, which relates to

*r*. So, only option is 10 lb. We ask, how many

*m*are there in 10 lb.? $$?{m}^{3}=10lb\left(\frac{453.6g}{1lb}\right)\left(\frac{c{m}^{3}}{2.70g}\right){\left(\frac{1m}{100cm}\right)}^{3}=1.67999\times {10}^{-3}{m}^{3}$$ For

^{3}*r*we need the value in

*m*. The radius of the circular area is 0.0404 in. So, $$?m=0.0404in\left(\frac{2.54cm}{1in}\right)\left(\frac{1m}{100cm}\right)=1.02616\times {10}^{-3}m$$ Substitute the numbers into

*l = V /(πr*, we get. $$l=\frac{1.67999\times {10}^{-3}{m}^{3}}{\pi {\left(1.02616\times {10}^{-3}m\right)}^{2}}=\frac{1.67999\times {10}^{-3}{m}^{3}}{3.30811\times {10}^{-6}{m}^{2}}=507.842m=508m$$

^{2})

** Example:** One acre-ft is defined as a volume of an acre of
land filled with water to a depth of one foot. What is 1 acre-ft in
ft

^{3}? and in

*m*, given that 640 acre = 1 mi

^{3}^{2}, 1 mi = 1.609344 km, 1 in. = 2.54 cm, and 12 in. = 1 ft?

The pictorial representation of the question is as follows.

We want to know the volume of acre-ft in, first, *ft ^{3}*.
If we convert the area of 1 acre in

*ft*, the volume is obtained by multiplying 1

^{2}*ft*at the end, so our question is, how many

*ft*are there in 1 acre? $$?f{t}^{2}=1acre\left(\frac{1m{i}^{2}}{640acre}\right){\left(\frac{1.609344km}{1mi}\right)}^{2}{\left(\frac{{10}^{3}m}{1km}\right)}^{2}{\left(\frac{{10}^{2}cm}{1m}\right)}^{2}{\left(\frac{1in}{2.54cm}\right)}^{2}{\left(\frac{1ft}{12in}\right)}^{2}$$ $$?f{t}^{2}=43560f{t}^{2}=4.36\times {10}^{4}f{t}^{2}\to 4.36\times {10}^{4}f{t}^{3}$$ The last arrow indicates that the height of 1 ft is multipled to get the volume.

^{2}

Let's calculate now the volume in *m ^{3}*.
$$?{m}^{3}=43560f{t}^{3}{\left(\frac{12in}{1ft}\right)}^{3}{\left(\frac{2.54cm}{1in}\right)}^{3}{\left(\frac{1m}{{10}^{2}cm}\right)}^{3}=1233.412{m}^{3}=1.23\times {10}^{3}{m}^{3}$$

** Example:** Taken from the Prelab question: What is the volume of
a room in L that hold 30.0 kg of oxygen (O

_{2}), if the density of O

_{2}is 1.300 g/L, and oxygen in air is 21.0 % by volume?

The pictorial representation of the problem is

The key is to understand what each of the givens and what we are looking for.
First, we are looking for the volume of the room. There seems to be no
conversion factor that relates O_{2} to the room. But, air does relate
to O_{2}. So, instead of asking for the volume of the room, we are
asking for the volume of air. Density is a conversion factor converting
mass to volume or vise versa. The percent by volume is also a conversion factor
because when expanded, it means that
$$21{\%}_{{O}_{2}(volume)}=\frac{21{L}_{{O}_{2}}}{100{L}_{air}}\times 100\%$$
So, we use the ratio of 21 *L _{O2}* /
100

*L*as the conversion factor. It must mean that the 30.0 kg is our given, so that we ask, "How many

_{air}*L*of air that contains a 30

*kg*of O

_{2}?" Then, $$?{L}_{air}=30.0k{g}_{{O}_{2}}\left(\frac{{10}^{3}{g}_{{O}_{2}}}{1k{g}_{{O}_{2}}}\right)\left(\frac{1{L}_{{O}_{2}}}{1.31{g}_{{O}_{2}}}\right)\left(\frac{100{L}_{air}}{21.0{L}_{{O}_{2}}}\right)=1.0905125\times {10}^{5}{L}_{air}=1.09\times {10}^{5}{L}_{air}$$

** Example:** 1.93 Sodium (Na) metal undergoes a chemical reaction
with chlorine (Cl) gas to yield sodium chloride, table salt. If 1.00 g of
sodium reacts with 1.54 g of chlorine, 2.54 g of sodium chloride is formed,
17.9 kJ of heat is released. How much sodium and how much chlorine in
grams would have react to release 171 kcal of heat

Pictorially, you have

You can see that 17.9 kJ of heat generated per 1.00 g Na. So, we can use
17.9 kJ/g_{Na} as a conversion factor.
We ask,
$$?{g}_{Na}=171kcal\left(\frac{1000cal}{1kcal}\right)\left(\frac{4.184J}{1ca;}\right)\left(\frac{1kJ}{1000J}\right)\left(\frac{1.00{g}_{Na}}{17.9kJ}\right)=39.970{g}_{Na}=40.0{g}_{Na}$$