Download 02/08/16 CHEM 4 Quiz 3 (20 points)

02/08/16
CHEM 4
ANSWER KEY
Name _________________________________
Quiz 3 (20 points)
1. (4 points) Define the following terms.
Atom is the smallest particle of a chemical element; an electrically neutral particle made of a
protons and neutrons located inside the nucleus and electrons located outside the nucleus.
Molecule is the smallest particle of a chemical substance; a group of atoms held together by
chemical bonds that carries the total electric charge of zero.
Element (type of atom) is a collection of atoms each of which contains the same specific
number of protons in the nucleus.
Element (elementary substance) is a chemical substance made of atoms of which contains
the same specific number of protons in the nucleus
Compound is a chemical substance made of atoms of at least two different elements.
Mixture is a material of variable chemical composition; it is made of particles of at least two
different chemical substances.
2. (3 points)
(a) Fill the blanks.
106
1 mm = ________
nm
1012 nm2
1 mm2 = ________
1018 nm3
1 mm3 = ________
(b) What is the edge of a cube in centimeters if the volume of that cube is 6.4×104 mm3?
Show work.
Vcube = a3
3
a = √Vcube = √64×103 mm3 = 4.0×101 mm
3
4.0×101 mm = 4.0 cm
3. (3 points) Determine the mass in kilograms of 2.00 L of rubbing alcohol which has a
density of 0.786 g/mL. Show work.
Density = 0.786 g/mL = 0.786 kg/L
Mass = Volume × Density = 2.00 L × 0.786 kg/L = 1.57 kg
PLEASE TURN OVER!!!
4. (2 points) In the annotated drawing of the three beakers, the cubes are missing. Add the cubes
to the drawing.
beaker with water
and aluminum cube
beaker with water
and ice cube
beaker with water
and maple wood cube
5. (4 points) Gold is the most malleable of all metals; a single gram of it can be beaten into a
sheet of one square meter. What would be the thickness of such sheet of gold in
millimeters? Show work.
Volume = Mass / Density = 1.00 g / (19.3 g/cm3) = 0.0518 cm3
Volume = Area × Thickness = 1.00×104 cm2 × Thickness = 0.0518 cm3
Thickness = 0.0518 cm3 / (1.00×104 cm2) = 5.18×10−6 cm = 5.18×10−5 cm
6. (4 points Perform the following calculation, determine the uncertainty associated with the
calculated value, and record the result as “value” ± ”uncertainty” accompanied by the
appropriate unit. Assume the uncertainty ±1 in the last significant digit in each of the
numbers given. Show work.
3.63 cm2 +
2677 cm3
= 256.17 cm3
2.4 cm + 8.2 cm
Value Max =
2678 cm3
3.64 cm +
= 261.14 cm2
2.3 cm + 8.1 cm
Value Min =
2676 cm3
3.62 cm +
= 251.398 cm2
2.5 cm + 8.3 cm
2
2
Value Interval = 261.14 cm2 − 251.398 cm2 ≈ 10 cm2
Uncertainty =
10 cm2
2
= 5 cm2
256
5
cm2
___________
± __________
_________
(value)
(uncertainty) (unit)