CHEM 1151 Study Guide Exam 1 Answers. You must complete the first 5 questions yourself to receive the extra credit.
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CHEM 1151 Exam 1 Study Guide Summer 2022. Chapter 1 and 2. Answer Questions 1-5 to receive up to 5 extra credit points on the first exam.
Your answers must include the question and the answer to be counted. You are welcome to work together on this extra credit assignment but please make sure that you learn how to do these questions since they may appear on the exam. If you need assistance from me to do these questions then please attend office hours, email me, or go on discord. You can also review the textbook, and PowerPoints to find information on how to solve them on your own.
Late Submissions will not be accepted. Submissions will be due by the last day of the exam at 11:59 PM at the latest. Turn these in early if you are worried about the deadline. For those of you in my in-person class you may turn it in as a hard or digital copy.
Question 1:
If a piece of metal has a mass of 1526 mg and a volume of 0.52 mL, what is the density of the metal? 1000 mg = 1 g. 1 cm3 = 1 mL
Question 2:
Convert 502 mL to L. 1000 mL = 1 L.
Convert 0.0027 km to cm. 100 cm = 1 m. 1000 m = 1 km.
Convert 2528.8 mm to meters. 1000 mm = 1 m.
Question 3:
Write 0.0027 km in scientific notation to 2 significant digits.
Write 12,000,000 in scientific notation to 2 significant digits.
Write the following in scientific notation.
102.2 mL
10,000 g
620 L
12,321 km
Question 4:
How many significant digits do the following figures have?
102.2 mL
10,000 g
620 L
9.27 x 104 mol
0.00025 mg
0.0010 cm
12,321 km
12.210 cm
9.23 x 103 mol
Write the following standard notation figures in standard form.
9.27 x 104
1.286 x 102
2.37 x 10-3
Question 5:
What would be the total volume dispensed from a burette if the initial reading was 1.28 mL and the final volume was 12.36 mL?
Convert 60 miles per hour to meters per second. 1 mile = 1.609 km. 1 km = 1000 m. 1 hour = 60 minutes 1 minute = 60 seconds.
Displacement can be used to determine the density of an object. What is the density of a 52.6 g piece of jewelry if it the graduated cylinder before adding the jewelry measures 16.2 mL, and after adding the jewelry it measures 41.6 mL? Density = g/mL
Question 6:
How many protons, neutrons, and electrons are in the following?
| Atom | Protons | Neutrons | Electrons |
| Be | 4 | 5 | 4 |
| Na | 11 | 12 | 11 |
| Cl | 17 | 18 | 17 |
| He | 2 | 2 | 2 |
| Cr | 24 | 28 | 24 |
| Fe | 26 | 30 | 26 |
| Si | 14 | 14 | 14 |
| Ca | |||
| Na+ | 11 | 12 | 10 |
| Cl- | 17 | 18 | 18 |
| Ca2+ |
Beryllium has 4 as the atomic number so it has 4 protons, and 4 electrons.
Protons = Electrons if the atom is neutral.
Charge = Protons – Electrons.
Atomic mass number of Beryllium = 9.012 = 9
Atomic mass number = Protons + Neutrons.
Atomic mass number – protons = Neutrons.
9 – 4 = 5 Neutrons.
Chlorine has 17 as the atomic number.
In this example Cl is neutral since it does not have a + or – charge. Therefore, Charge = Protons – Electrons
0 charge = 17 protons – 17 electrons.
Atomic mass number
35.4527 = 35.
35 atomic mass number – 17 protons = 18 Neutrons.
Na+ is a sodium with a (+1) charge.
Charge = Protons – Electrons.
1 = 11 – Electrons
10 = electrons
Na+ has 10 electrons. Everything else is the same since charge only affects the number of electrons.
Cl-1
-1 charge = protons – electrons
-1 = 17 – electrons
-1 – 17 = – electrons
-18 = – electrons
18 = electrons
Cl-1 has 18 electrons.
Check your work.
-1 charge = 17 protons – 18 electrons.
Question 7:
C-12 and C-13 are isotopes. What are the similarities and differences between these two isotopes?
List the number of protons, neutrons, and electrons in each.
C-12 is an atom that has 12 as the atomic mass number, disregard the atomic mass number on the periodic table when you have Carbon-12. Only the number of neutrons changes, the number of protons and electrons stays the same.
| Atom | Protons | Neutrons | Electrons |
| C-12 | 6 | 6 | 6 |
| C-13 | 6 | 7 | 6 |
| C-14 | 6 | 8 | 6 |
The atomic number does not change. Carbon should always have 6 protons as can be found on the periodic table. The number of neutrons did change.
There is no mention of the charge so you have to assume that the charge is neutral.
Atomic mass number – protons = Neutrons.
12 – 6 = 6
13 – 6 = 7
14 – 6 = 8
Question 8:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 …
Using this information about the order of the orbitals what is the electron configuration of the following?
C = 6 electrons. 1s 2 2s 2 2p 2 The bolded numbers are the number of electrons. You need to fill out the chart while using up all of the electrons from left to right using the order 1s2 2s2 2p6 3s2 3p6 4s2 … You will eventually need to memorize the order, but not for this Friday/Saturday’s exam.
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