Posted: September 20th, 2022

URGENT HOMEWORH

See attached

CHEM

11

5

1 Study Guide Exam 1 Answers. You must complete the first 5 questions yourself to receive the extra credit.

Na

me:________________________________________ Data________________________

CHEM 1151 Exam 1 Study Guide Summer

2

022. 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 15

2

6

mg and a volume of 0.52 mL, what is the density of the metal?

10

00 mg = 1 g. 1 cm3 = 1 mL

Question 2:

Convert 502 mL to L. 1000 mL = 1 L.

Convert 0.002

7

km to cm. 100 cm = 1 m. 1000 m = 1 km.

Convert 25

2

8

.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

6

6

6

6

6

6

C-12 6
C-13 7

C-14

8

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.

N = 7 electrons. 1s2 2s2 2p3. 2 + 2 + 3 = 7 electrons.

O = 8 electrons. 1s2 2s2 2p4.

F = 9 electrons. 1s2 2s2 2p5.

Fe = 26 electrons. 1s2 2s2 2p6 3s2 3p6 4s2 3d6 = 2 + 2 + 6 + 2 + 6 + 2 + 6 = 26 e-.

The noble gas electron configuration would be like this for Fe.

Fe = 26 electrons.

Ar

is the closest noble gas with less than 26 electrons.

Fe noble gas electron configuration = [Ar] 4s2 3d6 = 26 electrons.

Mg = 12 electrons. 1s2 2s2 2p6 3s2. 2 + 2 + 6 + 2 = 12 electrons.

Mg noble gas electron configuration = [Ne] 3s2 = 12 electrons.

Ca
Ar

He = 2 electrons = 1s2

Question 9:

Specific Heat. Change in temperature = final – initial temperature

How much energy in joules is required to heat up 250 mL of water from a temperature of 25 degrees C to 89 degrees C? The specific heat of water is 4.184 J/gC. Remember, the density of water is 1 g/mL.

The formula is Energy = specific heat x mass x change in temperature.

Energy = 250 grams x 4.184 J/gC x (89 C – 25 C)

Energy = 1,046 J/C x (64 C)

Energy = 66,944 J = 66.9 KJ

1 KJ = 1000 J

Remember, water has a density of 1 g / mL

1 mL water = 1 g of water since the density of water is 1 g / mL.

How much energy in KJ is required to heat up 65 L of water from a temperature of 20 degrees C to 92 degrees C? 1 Kj = 1000 J

65 L water.

1 L water = 1000 mL water

1 mL water = 1 g water.

Energy = 65,000 g water x 4.184 J/gC x (92 C – 20 C)

Energy = 271,960 J/C x 72 C

Energy = 19,581,120 J = 19,581 KJ = 20,000 KJ

Convert to KJ where 1 KJ = 1000 J

Question 10:

Convert 98 C to Kelvin.

Temperature in Kelvin = Temperature in Celsius + 273.15

98 + 273.15 =
371.15 Kelvin

Convert 373 K to C.

Temperature in Celsius = Temperature in kelvin – 273.15

373 K – 273.15 =
99.85 C

How much energy is required to melt an a 12 g ice cube?

The heat of fusion of water = 334 J/1 g or 80 cal/1 g

334 J/g x 12 g =
4,008 J

How much energy is required to vaporize (turn into steam) 2 L of water?

The heat of vaporization of water = 2260 J/1 g or 540 cal/1 g

2 L = 2000 mL

1 mL = 1 g if it is made of water.

2 L = 2000 g of water.

2000 g water x 2260 J/g = 4,520,000 joules are required to vaporize 2 L of water.

If you want to get the answer in calories do the following.

2000 g water x 540 cal/g = 1,080,000 calories are required to vaporize 2 L of water.

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