Assignment Help: SCIN131-Introduction to Chemistry-APUS-Midterm Assignment

Part 1 of 4 - Short Answer

15.0/ 15.0 Points

Please show as much work as possible for full credit.

Question 1 of 13

5.0/ 5.0 Points

A room contains 48 kg of air. How many KWh of energy are necessary to heat the air in the house from 7^oC to 28^oC? The heat capacity of air is 1.03 J/g^oC.

Energy in joules = mass air x specific heat air x (Tfinal-Tinitial) = 48000*1.03*(28-7) = 1038240 J

1 J = 2.78*10-7 KWh.

Energy necessary = 1038240 * 2.78 * 10-7 = 0.2886 KWh.

Question 2 of 13

5.0/ 5.0 Points

State the steps in the Scientific Method.

The scientific method to discover the answer to a scientific problem needs the following steps:

Research

Problem

Hypothesis

Project Experimentation

Project Conclusion

In simple words we can state something like below,

Ask a Question

Do Background Research

Construct a Hypothesis

Test Your Hypothesis by Doing an Experiment

Analyze Your Data and Draw a Conclusion

Communicate Your Results

Question 3 of 13

5.0/ 5.0 Points

Hexane boils at 156.2^oF. What is this temperature in Celcius? What is this temperature in Kelvin?

T in °C = (T°F - 32) x (5 / 9)
C=(156.2-32)x(5/9)
C=(124.2)x(5/9)
C= 69 degrees celcius
T in K = T°C + 273.15
K=69+273.15
K=342.15

Part 2 of 4 -

10.0/ 10.0 Points

Question 4 of 13

5.0/ 5.0 Points

Please write the names of the following: Ag₂O, P₂S_3,NaOH

Ag2O is Silver(I) oxide

P2S3 is Phosphorus Sulfide

NaOH is Sodium Hydroxide

Question 5 of 13

5.0/ 5.0 Points

Give the A)Element Symbol, B) Atomic Number, C) Number of Protons, D) Number of Neutrons, (E) Mass Number, and (F) total Number of Electrons for: Barium–138 and Phosphorous-32

You must show your work in detail (where applicable) to receive credit.

Barium 138

Element symbol Ba

Atomic number is 56

Number of protons 56

Number of neutrons = 82

Mass number = 138

Total number of electrons = 56

Phosphorus 32

Element Symbol is P

Atomic number is 15

Number of protons = 15

Number of neutrons = 17

Mass number= 32

Total number of electrons = 15

Number of protons = Atomic number

Number of neutrons= Mass number- atomic number

Feedback: See chapter 4

Part 3 of 4 -

30.0/ 30.0 Points

Question 6 of 13

5.0/ 5.0 Points

What is the theoretical yield in grams of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 gCuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

Molar Mass of ( Na2S ) = 78.06 g/mol

Given 15.5 g counts to 15.5 g / 78.06 = 0.198 moles of Na2S

Molar Mass of ( CuSO4) = 159.64 g / mol

12.1 g / 159.64 = 0.0758 moles CuSO4

the reaction is

Na2S + CuSO4 ------> Na2SO4 + CuS

the ratio between Na2S and CuSO4 is 1 : 1 so CuSO4 is the limiting reactant.

We would get 0.0758 mole CuS

Molar Mass of CuS = 95.54 g/mol)

0.0758 mole x 95.54 g/mol = 7.24 g of CuS

Question 7 of 13

5.0/ 5.0 Points

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

Molar mass of Al = 27g/mol

Molar mass of O2 = 32g/mol

number of moles of Al = 10g / 27g/mol = 0.37 mol Al

number of moles of O2 = 19g / 32 g/mol = 0.59 mol O2

Taking the moles of each substance and dividing it by the coefficient of the balanced equation:

4Al + 3O2 → 2Al2O3

For Al: 0.37 / 4 = 0.09

For O2: 0.59 / 3 = 0.197

The lowest number indicates the limiting reagent.

Hence Aluminum is limiting reagent.

Question 8 of 13

5.0/ 5.0 Points

Calculate the empirical formulas for the following compounds that have been decomposed in the laboratory. A) 1.651 g Ag, 0.1224 g O; B) 0.672 g Co, 0.569 g As, 0.486 g O

1.651 g Ag x (1 mol/108 g Ag) = .0153 mol Ag

0.1224 g O x (1 mol/16 g O) = 0.00765 Mol O
dividing by least

0.0153/.00765 = 1.998 (round to 2) mol Ag
0.00765/0.00765 = 1 mol O Empirical formula is: Ag2O,
0.672 g Co x (1 mol/28 g Co) = 0.024 mol Co
0.569 g As x (1 mol/75 g As) = 0.0076 mol As
0.486 g O x (1 mol/16 g O) = 0.030375 mol O

Divide by lowest value (mol As):
0.024 mol Co /.0076 = approx. 3 mol Co
0.030375 mol O/.0076 = approx. 4 mol O
0.0076 mol As/0.0076 = 1 mol As
Therefore Empirical formula is: Co3AsO4 with a ratio of 3:1:4

Feedback: Please see Chapter 6 for empirical formulas

Question 9 of 13

5.0/ 5.0 Points

Hydrogen, a possible future fuel, can also be obtained from other compounds such as ethanol. Ethanol can be made from the fermentation of crops such as corn. How much hydrogen in grams can be obtained from 1.0 kg of ethanol (C2H5OH)?

The molar mass of C2H5OH = 2x12 + 5x1 + 16 + 1 = 46g.
There are 6 H's in the fomula so 6/46 of the mass of C2H5OH is H.
1.0kg x 6/46 = 0.13kg of H2 is obtained.

Question 10 of 13

5.0/ 5.0 Points

An iron chloride compound contains 18.62 grams of iron and 17.75 grams of chlorine. What is the most likely empirical formula for this compound?

Molar mass of Fe= 56.

18.62/56 = 0.3325
Molar mass of Cl = 35.5

Cl: 17.75/35.5 = 0.5
dividing by smallest:

Fe 0.3325/0.3325 = 1;

Cl 0.5/0.3325 = 1.5

Multiplying by 2 to obtain whole numbers: 2:3

So Emperical Formula is Fe2Cl3

Feedback: Please see Chapter 6 for Empirical Formulas.

Question 11 of 13

5.0/ 5.0 Points

What is the limiting reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of Li3PO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2

According to the balanced equation

3Ca(NO3)2 + 2Li3PO4 ----> 6LiNO3 + Ca3(PO4)2
3 mol Ca(NO3)2 reacts with 2 mol Li3PO4
Assuming all the Ca(NO3)2 is used up:

3.4 mol Ca(NO3)2 x (2 mol Li3PO4 / 3 mol Ca(NO3)2 ) = 2.27 mol Li3PO4
Since there are more moles of Li3PO4 than the 2.27 moles required to use up all the Ca(NO3)2, Li3PO4 is present in excess. This implies Ca(NO3)2 is the limiting reagent.

Part 4 of 4 -

30.0/ 30.0 Points

Question 12 of 13

15.0/ 15.0 Points

Bornite (Cu₃FeS₃) is a copper ore used in the production of copper.

When heated, the following (unbalanced) reaction occurs:

Cu₃FeS₃(s) + O₂(g) → Cu(s)+ FeO(s) + SO₂(g)

If 2.50 kg of bornite is reacted with excess O₂ and the process has an 82.5% yield of copper, how much copper is produced (in kg)?

Balanced equation is 2 Cu3FeS3(s) + 7 O2(g)------> 6 Cu(s) + 2 FeO(s) + 6 SO2(g)
2 moles Boronite produces 6 moles Cu

Converting given amount of boronite into moles

Molar mass of boronite = 342.71 g/ mole

therefore 2500g counts to 7.31 mole

So amount of copper produced = 7.31 * 3 = 21.93 moles

Molar mass of copper = 63.55 g/mole

Therefore mass of copper = 21.93 * 63.55 = 1.39 kg

Given % yield of copper is 82.5%

So final amount of copper produced = 1.146 kg

Feedback: The problem requires many steps from chapters 6-8.

Question 13 of 13

15.0/ 15.0 Points

A sample was decomposed in the laboratory and found to have 38.67g C, 16.22g H, and 45.11g N.

a)Find the molecular formula of this compound if the Formula mass is 62.12 g/mole.
b)Determine how many H atoms would be in a 3.50g sample of this compound.

calculating empirical formula

38.67g C/ 12 = 3.225 mol C

16.22g H/ 1 = 16.22 mol H

45.11g N/ 14 = 3.22 mol N

Dividing by least

3.225/3.22 = 1 mol C

16.22/3.22 = 5 mol H

3.22 /3.22 = 1 mol N

emperical formula = CH5N

Emperical formula weight = 31g

A)Given formula weight = 62.12g

so formula of compound = C2H10N2

B) 62 g contain 10 moles H atoms

3.50 g compund contains 0.564 moles = 0.564 * 6.023 *10^23 = 3.4 * 10^23 atoms

Feedback: This problem uses many concepts learned in Lessons 1 - 3.

Part 1 of 4 - Short Answer

15.0/ 15.0 Points

Please show as much work as possible for full credit.

Question 1 of 13

5.0/ 5.0 Points

A hybrid SUV consumes fuel at a rate of 12.8 km/L. how many miles can the car travel on 22.5 gallons of gasoline?

Fuel consumption rate (FCR) = distance travelled/ volume of fuel convert to miles per gallon

FCR = 7.95355 miles/ 0.264 gal

Using FCR = distance/ volume of fuel

l then (7.95355)/ (0.264) = distance in miles/ 22.5 gal

(7.95355)(22.5)/ (0.264) = 677 miles

Question 2 of 13

5.0/ 5.0 Points

A 52.3 kg sample of ethanol is needed for a reaction. What volume in liters of ethanol should be used? The density of ethanol is 0.789 g/mL

Density = Mass/Volume

Given density = 0.789 gm/mL

Volume = mass/Density = 52300/0.789 = 66.2 L

Question 3 of 13

5.0/ 5.0 Points

Give the A)Element Symbol, Atomic Number, Mass Number, Number of Protons, Total electrons and Number of Neutrons for: Manganese – 55 and Carbon - 14

Manganese Mn

Atomic Number: 25

Mass Number: 55

Number of Protons: 25 (Same as atomic number)

Total electrons: 25 (Assuming it is now neutral, it should be the same as number of protons)

Number of neutrons: 30 (Mass number - atomic number)

Carbon is C

Atomic Number: 6

Mass Number: 14

Number of Protons: 6

Total electrons: 6

Number of neutrons: 14-6=8

Part 2 of 4 -

8.5/ 10.0 Points

Question 4 of 13

5.0/ 5.0 Points

Write the name of each of the following Ionic Compounds: Ba₃(PO₄)₂ , MgSO₄ , PbO₂

Ba₃(PO₄)₂ is Barium Phosphate

MgSO₄is Magnesium sulfate

PbO2 is Lead (IV) Oxide

Feedback: Please review making and naming ionic compounds

Question 5 of 13

3.5/ 5.0 Points

Please write the names of the following: Ag₂O, P₂S_3,NaOH

Ag₂O,is Silver Oxide

P₂S₃is Phosphorus Sulfide

NaOH is Sodium hydroxide

Comment: Diphosphorus trisulfide

Part 3 of 4 -

26.0/ 30.0 Points

Question 6 of 13

5.0/ 5.0 Points

Vitamin C is known chemically by the name ascorbic acid. Determine the empirical formula of ascorbic acid if it is composed of 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen.

H = 1 g/mol
C = 12 g/mol
O = 16 g/mol
Then you divide A.M. / %
For the carbon:
40.92 / 12 ~ 3.4
For the hydrogen:
4.58 / 1 = 4.58
For the oxygen:
54.5 / 16 ~ 3.4
Then you divide the smallest result to each element:
C = 3.4 / 3.4 = 1
H = 4.6 / 3.4 = 1.35
O = 3.4 / 3.4 = 1
But the number must be complete, so try multiplying by another number that will turn it into a complete number:
C = 1 x 3 = 3
H = 1.35 x 3 = 4,05 ~ 4
O = 1 x 3 = 3
The empirical formula for the ascorbic acid is:
C3H4O3

Question 7 of 13

5.0/ 5.0 Points

Determine the empirical formula of a compound containing 60.3% magnesium and 39.7% oxygen.

dividing it by its respective atomic weights:

60.3/24.3 = 2.48 Mg
39.7/16.0 = 2.48 O
The relative numbers of atoms are in a 1:1 ratio, so the formula is MgO

Feedback: Please see Chapter 6 for a problem like this.

Question 8 of 13

5.0/ 5.0 Points

Ethanol has a molar mass of 46.07 g/mol
So that mass is made of the carbon (12g per mole) and the oxygen (16 g/mol) and the hydrogen (1 g /mol)
1 kg = 1000 g
1000 g / 46.07 g/mol = 21.7 moles of ethanol in 1 kg
Hydrogen has a mass of 1 g/mol, but hydrogen gas has a mass of 2g/mol
There are six hydrogens in each molecule of ethanol, so each molecule will make 3 molecules of H2 gas or 3 moles of hydrogen per mole of ethanol.
21.7 moles * 3g per mole H2 = 65.12 grams of hydrogen gas H2

Question 9 of 13

2.5/ 5.0 Points

What is the limiting reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of Li3PO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2

3 mol Ca(NO3)2 reacts with 2 mol Li3PO4
Assuming all the Ca(NO3)2 is used up:
3.4 mol Ca(NO3)2 x (2 mol Li3PO4 / 3 mol Ca(NO3)2 ) = 2.27 mol Li3PO4
Since there are more moles of Li3PO4 than the 2.27 moles required to use up all the Ca(NO3)2,

Li3PO4 is present in excess. This means the Ca(NO3)2 is the limiting reagent.

Comment: See attached document for the correct answer.

MT 2.docx

(70 KB)

Question 10 of 13

5.0/ 5.0 Points

What is the limiting reagent and percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actual amount of CuS produced was 3.05. Reaction: Na2S + CuSO4 → Na2SO4 + CuS

Molar Mass of ( Na2S ) = 78.06 g/mol

15.5 g / 78.06 = 0.198 mole Na2S

MM of ( CuSO4) = 159.64 g / mol
12.1 g / 159.64 = 0.0758 mole CuSO4
the reaction is
Na2S + CuSO4 ---> Na2SO4 + CuS
the ratio between Na2S and CuSO4 is 1 : 1 so CuSO4 is the limiting reactant.
We would get 0.0758 mole CuS ( MM= 95.54 g/mol)
0.0758 mole x 95.54 g/mol = 7.24 g
3.05 : 7.24 = x : 100
x = 42.1 %

Question 11 of 13

3.5/ 5.0 Points

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

Formula Weight of Al = 27g/mol
Formuls Weight of O2 = 32g/mol
nAl = 10g / 27g/mol = 0.37 mol Al
nO2 = 19g / 32 g/mol = 0.59 mol O2
By taking the moles of each substance and dividing it by the coefficient of the balanced equation:
For Al: 0.37 / 4 = 0.09
For O2: 0.59 / 3 = 0.197
The lowest number indicates the limiting reagent. Aluminum will run out first.

Comment: See attached document for the correct answer.

Part 4 of 4 -

30.0/ 30.0 Points

Question 12 of 13

15.0/ 15.0 Points

38.67g C/12 =3.225 mol

C16.22g H/ 1 = 16.22 mol H

45.11gN / 14 = 3.22 mol N

Dividing by least

3.225/3.225 = 1 mol C

16.22/3.22 = 5 mol H

3.22/3.22 = 1 mol N

emperical formula = CH5N

Emperical formula weight = 31 g

A) given formula weight = 62.12 g

So molecular formula of compound = C2H10N2

B) 62 g contain 10 moles H atoms

3.50 g compound contains 0.564 moles = 0.564 * 6.023*10^23 atoms = 3.4 * 10^23 atoms

Feedback: This problem uses many concepts learned in Lessons 1 - 3.

Question 13 of 13

15.0/ 15.0 Points

Bornite (Cu₃FeS₃) is a copper ore used in the production of copper.

When heated, the following (unbalanced) reaction occurs:

Cu₃FeS₃(s) + O₂(g) → Cu(s)+ FeO(s) + SO₂(g)

If 2.50 kg of bornite is reacted with excess O₂ and the process has an 82.5% yield of copper, how much copper is produced (in kg)?

From balanced equation

2 moles of bornite produces 6 moles of copper

Converting given amount of bornite into moles

MM of bornite = 342.71g/mol

2500 g counts to 7.31 mole

So amount of copper produced = 7.31 * 3 = 21.93 moles

MM of copper = 63.55 g/mol

Mass of copper = 21.93*63.55 = 1.39 kg

Given % yield of copper is 82.5%

Final amount of copper produced = 1.146 kg

Feedback: The problem requires many steps from chapters 6-8.

Part 1 of 4 - Short Answer

15.0/ 15.0 Points

Please show as much work as possible for full credit.

Question 1 of 13

5.0/ 5.0 Points

A log is either oak or pine. It displaces 2.7 gal of water and weighs 19.8 lb. Is the log oak or pine? (Density of oak = 0.9 g/cm³; Density of pine = 0.4 g/cm³)

The density of an object is mass/ volume, so we convert gallons to liters (2.7x3.87)= 10.206 liters.

That is the volume.

Then calculating the weight in metrics (8.98112 kg) and 8.98112 kg/10.206liters to equal 0.9

The answer is oak

Question 2 of 13

5.0/ 5.0 Points

A room contains 48 kg of air. How many KWh of energy are necessary to heat the air in the house from 7^oC to 28^oC? The heat capacity of air is 1.03 J/g^oC.

We need 1.03 kJ for 1 kg air and a temperature rise by 1°. The heat energy needed for 48 kg air and a temp. rise by 21° is then

E = 1.03*48*21 kJ = 1038.24 kJ = 1038.240 kW*s (kWs). with 1 J = 1 Ws

To find this amount of energy expressed in kWh, we have to divide by the number of seconds in an hour, which is 60^2 = 3600:

1038.24 kWs = 1038.24/3600 kWh = 0.2884 kWh

Question 3 of 13

5.0/ 5.0 Points

Give 2 examples of a Physical Property. Give 2 examples of a Chemical Property.

Examples of Physical Properties

A physical property is any property of matter or energy that can be measured. It is an attribute of matter that can be observed or perceived.

Common Physical Properties

Absorption of electromagnetic - The way a photon’s energy is taken up by matter

Absorption (physical) - Absorption between two forms of matter

Color - Hue of an object as perceived by humans

Concentration - Amount of one substance in a mixture

Chemical properties can only be established by changing a substance’s chemical identity, and are different from physical properties, which can be observed by viewing or touching a sample.

The internal qualities of a substance must be altered to determine its chemical properties. For example:

Flammability - How easily something will burn or ignite, is a chemical property because you can’t tell just by looking at something how easily it will burn. Fire testing is done to determine how difficult or easy it will be to get a certain material to burn.

Information about flammability is used in building codes, fire codes, insurance requirements, and storing, handling, and transporting highly flammable materials.

Heat of Combustion - This chemical property is the amount of energy that is released as heat when a substance is burned with oxygen.

Part 2 of 4 -

10.0/ 10.0 Points

Question 4 of 13

5.0/ 5.0 Points

Write the chemical formula for each of the following ionic compounds: Copper (II) Bromide, Silver (II) Nitrate, Sodium Sulfate

Copper (II) Bromide CuBr2

Silver (II) Nitrate Ag(NO3)2

Sodium Sulfate Na2SO4

Feedback: CuBr2
Ag(NO3)2
Na2SO4

Question 5 of 13

5.0/ 5.0 Points

Write the chemical formulas for the following: Calcium Nitrate, Potassium Sulfate, Ammonium Hydroxide.

Calcium Nitrate is Ca(NO₃)₂

Potassium Sulfate is K₂SO₄

Ammonium Hydroxide is NH₄OH

Feedback: These can be found in chapter 5.

Part 3 of 4 -

30.0/ 30.0 Points

Question 6 of 13

5.0/ 5.0 Points

Sodium peroxide (Na2O2) reacts with water to form sodium hydroxide and oxygen gas. Write a balanced equation for the reaction and determine how much oxygen in grams is formed by the complete reaction of 35.23 g Na2O2.

First converting grams of sodium peroxide to moles of sodium peroxide using the molar mass:
35.23g Na2O2 x (1mole Na2O2 / 77.98g Na2O2)= 0.4518moles Na2O2
We know that for every 2 moles of Na2O2 in the reaction, there is one mole of O2.
0.4518 moles Na2O2 x (1 mole O2 / 2 moles Na2O2) = 0.2259 moles O2
Converting to grams using molar mass:
0.2259 moles O2 x (32g O2 / 1 mole O2) = 7.228g O2

Question 7 of 13

5.0/ 5.0 Points

A 9.56-g sample of aluminum completely reacts with oxygen to form 13.34 g of aluminum oxide. Use this data to calculate the mass percent composition of aluminum in aluminum oxide.

If all the Al completely reacts then it is all now present in the Aluminium oxide. So there are 9.56 g Al in the aluminium oxide.

mass % Al = mass Al / total mass x 100/1
mass % Al = 9.56 g / 13.34 g x 100
= 71.66 %

Feedback: Please see Chapter 6 for help with calculating mass percents of compounds.

Question 8 of 13

5.0/ 5.0 Points

Determine the correct empirical formula of a compound containing 26.68% carbon, 2.24% hydrogen, and 71.08% oxygen.

Carbon = 12g/mol
Hydrogen = 1.001g/mol
Oxygen = 16g/mol
Moles of carbon = 26.68g*1mol/12g which gives 2.22 moles
Hydrogen = 2.24g*1mol/1.001 g which gives about 2.24 moles
Oxygen = 71.08*1mol/16g which gives 4.44 moles
Then we divide all the values by the fewest number of moles you got to find out how much of each is in the empirical formula which is carbon so divide everything by 2.22.
2.22/2.22 = 1 C
2.24/2.22 = ~1 H
4.44/2.22 = 2 O
So the empirical formula for oxalic acid is CHO2.

Feedback: Please see Chapter 6 for a problem like this.

Question 9 of 13

5.0/ 5.0 Points

What is the limiting reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of Li3PO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2

3 mol Ca(NO3)2 reacts with 2 mol Li3PO4
Assume all the Ca(NO3)2 is used up:
3.4 mol Ca(NO3)2 x (2 mol Li3PO4 / 3 mol Ca(NO3)2 ) = 2.27 mol Li3PO4
Since there are more moles of Li3PO4 than the 2.27 moles required to use up all the Ca(NO3)2, Li3PO4 is present in excess. This means the Ca(NO3)2 is the limiting reagent.

Question 10 of 13

5.0/ 5.0 Points

Determine the empirical formula of a compound containing 60.3% magnesium and 39.7% oxygen.

To get from weight % to relative numbers of atoms each weight % is divided by its respective atomic weight:
60.3/24.3 = 2.48 Mg
39.7/16.0 = 2.48 O
The relative numbers of atoms are in a 1:1 ratio, so the formula is MgO

Feedback: Please see Chapter 6 for a problem like this.

Question 11 of 13

5.0/ 5.0 Points

Determine the empirical formula of a compound containing 83% potassium and 17.0% oxygen.

83g/39.0983g/mol = 2.1282mol
17g/15.9995g/mol = 1.0625 mol
Number of Potassium = 2.1282mol/1.0625 mol = 2
Number of Oxygen = 1.0625mol/1.0625mol = 1.
Emperical Formula = K2O

Feedback: Please see Chapter 6 for a problem like this.

Part 4 of 4 -

30.0/ 30.0 Points

Question 12 of 13

15.0/ 15.0 Points

Bornite (Cu₃FeS₃) is a copper ore used in the production of copper.

When heated, the following (unbalanced) reaction occurs:

Cu₃FeS₃(s) + O₂(g) → Cu(s)+ FeO(s) + SO₂(g)

If 2.50 kg of bornite is reacted with excess O₂ and the process has an 82.5% yield of copper, how much copper is produced (in kg)?

From the balanced equation

2 moles of bornite produces 6 moles of copper

Converting given amount of bornite into moles

MM of bornite = 342.71 g/ mol

2500 g counts to 7.31 mles

So amount of copper produced = 7.31 * 3 = 21.93 moles

MM of copper = 63.95 g/mol

Mass of copper = 21.93 * 63.95 = 1.39 kg

Given % yield = 82.5 %

So amount of copper produced = 1.146 kg

Feedback: The problem requires many steps from chapters 6-8.

Comment: excellent!

Question 13 of 13

15.0/ 15.0 Points

Moles of C in the compound = 38.67 / 12 = 3.223 moles
moles of H = 16.22 / 1 = 16.22 moles
moles of N = 45.11 /14 = 3.223
molar ratios in the compound C:H:N = 3.223 :16.22: 3.223
divide by the smallest number 3.223 and we get 1:5:1
empirical formula is CH5N
empirical mass is ( 12+5+14) = 31 g
molar mass is 62.12 g so there are 2 empirical units in a mole of compound
molecular formula is C2H10N2
moles of compound in 3.50 g = 3.50 / 62.12 = 0.056342563 moles
there are 10 moles of H in a mole of the compound so we have 0.56342563 moles of H in the mass
so number of atoms = 6.02 *10 ^23 atoms / mol * 0.56342563 mole
= 3.392 *10^23 atoms of H in the sample mass

Feedback: This problem uses many concepts learned in Lessons 1 - 3.

Comment: great!

Part 1 of 4 - Short Answer

0.0/ 15.0 Points

Please show as much work as possible for full credit.

Question 1 of 13

0.0/ 5.0 Points

`My husband and I always argue about the temperature in our house. I like it at 294.0 K, while he likes to set the thermostat to 300.0 degrees K. As we take turns constantly changing the setting, our poor heating and cooling system tries to keep up. I argue that it causes us to waste money. My living room contains 50 kg of air. In the summer time, I pay about $0.14 per kWh of energy used. How much does it cost me every time we heat the air in the living room from 294.00 K to 300 K? (HINT: How many kWh does it take?) The heat capacity of our air is 1.01 J/(gC).

All work must be shown, labeled, and explained in detail to receive credit. Note that where applicable, superscripts (X²) can be written as X^2 and subscripts (X₂) can be written as X_2 if it saves you time.

First lets convert the given temperatures into C from K, so as 0C = 273K

294K = 294-273 = 21C

300K = 300-273 = 27C

and as we need to use the mass in grams, 50kg = 50000 g

q=mc(dt),

q= amount of heat energy gained or lost by the substance

m= mass of sample

C= heat capacity

so by plugging the given values, we get,

q = 50000(1.01)*(27-21)

q = 303000 J = 0.0841666667 kWh.

As it is given that it costs $0.14 per kWh, we will be charged 0.14*0.084167 = $ 0.01178

so it costs $0.01178 every time to heat the room from 294K to 300 K

Feedback: This is a combination of Chapter 2 and section 3.12

Question 2 of 13

0.0/ 5.0 Points

You go to turn on the TV using the remote control, but the TV does not turn on. Use the scientific method to develop a strategy by which you would determine what caused the TV to not turn on. Include in your answer what you would do in the major steps of your experiment from beginning to the end.

The scientific method to discover the answer to a scientific problem needs the following steps:

Research : How a TV remote control works ? Battery and obstrucals if any between TV and remote

Problem : TV doesnt turn on using Remote Control

Hypothesis: Are there any obstrucals in the path of rays ? or batteries are dead ?

Project Experimentation: Will remove the obstrucles and will try to use the same baterries on a different device to check if they work or no.Project Conclusion: Based on these experimentation, I can determine what might be the reason the TV is not switching on.

Question 3 of 13

0.0/ 5.0 Points

The transformation of raw materials into glass happens at 2400*F. This is when the glass appears to be bright orange, as seen in glass blowing.

(1) What is this temperature in C?
(2) What is this temperature in K?
All work must be shown, labeled, and explained in detail to receive credit. Note that where applicable, superscripts (X²) can be written as X^2 and subscripts (X₂) can be written as X_2 if it saves you time

Its given that the transformation of the raw materials into glass happens at 2400F.
The relation between the Temperature in F and C is given by the following formula,
T_(°C) = (T_(°F) - 32) × 5/9, By plugging in the given values into the relation, we get,
T_{(°C) = (2400-32) x 5/9 = 1315.56 C}
The relation between the temperature in C and K is given by the following formula,
T_(°K) = T_{(°C)+273, By plugging in the given values into the relation, we get,}
T_{(°K) = 1315.56+273 = 1588.56 K}

Feedback: See section 3.10

Part 2 of 4 -

0.0/ 10.0 Points

Question 4 of 13

0.0/ 5.0 Points

For Radon-222 and Carbon-14 answer each of the following.

A) Element Symbol
B) Atomic Number
C) Number of Protons
D) Number of Neutrons
E) Mass Number
F) total Number of Electrons
You must show/explain your work in detail to receive credit.

Number of protons = Atomic number
Number of neutrons= Mass number- atomic numbe
In a neutral atom, number of electrons = number of protons.
Radon-222
Element symbol Rn
Atomic number is 86
Number of protons 86
Number of neutrons = 136
Mass number = 222
Total number of electrons = 86
Carbon-14
Element symbol C
Atomic number is 6
Number of protons 6
Number of neutrons = 8
Mass number = 14
Total number of electrons = 66

Feedback: this is covered in chapter 4

Question 5 of 13

0.0/ 5.0 Points

`Chemistry really is another language sometimes. It abbreviates a lot, and chemical formulas are a way to write a little and still convey a lot of information. For example, consider the chemicals mercury, cadmium (II) phosphate, Lead (II) hydroxide, and pentanitrogen decaoxide.
Translate these names into chemical formulas using the appropriate procedure described in the course text and my lecture material. Answer the following:
(1) What is the correct formula of each of these compounds?
(2) Explain how exactly (i.e. the procedure used) you arrived at your answer...in other words, walk me through your thinking. You will not receive credit if you don't show your work in detail.

mercury : Was previously known as hydrargyrum and thus named as Hg
cadmium (II) phosphate : Cadmium (II) is a +2, phosphate is a -3 so we get Cd₃(PO₄)₂
Lead (II) hydroxide: Lead (II) is a +2 and OH has -1 so we get, Pb(OH)₂
pentanitrogen decaoxide : Penta represents 5 and Deca represents 10, so the formula would be N₅O₁₀

Feedback: chapter 5 discusses naming

Part 3 of 4 -

0.0/ 30.0 Points

Question 6 of 13

0.0/ 5.0 Points

We have covered quite a bit so far in the course. You should be able to predict the products and write the full equation for each of the reactions below, predicting the products and balancing each equation.
a) Lithium carbonate + hydrochloric acid
b) magnesium hydroxide + phosphoric acid
c) sodium sulfate + aluminum nitrate
d) iron (III) oxide + barium chloride

a) Lithium carbonate + hydrochloric acid
2HCl + Li₂CO₃ ---> 2LiCl + H₂O+ CO₂

Products are lithium chloride + hydrogen gas + carbon dioxide gas
b) magnesium hydroxide + phosphoric acid
3Mg(OH)₂+ 2H₃PO₄ -----> Mg₃(PO₄)₂ + 6H₂O
Products are magnesium phosphate + water
c) sodium sulfate + aluminum nitrate
3Na2SO4 (aq) + 2Al(NO3)3 (aq) → Al2(SO4)3 (s) + 6NaNO3 (aq)
Products are Aluminium sulfate and Sodium nitrate
d) iron (III) oxide + barium chloride
3Fe2O3+ 3BaCl2 ---> 3FeCl2+ 3BaO

Feedback: This deals with writing chemical equations using chemical formulas and using the law of conservation to balance it.

Question 7 of 13

0.0/ 5.0 Points

If you have a compound that is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass and has a molecular weight of about 160 g/mol, determine the molecular formula. You must show all work for credit.

Molar masses of
Carbon : 12.011 g/mol
Hydrogen: 1.008 g/mo
Nitrogen: 14.007 g/mol
When we divide the given percentages by respective molar masses we get,
Carbon: 74.1/12.011 = 6.1693 mol
Hydrogen: 8.6/1.008 = 8.5317 mol
Nitrogen: 17.3/14.007 = 1.2351 mol
Now dividing each of these moles with the least one we get,
Carbon: 6.1693/1.2351 = 5
Hydrogen : 8.5317/1.2351 = 7
Nitrogen: 1.2351/1.2351 = 1
So empirical formula is C₅H₇N with a empirical mass of (60+7+14) = 81
Molecular mass = 160
so there are two empirical units which makes the molecular formula as C₁₀H₁₄N₂

Feedback: Please see calculating mass percents of compounds in Chapter 6

Question 8 of 13

0.0/ 5.0 Points

Determine the empirical formula a compound that consists of 59% C, 7.1% H, 26.2% O, and 7.7% N.

You must show all work for credit.

Molar masses of
Carbon : 12.011 g/mol
Hydrogen: 1.008 g/mol
Nitrogen: 14.007 g/mol
Oxygen : 15.999 g/mol
When we divide the given percentages by respective molar masses we get,
Carbon: 59/12.011 = 4.9122 mol
Hydrogen: 7.1/1.008 = 7.0437 mol
Nitrogen: 7.7/14.007 = 0.5497 mol
Oxygen: 26.2/15.999 = 1.6376 mol
Now dividing each of these moles with the least one we get,
Carbon: 4.9122/0.5497 = 9
Hydrogen : 7.0437/0.5497 = 13
Nitrogen: 0.5497/0.5497 = 1
Oxygen: 1.6376/0.6597 = 3
So the Empirical formula would be : C₉H₁₃NO₃

Question 9 of 13

0.0/ 5.0 Points

You are in a lab analyzing samples from a crime scene. One of the powders you are analyzing has a composition of 63.57% C, 6.00% H, 9.27% N, and 21.17% O. The molar mass of your sample is 151.16 g/mol. Determine the molecular formula for this substance.

You must show all work and calculations to receive credit.

Molar masses of
Carbon : 12.011 g/mol
Hydrogen: 1.008 g/mol
Nitrogen: 14.007 g/mol
Oxygen : 15.999 g/mol
When we divide the given percentages by respective molar masses we get,
Carbon: 63.57/12.011 = 5.2926 mol
Hydrogen: 6.00/1.008 = 5.9523 mol
Nitrogen: 9.27/14.007 = 0.6618 mol
Oxygen: 21.17/15.999 = 1.3232 mol
Now dividing each of these moles with the least one we get,
Carbon: 5.2926/0.6618 = 8
Hydrogen : 5.9523/0.6618 = 9
Nitrogen: 0.6618/0.6618 = 1
Oxygen: 1.3232/0.6618 = 2 So the Empirical formula would be : C₈H₉NO₂
So empirical formula is C₈H₉NO₂ with a empirical mass of (96+9+14+32) = 151
Molecular mass = 151.16
so there are two empirical units which makes the molecular formula as C₈H₉NO₂

Question 10 of 13

0.0/ 5.0 Points

Determine the empirical formula for a compound if it is composed of 63.56% carbon, 6.00% hydrogen, 9.27% nitrogen, and 21.17% oxygen.

Molar masses of
Carbon : 12.011 g/mol
Hydrogen: 1.008 g/mol
Nitrogen: 14.007 g/mol
Oxygen : 15.999 g/mol
When we divide the given percentages by respective molar masses we get,
Carbon: 63.56/12.011 = 5.2918 mol
Hydrogen: 6.00/1.008 = 5.9523 mol
Nitrogen: 9.27/14.007 = 0.6618 mol
Oxygen: 21.17/15.999 = 1.3232 mol
Now dividing each of these moles with the least one we get,
Carbon: 5.2918/0.6618 = 8
Hydrogen : 5.9523/0.6618 = 9
Nitrogen: 0.6618/0.6618 = 1
Oxygen: 1.3232/0.6618 = 2
So the Empirical formula would be : C₈H₉NO₂

Feedback: This deals with writing chemical equations using chemical formulas and using the law of conservation to balance it.

Question 11 of 13

0.0/ 5.0 Points

You go to the drug store and buy a 480 mL bottle of Milk of Magnesia, which has magnesium hydroxide as the active ingredient. If the density of magnesium hydroxide is 2.34 g/mL, determine the number of molecules in the bottle of Milk of Magnesia.

In order to find the number of molecules, we will have to find the total number of moles available and then multiply it with the Avogadro's number.
So to determine, number of moles, we will have to use the below formula and assume that it is in STP conditions.
To find volume, we need to make use of the mass and density given here.
As Density = mass / Volume, Volume = mass/density
so V = (480/2.34) = 205.1282 mL = 0.2051282 L
n = PV/RT = (1 atm)(0.2051282 L)/(0.08206 L atm mol^-1 K^-1 × 273.15 K) = 0.00915151 moles
To find the number of moles we need to multiply this with 6.02×10²³ molecules/mol
0.00915151 mol × 6.02×10²³ molecules/mol = 5.51 × 10²¹ molecules

Part 4 of 4 -

0.0/ 30.0 Points

Question 12 of 13

0.0/ 15.0 Points

A patient presented to the Emergency Room sweating with an elevated blood pressure and heart rate. Lab results determined that the compound used in excess had the following elemental composition and molar mass.

Element	Percent
C	59.0 %
H	26.2 %
O	26.2 %
N	7.7 %

Molar Mass 366.40 g/mol.

(1) What are the empirical formula and molecular formula for this compound?
(2) Determine how many N atoms would be in a 15.00 g sample of this compound.
All work must be shown, labeled, and explained in detail to receive credit. Note that superscripts (X²) can be written as X^2 and subscripts (X₂) can be written as X_2 if it saves you time. Be careful to account for significant figures in your final answers.

1) Molar masses of Carbon : 12.011 g/mol
Hydrogen: 1.008 g/mol
Nitrogen: 14.007 g/mol
Oxygen : 15.999 g/mol
When we divide the given percentages by respective molar masses we get,
Carbon: 59/12.011 = 4.9122 mol
Hydrogen: 26.2/1.008 = 25.9921 mol
Nitrogen: 7.7/14.007 = 0.5497 mol
Oxygen: 26.2/15.999 = 1.6376 mol
Now dividing each of these moles with the least one we get,
Carbon: 4.9122/0.5497 = 9
Hydrogen : 25.9921/0.5497 = 48
Nitrogen: 0.5497/0.5497 = 1
Oxygen: 1.6376/0.5497 = 3
So the Empirical formula would be : C₉H₄₈NO₃with a empirical mass of 218.
Its given that the molar mass is = 366.40 so after rounding, we can say that there are 2 units
so the Molecular formula : C₁₈H₉₆N₂O₆
2)
As we can see from the formula we have a total of 2 moles of nitrogen, so total number of N atoms would be 2x6.022x10^23
so it would be = 1.2044x10^24

Feedback: This problem uses many concepts learned in Lessons 1 - 3.

Question 13 of 13

0.0/ 15.0 Points

This past June, my husband and I bought a VW Passat TDI.The diesel tank on my car can hold 18.5 gallons. The average density of diesel fuel is 0.832 kg/L, which is more dense than that of gasoline.
Conversion factors that may help you: 1 gallon = 3.785L; 1 lb = 453.59 g; 1 kg = 2.2 pounds
1. When my tank is full, what is the mass of the fuel in pounds?
2. Fuels are made from long chain hydrocarbon alkanes, ranging from 8-21 carbon atoms. The average diesel fuel formula is C₁₂H₂₃ (12 and 23 are subscripts). Looking at my gas gauge, I have about 1/4 of a tank of fuel remaining (4.625 gal). How many moles of diesel fuel are in my tank?
All work must be shown, labeled, and explained in detail to receive credit. Note that superscripts (X²) can be written as X^2 and subscripts (X₂) can be written as X_2 if it saves you time. Be careful to account for significant figures in your final answers.

1. As it is given that the car tank can hold 18.5 gallons all we have to do is to convert the gallons to pounds.
as 1 gallon = 3.785 L
18.5 gallons = 3.785*18.5 L = 70.0225 L
then L is to be converted to Kg as 1L = 1Kg, we get a total of 70.0225 Kg
Finally we need to convert this Kg to Pounds as 1 Kg = 2.2 Pounds
we get 70.0225 Kg = 2.2*70.0225 = 154.0495 Pounds.
2.
Its given that only 1/4th of fuel is remaining, which is 4.625 gal = 38.512375 pounds. = 17505.625 g
Molar mass = 167 g
so number of moles of diesel fuel in the tank = 17505.625/167 = 104.824
Approximated to 105 moles.

Feedback: The problem requires many steps from Lessons 1-3

Question 1 of 13

0.0/ 5.0 Points

A room contains 48 kg of air. How many KWh of energy are necessary to heat the air in the house from 7^oC to 28^oC? The heat capacity of air is 1.03 J/g^oC.

Energy in joules = mass air x specific heat air x (Tfinal-Tinitial) = 48000*1.03*(28-7) = 1038240 J

1 J = 2.78*10^-7 KWh.

Energy necessary = 1038240 * 2.78 * 10^-7 = 0.2886 KWh.

Question 2 of 13

0.0/ 5.0 Points

Give 2 examples of a Physical Property. Give 2 examples of a Chemical Property.

A physical property is any property of matter or energy that can be measured. It is an attribute of matter that can be observed or perceived.

Common Physical Properties

Absorption of electromagnetic - The way a photon’s energy is taken up by matter

Absorption (physical) - Absorption between two forms of matter

Color - Hue of an object as perceived by humans

Concentration - Amount of one substance in a mixture

Chemical properties can only be established by changing a substance’s chemical identity, and are different from physical properties, which can be observed by viewing or touching a sample.

The internal qualities of a substance must be altered to determine its chemical properties. For example:

Information about flammability is used in building codes, fire codes, insurance requirements, and storing, handling, and transporting highly flammable materials.

Heat of Combustion - This chemical property is the amount of energy that is released as heat when a substance is burned with oxygen.

Question 3 of 13

0.0/ 5.0 Points

Hexane boils at 156.2^oF. What is this temperature in Celcius? What is this temperature in Kelvin?

T in °C = (T°F - 32) x (5 / 9)

C=(156.2-32)x(5/9)

C=(124.2)x(5/9)

C= 69 degrees celcius

T in K = T°C + 273.15

K=69+273.15

K=342.15

Part 2 of 4 -

0.0/ 10.0 Points

Question 4 of 13

0.0/ 5.0 Points

Write the chemical formula for each of the following ionic compounds: Copper (II) Bromide, Silver (II) Nitrate, Sodium Sulfate

Copper (II) Bromide CuBr₂

Silver (II) Nitrate Ag(NO₃)₂

Sodium Sulfate Na₂SO₄

Feedback: CuBr2
Ag(NO3)2
Na2SO4

Question 5 of 13

0.0/ 5.0 Points

Write the name of each of the following Ionic Compounds: Ba₃(PO₄)₂ , MgSO₄ , PbO₂

Ba₃(PO₄)₂ is Barium Phosphate

MgSO₄is Magnesium sulfate

PbO2 is Lead (IV) Oxide

Feedback: Please review making and naming ionic compounds

Part 3 of 4 -

0.0/ 30.0 Points

Question 6 of 13

0.0/ 5.0 Points

An unknown acid has a molar mass of 60.05 g/mol. Given the following percent composition, what is the molecular formula? 40%C, 6.7% H, 53.3% O

H = 1 g/mol

C = 12 g/mol

O = 16 g/mol

Then we divide A.M. / %

For the carbon:

40 / 12 ~ 3.33

For the hydrogen:

6.7 / 1 = 6.7

For the oxygen:

53.3 / 16 ~ 3.33

Then we divide the smallest result to each element:

C = 3.33 / 3.33 = 1

H = 6.7 / 3.33 = 2

O = 3.33 / 3.33 = 1

The empirical formula is:

CH₂O<o:p></o:p>

Empirical Weight = 12.011+2(1.008)+15.999= 30.026<o:p></o:p>

Molar Weight = 60.05<o:p></o:p>

So, its almost 2 times, which makes the Molecular formula as C₂H₄O₂(Molar mass = 60.052)<o:p></o:p>

Question 7 of 13

0.0/ 5.0 Points

When Sodium Peroxide reacts with water to form sodium hydroxide and oxygen gas, the balanced equation is given as follows.

2Na₂O₂ + 2 H₂O ---> 4 NaOH + O₂

We need to first convert grams of sodium peroxide to moles of sodium peroxide using the molar mass:

35.23g Na₂O₂ x (1mole Na₂O₂/ 77.98g Na₂O₂)= 0.4518 moles Na₂O₂

We know that for every 2 moles of Na₂O₂ in the reaction, there is one mole of O₂.

0.4518 moles Na₂O₂ x (1 mole O₂ / 2 moles Na₂O₂) = 0.2259 moles O₂

Convert to grams using molar mass:

0.2152 moles O₂ x (32g O₂ / 1 mole O₂) = 7.2288 g O₂

Question 8 of 13

0.0/ 5.0 Points

A chromium oxide compound contains 52.0 grams of chromium and 24.0 grams of oxygen. What is the most likely empirical formula of this compound?

The atomic mass of O is 16, and the atomic mass of Cr is 52.

the most safe way to solve this problem, is to find the moles of Cr atoms and O atoms:

moles of Cr atoms= mass of Cr/atomic mass of Cr=52/52=1mol

moles of O atoms= mass of O atoms/atomic mass of O = 24/16=1.5.

So, for every 1 atom of Cr you should have 1.5 atoms of O.

To get rid of decimals, for every 2 atoms of Cr we need 3 atoms of O.

This means that the formula should be: Cr₂O₃

Feedback: Please see Chapter 6 for empirical formulas.

Question 9 of 13

0.0/ 5.0 Points

An iron chloride compound contains 18.62 grams of iron and 17.75 grams of chlorine. What is the most likely empirical formula for this compound?

Fe 18.62/55.845 = 0.333422

Cl: 17.75/35.45 = 0.5007

divide by smallest:

0.333422/0.333422 = 1; 0.5007/0.333422 = 1. 5017

Multiply by 2 to obtain whole numbers: 2:3

i.e. Empirical Formula is Fe2Cl3

But there is no such Iron chloride exists, so I would go with FeCl3.

Feedback: Please see Chapter 6 for Empirical Formulas.

Question 10 of 13

0.0/ 5.0 Points

A 7.96 gram sample of silver reacts with oxygen to form 8.55 gram of the metal oxide. What is the formula of the oxide?

mass of oxide - mass of silver = mass of oxygen= 8.55g AgO - 7.96g Ag = 0.59g O

7.96g Ag x (1 mol Ag/107.9g Ag) = 0.074 mol Ag

0.59 O x (1 mol O/16.0g O = 0.04 mol O

Ag 0.074 O 0.04 (divide both by 0.04)

Ag 1.85 O = Ag₂O

Feedback: Please see calculating mass percents of compounds in Chapter 6

Question 11 of 13

0.0/ 5.0 Points

Determine the empirical formula of a compound containing 60.3% magnesium and 39.7% oxygen.

dividing it by its respective atomic weights:

60.3/24.3 = 2.48 Mg

39.7/16.0 = 2.48 O

The relative numbers of atoms are in a 1:1 ratio, so the formula is MgO

Feedback: Please see Chapter 6 for a problem like this.

Part 4 of 4 -

0.0/ 30.0 Points

Question 12 of 13

0.0/ 15.0 Points

calculating empirical formula

38.67g C/ 12 = 3.225 mol C

16.22g H/ 1 = 16.22 mol H

45.11g N/ 14 = 3.22 mol N

Dividing by least

3.225/3.22 = 1 mol C

16.22/3.22 = 5 mol H

3.22 /3.22 = 1 mol N

emperical formula = CH₅N

Emperical formula weight = 31g

A)Given formula weight = 62.12g

so formula of compound = C₂H₁₀N₂

B) 62 g contain 10 moles H atoms

3.50 g compund contains 0.564 moles = 0.564 * 6.023 *10^23 = 3.4 * 10^23 atoms

Feedback: This problem uses many concepts learned in Lessons 1 - 3.

Question 13 of 13

0.0/ 15.0 Points

Bornite (Cu₃FeS₃) is a copper ore used in the production of copper.

When heated, the following (unbalanced) reaction occurs:

Cu₃FeS₃(s) + O₂(g) → Cu(s)+ FeO(s) + SO₂(g)

If 2.50 kg of bornite is reacted with excess O₂ and the process has an 82.5% yield of copper, how much copper is produced (in kg)?

From balanced equation

2 moles of bornite produces 6 moles of copper

Converting given amount of bornite into moles

MM of bornite = 342.71g/mol

2500 g counts to 7.31 mole

So amount of copper produced = 7.31 * 3 = 21.93 moles

MM of copper = 63.55 g/mol

Mass of copper = 21.93*63.55 = 1.39 kg

Given % yield of copper is 82.5%

Final amount of copper produced = 1.146 kg

Feedback: The problem requires many steps from chapters 6-8.

Assignment Help

Monday 29 February 2016

SCIN131-Introduction to Chemistry-APUS-Midterm Assignment

Question 1 of 13

Question 2 of 13

Question 3 of 13

Question 4 of 13

Question 5 of 13

Question 6 of 13

Question 7 of 13

Question 8 of 13

Question 9 of 13

Question 10 of 13

Question 11 of 13

Question 12 of 13

Question 13 of 13

Question 1 of 13

Question 2 of 13

Question 3 of 13

Question 4 of 13

Question 5 of 13

Question 6 of 13

Question 7 of 13

Question 8 of 13

Question 9 of 13

Question 10 of 13

Question 11 of 13

Question 12 of 13

Question 13 of 13

No comments:

Post a Comment