Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

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Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110
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Transcript of Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

Page 1: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

Exam Tomorrow (R)

10-11:30 am

Review/Problem Session

Today, 11 am, SL 110

Page 2: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.
Page 3: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

•hypochlorous acid

•HClO•Hypo-chlorite ion

•ClO-

•chlorous acid

•HClO2•chlorite ion

•ClO2-

•chloric acid

•HClO3•chlorate ion

•ClO3-

•perchloric acid

•HClO4•Per-chlorate ion

•ClO4-

•Oxo-acid• Oxo-anion

Page 4: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• What are the names of the following compounds?

– OF2

– S4N4

– BCl3 OF2 is oxygen difluoride

S4N4 is tetrasulfur tetranitride

BCl3 is boron trichloride

Page 5: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• What are the formulas for the following binary molecular compounds?

– carbon disulfide

– nitrogen tribromide

– dinitrogen tetrafluorideThe formula for carbon disulfide is CS2.

The formula for dinitrogen tetrafluoride is N2F4.

The formula for nitrogen tribromide is NBr3.

Page 6: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Bromine has an oxoacid, HBrO2, bromous

acid (compare to HClO2, chlorous acid). What are the name and formula of the corresponding anion?

The anion corresponding to HBrO2 isbromite, BrO2

-.

Page 7: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Organic Compounds

• An important class of molecular substances; they contain carbon combined with other elements – notably hydrogen, oxygen, and nitrogen.

• Hydrocarbons contain only carbon and hydrogen.

Page 8: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• A functional group is a reactive portion of a molecule that undergoes predictable reactions.

•Examples•Name of Group

•Functional Group

•Methyl alcohol•AlcoholOH •Dimethyl ether•EtherO

•Acetic acid•Carboxylic acid

C

O

OH

Page 9: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Hydrate

• A compound that contains water molecules weakly bound in the crystals.

• The formula of a hydrate is written with a dot before the water molecule(s) included.

• For example:CuSO45H2O

Page 10: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Hydrates are named using the anhydrous (without water) compound name followed by the prefix for the number of water molecules included and the word “hydrate.”

• For example:CuSO45H2O is named

copper(II) sulfate pentahydrate.

Page 11: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• A compound whose common name is green vitriol has the chemical formula FeSO47H2O. What is the chemical name of this compound?

FeSO47H2O is iron(II) sulfate heptahydrate.

Page 12: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Calcium chloride hexahydrate is used to melt snow on roads. What is the chemical formula of the compound?

The chemical formula for calcium chloride hexahydrate is CaCl26H2O.

Page 13: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas.

• For example: 2Na + Cl2 2NaCl

• Reactants are the starting materials; they are written on the left of the equation.

• Products are the materials at the end of the reaction; they are written on the right of the equation.

Page 14: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Because a reaction must accurately describe the chemical reaction, it must be consistent with the law of conservation of mass.

• When this is not the case, after correct formulas are written for each reactant and product, the coefficients are adjusted so that the same number of each atom is present in both the reactants and the products.

• This is called balancing the equation.

Page 15: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

Figure 2.25: Representation of the reaction of methane with

oxygen.

Write a balanced chemical reaction for this combustion reaction.

CH4(g) + 2 O2 (g) 2 H2 O(l) + CO2(g)

Page 16: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Balance the following equation:

• CS2 + O2 CO2 + SO2

Tally the number of each atom on each side:C 1 on reactant side; 1 on product sideS 2 on reactant side; 1 on product sideO 2 on reactant side; 4 on product side

Begin by inserting the coefficient “2” before SO2 on the product side. We leave O2 until later because it is an element.

Page 17: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• CS2 + O2 CO2 + 2SO2

• Tally the atoms again:• C 1 on reactant side; 1 on product side• S 2 on reactant side; 2 on product side• O 2 on reactant side; 6 on product side

• Insert a “3” before O2:

• CS2 + 3O2 CO2 + 2SO2

Page 18: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• CS2 + 3O2 CO2 + 2SO2

• Tally the atoms again:• C 1 on reactant side; 1 on product side• S 2 on reactant side; 2 on product side• O 6 on reactant side; 6 on product side

• The reaction is now balanced!

Page 19: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Balance the following equation:

• NH3 + O2 NO + H2O

Tally the number of each atom on each side:N 1 on reactant side; 1 on product sideH 3 on reactant side; 2 on product sideO 2 on reactant side; 2 on product side

Begin by inserting the coefficient “2” before NH3 on the reactant side and the coefficient “3” before H2O on the product side. We leave O2 until later because it is an element.

Page 20: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• 2NH3 + O2 NO + 3H2O

• Tally the atoms again:• N 2 on reactant side; 1 on product side• H 6 on reactant side; 6 on product side• O 2 on reactant side; 4 on product side

• To balance N, insert a “2” before NO:

• 2NH3 + O2 2NO + 3H2O

Page 21: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

•2NH3 + O2 2NO + 3H2O

•Tally the atoms again:N 2 on reactant side; 2 on product side

H 6 on reactant side; 6 on product side

O 2 on reactant side; 5 on product side

•Since this gives us an odd number oxygens, we double the coefficients on NH3, NO, and H2O and to balance O, insert a “5” before O2.

Page 22: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

Tally the atoms again to double check:

4NH3 + 5O2 4NO + 6H2O

N 4 on reactant side; 4 on product side

H 12 on reactant side; 12 on product side

O 10 on reactant side; 10 on product side

•The reaction is now balanced!

Page 23: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• Balance the following equation:

• C2H5OH + O2 CO2 + H2O

Tally the number of each atom on each side:C 2 on reactant side; 1 on product sideH 6 on reactant side; 2 on product sideO 3 on reactant side; 3 on product side

Begin by balancing H. Insert the coefficient “3” before H2O on the product side. We leave O2 until later because it is an element.

Page 24: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• C2H5OH + O2 CO2 + 3H2O

• Tally the number of each atom on each side:

• C 2 on reactant side; 1 on product side

• H 6 on reactant side; 6 on product side

• O 3 on reactant side; 5 on product side

• To balance C, insert a “2” before CO2.

Page 25: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• C2H5OH + O2 2CO2 + 3H2O

•Tally the number of each atom on each side:

•C 2 on reactant side; 2 on product side

•H 6 on reactant side; 6 on product side

•O 3 on reactant side; 7 on product side

•To balance O, insert a “3” before O2.

Page 26: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

• C2H5OH + 3O2 2CO2 + 3H2O

•Tally the number of each atom on each side:

•C 2 on reactant side; 2 on product side

•H 6 on reactant side; 6 on product side

•O 7 on reactant side; 7 on product side

•The reaction is now balanced!

Page 27: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

•Polymer

•A very large molecule that is made up of a number of smaller molecules repeatedly linked together.

•Monomers

•The small molecules that are linked together to form a polymer.

C C

H

HH

H

C C

H

HH

H

C C

H

HH

H

C C

H

HH

H

C C

H

HH

H

Page 28: Exam Tomorrow (R) 10-11:30 am Review/Problem Session Today, 11 am, SL 110.

Other Resources

• Visit the student website at college.hmco.com/pic/ebbing9e