3ca2+ 2130430 - WELCOME TO MS. YAJIMA'S CLASS!chembymsyajima.weebly.com/uploads/3/8/0/2/38026837/chemistry_11... · Chemistry 11 Course Review Chemistry 11—Course Review Page 15

Chemistry 11 Course Review

Chemistry 11—Course Review Page 15

Unit 5. Solution Chemistry

Class Assignments Hand-In Assignment # 15—Electrolyte and Ion Concentration Hand-In Assignment # 16—Polarity Hand-In Assignment # 17—Like Dissolves Like and Intermolecular Force

1. Determine whether the following is electrolyte or not. Provide explanation for each.

Compound( Electrolyte(or(Non1electrolyte( Reason(

Sugar& & &Table&Salt& & &Acetic&acid&(Vinegar)& & &

2. Consider 0.250 M of Ca3(PO4)2 solution.

a. Write a balanced dissociation equation for Ca3(PO4)2

b. Calculate the concentration of each ion.

[Ca2+]= ___________________M

[PO43-]= ___________________M

c. 20.0mL of 1.50M Ca3(PO4)2 solution is mixed with 80.0mL of 1.00M CaBr2. Calculate the final concentration of each ion. (Show all steps and appropriate dissociation equations.)

NE It is organic ompnd

Nacl E soluble Ionic cmpnd

CTBCOOH E soluble organic Acid

Ca3(PO4)z → 3ca2+ +

2130430.7500.5000,250

M¥-32 "

0,750

0.500

Cars ( Pkk → 3Ca2+t 213043'

Cabrz → Ca#

+ 2Br-

i 1.50N i 1. OOM

/use MiVi=MfVf

fuseMiVi-MfVffO.zooM0.900M0i600MfO.qooMO.80OM1.60MExtEa-LCa3@041zTi-1.5OMTCaCPO4kTf-7.M,

[ CaBr2]i= 1. OOM [caBrz]f= ?M

Vj = 20.0mL Vf '

20.0mL+

✓ ; = 80.0mL Vf=

80.01mL+

80.01mL20.01mL--100.04mL100.01mLIDP

IDP 3ft 3SF

35T 35T [ Cabrz ]f= (lo0M)(80'0m# = 0,800N

µF = Mi . Vi=

( 1.50N ) ( 20.0mL ) ( ioo -

0mL ) 3SF⇐ 00.0mL ) 45T

4SF

Mf = [ Ca(PO4)z]f = 01300N3SF

There are two sources of Ca2Iso[Ca2' Tf

must be added .0.900N

0.800Mt-700 M

Chemistry 11 Course Review

Chemistry 11—Course Review Page 16

[Ca2+]= ___________________M

[PO43-]= ___________________M

[Br-]= ___________________M

3. Classify whether the following molecules are polar or not by drawing Lewis diagram based on VSEPR theory.

a. H2O b. CH3CH2CONH2

(amide) c. CF4

4. Draw H2O and CH3CH2CONH2 molecules and label hydrogen bond in between.

5. What is the strongest intermolecular attraction found between the following? a) Two molecules of BH3 (Boron is electron deficient)

Lewis Diagram Type of Compound Possible Intermolecular

Attractions Strongest Intermolecular

Attraction

b) Two molecules of CH3COOH

Lewis Diagram Type of Compound Possible Intermolecular

Attractions Strongest Intermolecular

Attraction

1.700

0.600

1. 60

H : :if :

NonpolmH 11 1

Hioihµ ,!hlg%?n¥H ÷ - C - ÷ :

..

1. .

bent H : .F :

polar H polar

*¥itII¥:*'¥Ht¥.

H

B Nonpolar Covalent London London, \

H H

H - bond

Hr¥p,cy÷0iµ Polar Covalent dipole - dipole H - bond

London

Chemistry 11 Course Review

Chemistry 11—Course Review Page 17

Unit 6. Organic Chemistry

Class Assignments Hand-In Assignment # 18—Electrolyte and Ion Concentration Hand-In Assignment # 19—Polarity Hand-In Assignment # 20—Like Dissolves Like and Intermolecular Force

1. Fill in the table to summarize characteristics of all families.

Family' Name'Ending' What'it'looks'like'(Use'3'carbons)'

alkane& & &

cycloalkane& & &

alkene& & &

alkyne& & &

benzene& & &

aldehyde& & &

keton& & &

amine& & &

amide& & &

alcohol& & &

carboxylic&acid& & &

ether& & &

ester& & &

~ ane CHs- CHZ - CHZ (C3H8)

cydonane ah"Iiµ .(GHG )

structurallyFD isomer

rene ¥ 'c=¥-YEH ( Costa )@

nyne CHIC- Ctb ( C3H4 )

benzene OI ( Catto )

0~ anal

CHs.

CHZ-

EH(60+16)^

0 structuralWismer~anone#CHs'd-CHs( GOHA )NHZ

#O 1

~ anaminl ctb.CH.

CHs¢3H'7M¥~ Anamidl ctb.CH#-NHz(C3OH5NH2)

# ft~ ANOI CHs-CH-CHs ( C3HmOH )

o~anoicwacid.cat#E.a(cF5iEfFMD

^

ether Ctb - 0 - CHZCHS ( C3OH8 )

nanoate

MEHMET.su#csozHa)

Chemistry 11 Course Review

Chemistry 11—Course Review Page 18

2. Draw&the&following&organic&molecules&using&condensed&structure.'

a. 2;fluorohexanal& b. 3,4;dimethylcyclooctyne&

c. cis;2;heptene& d. diethyl&ether&

e. 3;methyl;4;chlorobutanoic&acid& f. 1,3,5;triethylbenzene'

g. methyl&ethanoate& h. cyclopentanamine'

i. butyl&propyl&ether& j. 2,2;dimethylpropanamide'

c = c

F 0

cflz¥at3I 11I ICH3-CHRCHZ-CHZ-CH -

C-µ CHZchats

\ /CHZ

-

CHZH ftCtblttatratrcttz

.

CHZ ctbqtzl°

'Atzctb

( AtzatzU

,at }

0

111CHZ- CH-CHTC-

OH #atzatz

CHZCHZ

,NHZ

$ CHC / l

a,↳/ \o/ OHZ CHZCHZ1 1CH-CHZ

Atm10\Atzutzctb

CH3' C-C-Nµz

Atzutzcttzctk tttz

Chemistry 11 Course Review

Chemistry 11—Course Review Page 19

3. Name&the&following&organic&molecules.'

a. &

b. '

c. '

d. '

e. CH3COOCH2CH2CH3&

H3C

H2C

CC

CH2

CH3

H

H

HC

CHCH2

CH2

HC

CH OHH2C

H3C

Br

Br

CH3

CH

CH2

CH

C

HC

CH

H3C

H3C

H3C

CH3

H2C

CH

H2C

CH3

C

OH

O

I

3. 5,6 - trimethyl - 2

3

1- heptyne 4

76

5

2- ethylbutauoic acid ,

2

3

4

trans - 3- hexene

2,4 - dibromo - 3- ethylcycwhexanol a

3 1

4

g6

-propyl ethauoatlparent side