New Way Chem 1

8/10/2019 New Way Chem 1

1/22

Errata to B ook 1 (First Ed ition, 1997)

Original

P.4 2nd paragraph, l ine 1

... the mass/charge ( m /e ) ratio and the velocity of the particles ...

P.9 Bot tom of Table 1-3

*Note: Fluorine has no isotope.

P.11 Table 1-4

Amendment

P.12 1st paragraph, line 2

... nucleus of the uranium ( ) atom 92U ...

P. 13 Fig. 1-23 (b)

P.14 2nd paragraph, line 1 It has no mass and ...

P.15 Last paragraph, l ine 1

When the tritium nucleus loses an electron,one of the ... into a proton (Fig. 1-25).

P.17 Fig. 1-27

y-axis: Atomic mass

P.17 Last paragraph, line 4

... unknown in nature, i.e. all the presently knownelements which are not found in nature. Most of them with atomic numbers greater than 92, wereformed in bombardment reactions withaccelerated particles.

... nucleus of the uranium ( ) atom 92238 U ...

... the mass/charge ( m /e ) ratio of the particles ...

*Note: Beryllium and fluorine have no isotopes .

It has negligible mass and ...

During the reaction , one of the ... into a proton(i.e. 0

111

10n p e + ). At the same time, an

electron is emitted (Fig. 1-25).

y-axis: Mass number

** There should not be a white spot in the middleof the photograph.

Neutron n or n Neutron 0

1n

orn

... unknown in nature. Most of these isotopes formedare of atomic numbers greater than 92.


2/22

63Part 1

P.19 Check point 1-1, par t (b)

92232 Th ...

P.24 Fi g. 1-36(b)

90232 Th ...

P.26 Lower middle part of page

The relative atomic mass ... can be found in

the Periodic Table ... of this book.

P.29 Sect ion 1.5, 3., table

The relative atomic mass ... can be found in

Appendix I of this book.

The volatile liquid is stored in the ... with steam.

P.46 M i ddl e of page

The volatile liquid, trichloromethane(chloroform), is stored in the ... with steam.

P.55 Tabl e 2-2

35 37 50 5251

93

123

40

2

mass/charge

R e l a t i v e a b u n d a n c e

Type of radiation

... -parti cle ... Type of radiation

... - radiation ...

. . .. . .. . .Silver AgNO 3(aq ) 96 500Sodium NaCl( aq ) 96 500

. . .. . .. . .Silver AgNO 3(aq ) 96 500

**Delete the row of Sodium.

35 37 50 5251

93

123

40

2

mass/charge

R e l a t i v e a b u n d a n c e


3/22


P.61 Q23

How much ammonia could ... conditions? Assuming that the reaction between nitrogenand hydrogen goes to completion, how muchammonia could ... conditions?

P.77 Check Point 3-7, part (c)

(c) 20 cm 3 of a gaseous hydrocarbon and150 cm 3 of oxygen were exploded ...

P.78 Bottom of page

... the end point ( ) of the titration. ... the equivalence point ( ) of thereaction .

Titration with an Indicator

The equivalence point of a titration ... from theacid. The point at which the indicator changescolour is called the end point of the titrationwhich indicates the completion of thereaction. We should choose an indicator

whose end point matches with equivalencepoint of the reaction.

P.67 Example 3-3, Solution

Therefore, the molecular ... is C2H6.

P.69 Bottom of Example 3-5

Therefore, the molecular ... is either (CH 3)2 orC2H6.

P.70

(c) 20 cm 3 of a gaseous hydrocarbon and 150 cm 3

of oxygen (in excess) were exploded ...

Relative number of moles ...25

6.25 4 Relative number of moles ...

256.25

= 4

49 38

16 0

.

. =3.08

3 70

3 08

.

.

1.2 308

308

.

. =16 18

3 08

.

.

2

Numberof ... ...moles

Relative

numberof moles

Numberof ... ...moles

Relative

numberof moles

49 38

16 0

.

. = 3.09

3 70.

3.09

1.2

6 18.

3.09 = 2 3.09

3.09 =1

Titration with an Indicator

The end point of a titration ... from the acid. The indicator is a dye that changes colour atthe end point.


4/22

65Parts 1 , 2

P.79 Table 3-3

1. (a) 7 214 H He + 4 8

17 O +_________

(b) 13 H 2

3 He +_________

(c) 11 223 Na He + 4 12

26 Mg +_________

(d) 1 12 H H + 2 2

3 He +_________

P.84 Fig. 3-8(a) , lef t-hand side

(b) 8.54 g of hydrated iron (II) sulphate ...

Acid-base Pair pH at End Point Indicator Acid-base Pair

pH at Indicator

Equivalence Point

P.86 Check Point 3-8, par t (b)

(b) 8.54 g of iron (II) sulphate ...

P.87 Secti on 3.6, 4

The end point of an acid-base ...

P.91 Q1

1. (a) 14N + 4He 17 O +_________

(b) 3H 3He +_________

(c) 23Na + 4He 26Mg +_________

(d) 2H + 2H 3He +_________

P.91 Q3

3. (c) Sulphur has the formula S 8 underaverage conditions.

P.93 Q13

13. (c) ... crystallization per molecule of ...

P.106 Fi g. 4- 4

P.107 Fi g. 4- 6

P.108 Table 4-2, caption

Table 4-2 Wavelengths ... atomic hydrogen

3. (c) Sulphur has the formula S 8 under roomconditions.

13. (c) ... crystallization per unit of ...

**Delete the column of "Cosmic rays".

12 V d.c.

The equivalence point of an acid-base ...

**Exchange the diagrams in Fig. 4-6(a) andFig. 4-6(b).

Table 4-2 Wavelengths ... atomic hydrogen(measured in nm)

12 V a.c.


5/22


P.110 Top of page

Fig. 4-10 shows how a line ... betweendifferent energy levels.

P.111 Fig. 4-11

Fig. 4-10 shows the possible transitions of anelectron between different energy levels.

P.113 1st paragraph, li ne 6

... brought to higher energy levels. The unstableelectrons ... return to the ground state. This results

in ...

P.114 1st paragr aph

By observing ... (Fig. 4-14).

P.114 Fi g. 4-14

P.118 3rd paragraph, li ne 4

... that it is easier to remove the third and ... havedifferent energy states. This leads to ...

** Delete the paragraph.

... brought to higher energy levels. These electronsare unstable and will return to the ground state.

The excess energy is given out as radiation of definite wavelengths. This results in ...

** Delete Fig. 4-14.

n =

n = 7n = 8

n = 6n = 5n = 4

n = 3

n = 2

n = 1

Lymanseries

Balmerseries

Paschenseries

rgyls

ogenm

...

n =

n = 7n = 8

n = 6n = 5n = 4

n = 3

n = 2

n = 1

Lymanseries

Balmerseries

Paschenseries

Energylevelsof ahydrogenatom

...

... that it is easier to remove the outermost shellelectrons of Group III and Group VI elements.

This means electrons in the same shell of differentelements have different energy states. This leads to...


6/22

67Part 2

P.120 Middle of page

The symbols given for ... respectively. Thenumber of orbitals in ...

P.134 Tabl e

The symbols given for ... respectively. Theycorrespond to the spectral lines in the atomicspectra. They are:

s for sharp, p for principal,

d for diffused, f for fundamental

The number of orbitals in ...

P.135 Check Point 5-2, part (a)

(a) (i) Gold 79Au; and

(ii) Uranium 238 U.

P.138 4th paragraph They have either ... (see Groups IB & IIB).

P.139 Fig. 5-7

: The general formula ... f irst 36 elements inPeriodic Table.

P.140 Bottom of page and P.141 top of page

... enthalpy is the size of their atoms. As the size

... of each element decrease down a group.

(a) (i) Silicon; and

(ii) Copper.

They have either ... (see Groups IB & IIB). It shouldbe noted that the electronic configuration of Group VIB is [ ]ns 1(n-1)d 5 but not [ ] ns 2(n-1)d 4.

The general formula ... first 36 elements in Periodic Table. Note that the electronic configuration of

Group VIB is [ ]ns 1(n-1)d 5 but not [ ]ns 2(n-1)d 4.

Atomic ElementNumber

33 Antimony

Atomic ElementNumber

33 Arsenic

... enthalpy is the size of their atoms. Going down agroup, there is an increase in the number of electron shells. As the size of atoms increases downthe group, the outermost electrons of these atomswould be further away from the nucleus.Therefore, the ionization enthalpies of individual

atoms of each element decrease down a group.


7/22


Mg+(g ) Mg2+(g ) + e

Mg2+ (g ) Mg3+(g ) + e

Mg3+ (g ) Mg4+(g ) + e

P.144 Bottom of page

Mg+(g ) Mg 2+(g ) +2 e

Mg2+ (g ) Mg3+ (g ) +3 e

Mg3+ (g ) Mg4+ (g ) +4 e

P.146 Example 5-1

For the element carbon 1C ,

P.147 Check Point 5-4, part (a)

(a) Give ... 26Fe.

P.148 last paragraph, line 5

... electrons originally present in the atom(screening effect). This would ...

P.150 Point 5.4, line 6

4. ... orbitals have extra stability.

P.152 Q19

(b) the alkali metal without 4p electrons;

P.154 Q8

(c) Explain the general trend of the f irstionization enthalpies with respect to ...

(d) Explain the irregularity of ... the f irst

ionization enthalpies.

P.154 Q11

(d) A doubly ... 1 s 22s 22p 23s 23p 63d 5.

P.155 Q12

Suppose that the Pauli principle indicated ...

For the element 612 C ,

(a) Give ... 26 Fe.

... electrons originally present in the atom. Thiswould ...

4. ... degenerate orbitals have extra stability.

(b) the alkali metal with 19 electrons;

(c) Explain the general trend of the first ionizationenthalpies from Li to F.

(d) Explain the irregularity of ... the first ionization

enthalpies from Li to F.

(d) A doubly ... 1 s 22s 22p 63s 23p 63d 5.

Suppose that the Pauli's Exclusion Principleindicated ...


8/22

69Part 3


... Its value depends on the local environment ...of molecule is called the mean bond enthalpy ...

P.171 Table 6-3

... I ts value depends on the specific local environment

... of molecules having the particular bond is calledthe mean bond enthalpy ...

NO 3 SO 4

2

Zn2+ 839 814

NO 3 SO 4

2

Zn2+ 83.9 81.4

Substance H f o (kJ mol 1)

Cu ( s ) 48.5

Substance H f o (kJ mol 1)

CuS ( s ) 48.5

The standard enthalpy ... of hydrogen ( H c o [H2(g )])

is ...

P.173 Table 6-4

P.174 Upper part of page

2. ... The standard enthalpy ... of hydrogen ( Hc o )

[H2( g ) ], is ...

P.181 Check Point 6-4, part (a), li ne 8

(a) ... aqueous ammonia

P.182 Check Point 6-4, part (c)

(c) A student used ... in Fig. 6-13 ...

P.187 Example 6-5

C(s )(graphite) +O 2(g ) CO 2(g ) Hf

o [CO 2(g )]

H2(g ) +1

2O2(g ) H2O(l )

Hf o [H2O(l )]

(a) ... aqueous ammonia. (Density of water= 1 g cm -3)

(c) A student used ... in Fig. 6- 14 ...

C(s )(graphite) +O 2(g ) CO 2(g ) H c

o [C(s )]

H2(g ) +1

2O2(g ) H2O(l )

H c o [H2(g )]


9/22


P.188 Top of page

Note: H1 = Hf o [CO 2(g )]

H2 =2 Hf o [H2O(l )]

.

..= Hf

o [CO 2(g )] +2 Hf o [H2O(l )]

P.190 Example 6-9

Making use of Hess's law,

H o =2 H2 H1

P.194 Q 20, l ine 3 ... of 1 560 J kg 1 K 1. The ...

P.197 Q 13( c)

NaCl( s ) Na +(g ) +Cl (g )

P.211 Fig. 7-8

Note: H1 = H c o

[C( s )]

H2 = 2 H c o [H2(g )]

.

..= H c

o [C( s )] + 2 H c o [H2(g )]

Making use of Hess's law,

H o = H2 H1

... of 1 560 J K 1. The ...

H72

Li52

Na71

B29

C120

Si180

N3

P70

O142

S200

O+844

S+532

F348

Cl364

Br342

I314

Be

Na +(g ) + Cl (g ) NaCl( s )

H72

Li

Na

B C

Si

N

P

O

S

O

S

F

Cl

Br

I

Be59.8 1240 29 122 3 142 +780 322

52.9 120 74 200 +590 348

324

295

P.212 Upper part of page

... H o lattice [Na+Cl (s )] ... H o = H o lattice [Na

+Cl-(s )]


10/22

71Part 4

P.219 Check point 7-2

(a) List two ... of the enthalpy change of formation of the ionic compound.

P.229 Q15

15. Explain the following terms with suitableequation(s).

P.215 Example 7-1

(b) Calculate a lattice enthalpy for this oxide of boron.

P.215 Solution

(a)

(a) List two ... of the lattice enthalpy of an ioniccompound.

3B ( ) , 3O ( )3+ 2g g

3O( ), 2Al ( )g g 3+ H 4 = 3 ( 142 + 884)

= 2 226 kJ mol 1

(b) Calculate the lattice enthalpy for aluminiumoxide.

( ) , 3O ( )2g g

2AI3+

2AI3+

( ) , 3O( )g g H 4 = 3 ( 142 + 884)

= 2 226 kJ mol 1

(b) ... =15 541 kJ mol -1

P.218 Lower middle part of page

+3 H o atom [12 Cl2(g )] +3 HE.A.

P.219 Fig. 7-10, bottom left

(b) ... = -15 541 kJ mol -1

+3 H o atom [12 Cl2(g )] +3 ..

1 000

1 000

0

H [MgCl( )]f s

Formation of MgCl

H [Cl( )]E.A. g

H [Mg( )]atom s

1 H [Mg( )]st I.E. g

12

H [ Cl ( )]atom 2 g

H [MgCl( )]lattice s

1 000

1 000

0

H [MgCl( )]f s

Formation of MgCl

H [Cl( )]E.A. g

H [Mg( )]atom s

1 H [Mg( )]st I.E. g

12

H [ Cl ( )]atom 2 g

H [MgCl( )]lattice s

15. Explain the following terms with a suitableexample .


11/22


P.230 Q17

P.230 Q19

(b) From the result in (a), deduce ... not.

P.236 Last paragraph, li ne 4

... lying 3 d orbital for bond ...

P.237 First paragraph, line 2

... The presence of the low-lying d -orbital ...

P.240 Fig. 8-14

(b) From the result in (a), comment on the stabilityof the compound XY 2(s).

FCa 2+ FCa 2+

... lying vacant 3d orbital for bond ...

000

000

000

000

000

0C H2 6 C H6 14C H5 12C H4 10C H3 8CH 4

CH OH3C H OH2 5

C H OH3 7C H OH4 9

C H OH5 11C H OH6 13


... The presence of the low-lying vacant d -orbital ...

1 000

2 000

3 000

4 000

5 000

0C H2 6 C H6 14C H5 12C H4 10C H3 8CH 4

CH OH3C H OH2 5

C H OH3 7C H OH4 9



S t a n d a r d e n t h a l p y c h a n g e o f c o m b u s t i o n (

k J m o l )

1


12/22

73Part 4

P.240 Last paragraph, line 3

... additional t CH2 t group. This occurs as ...with molar mass.

P.242 Table 8-2

... additional t CH 2 t group. This shows that theamount of energy associated with a particularbond is more or less constant.

Bond

...C v CC t C...

Bond

...C v CC t Cl...

P.243 Bottom of page

C(graphite)( s ) +2H 2(g ) ...

C(g ) ...

2. From energy cycle and Hesss law

Given: H1 ...

P.245 Mi ddle of page

Energy required Energy released

=E(C u C) +E(H t H) =E(C t C) +2E(C t H)

=606 +431 =347 +2(413)

=1 037 kJ mol 1 = 1 173 kJ mol 1

Hf o =1 073 +(1 173)

=136 kJ mol 1

P.248 Fir st paragraph, li ne 1

The covalent radius is the space occupied by anatom in a ...

2. From energy cycle and Hess's law

C(graphite)( s ) +2H 2(g ) ...

C(g ) ...

Given: H1 ...

Energy required Energy released

=E(C u C) +E(H t H) =E(C t C) +2E(C t H)

= 612 + 436 = 348 + 2(412 )

= 1 048 kJ mol 1 = 1 172 kJ mol 1

Hf o = 1 048 +( 1 172 )

= 124 kJ mol 1

The covalent radius is the distance between thenucleus and the bonding electrons in a ...


13/22


P.249 Example 8-1, Solution

(a) Both CCl 4 and ... is resulted.

**Move Example 8-1 to P.254, before Example8-2.

(a) Explain why CCl 4 is tetrahedral in shape butNCl

3 is not.

(a) **Delete the sentence "The following data ...Cl 2."

(i) Predict the approximate bond length of SiH4 ......

(Hint: Assume that bond length isadditive. )

(ii) Assume ... observed ?

(a) In CCl 4, there are four bond pairs of electrons. The bond pairs have to stay as faraway as possible. They take up the shape of atetrahedron and so CCl 4 is tetrahedral inshape. The four electron pairs in NCl 3 takethe shape of a tetrahedron as well. However,one of the electron pairs is a lone pair and theother three are bond pairs. To account for theshape of NCl 3, we do not consider the lonepair. The shape of NCl 3 is trigonal pyramid.

(b) BCl 3 has six electrons round the central atom...

... orbitals and pairing of electrons. The strongestbond will be formed when the atoms approacheach other in such a way that there is an overlapof their atomic orbitals and the potential energy

of the system is a minimum. Hence, a covalent bond...

P.249 Example 8-1

(a) Explain the fact ... shapes.

(b) BCl 3 has six electrons (boron atom is sp 2

hybridized) round the central atom ...

P.250 Check Point 8-4 (a) The following data ... Cl 2.

(i) Predict the approximate bond lengthand bond enthalpies of SiH 4 ......

(Hint: Assume both bond length andbond enthalpy are addictive)

(ii) Assume ... observed.

P.250 Lower middle part of page, li ne 3 of paragraph

... orbitals and pairing of electrons. The more ...their atomic orbitals. Hence, a covalent bond ...


14/22

75Part 4

P.251 Mi ddle and lower part s of page


... the lone pair and bond pair will stay furtheraway than the two bond pairs. ...

P.256 Fig. 8-26

** Change all "bond pair electrons" to "bondpairs of electrons".

120

120C C

H

H

H

H

... the lone pair will stay further away than thethree bond pairs. ...

C C

H

H

H

H

121 - 122 (>120 )

118 - 119 (


15/22


P.269 F ig. 9- 6

________ Ionic bond

________ Ionic bond with covalent character

________ Polar covalent bond

________ Covalent bond

P.275 Section 9.1, point (3), l ine 2

... implies the presence of pure ionic bonds whilea large ...

P.277 Q14 (a) ... ionic radii of the Group IIA ions?

P.277 Q15

15. Fine jets of liquid PCl 3 and ...

P.277 Q16

16. Compare the sizes of the ions:

F 0.136 nm Al 3+ 0.050 nm

Cl 0.181 nm...

Based on the ... halides is rare.

P.277 Q19

... the theoretical value of lattice enthalpy of calcium fluoride with the ... value?

P.277 Q21

21. Write a short essay on electroregativity.


... The strength of the weak intermolecular forces

(or van der Waal's forces) affects ...

________ Covalent bond

________ Polar covalent bond

________ Ionic bond with covalent character

________ Ionic bond

... implies that the compound is almost purely ionicwhile a large ...

(a) ... ionic radii of the Group IIA elements ?

15. Fine jets of liquid CHCl 3 and ...

16. Consider the sizes of the ions:

Cl 0.181 nm Al3+ 0.050 nm...

Based on the ... halides do not exist .

... the theoretical lattice enthalpy of a substancewith the experimentally determined value?

21. Write a short essay on electronegativity .

... The strength of the weak intermolecular forces

affects ...


16/22

77Part 4

P.290 Fi g. 11- 3

P.290 Fi g. 11- 2

P.291 Fi g. 11- 5

He

H H

+ -

+ -

Instantaneousdipole

Instantaneousdipole

He

H H

+

+

Instantaneousdipole

Instantaneousdipole

-

-

d+ d-

He

Induceddipole

Permanent dipolein polar molecule

Induced

dipole innon-polarmolecule

+

Induceddipole

He

-

Permanent dipolein polar molecule

Induced

dipole innon-polarmolecule


17/22


P.291 Last paragraph, l ine 5

... which hold the noble gas atoms together ...

P.292 Fig. 11-6(b)

... which hold these non-polar molecules together ...

... For example, the two isomers with molecularformula C 5H12 have ...

+ ++ +


... For example, the two isomers of pentane(C5H12) have ...

P.293 1st pargraph, li ne 3

... a definite interval of time, which is largecompared with the instantaneous time. But theaverage ...

P.294 Middle of page

Fig. 11-7 Two isomers of pentane


(b) Name a substance which ... in this substance.


... include iodine, napthalene, solid ...

P.300 2nd paragraph, line 3

... hydrogen. But there are three C t Cl bonds ...polarization of the C t H bond increases. The ...

... a definite internal of time, but the average ...

Fig. 11-7 Two isomers of C5H12

(b) Plastics are substances which have extra-ordinarily strong van der Waal's forces. Explainwhy the van der Waal's forces are so strong inplastics.

... include iodine, naphthalene , solid ...

... hydrogen. Therefore, C H bond can beconsidered as non-polar. However, there are three

C Cl bonds ... polarization of the C H bondincreases. The ...


18/22

79Part 4

R: A hydrogen atom or anorganic group

n : a large positive integer

P.300 Fig. 11-16

Cl C H O C

H

H

O C H2 5

CH 3

Hydrogen bond

Cl C H O C

Cl

Cl

O C H2 5

CH 3

Hydrogen bond

P.304 Diagram at bottom of page

R: A hydrogen atom or anon-polar organic group

n : A large positive integer


(a) Name the types of bonding that are ...

P.309 Section 11.2, 2

2. Dipole-dipole interaction ... whileinstantaneous ...

P.312 Q20

(c) Account for the ...


... and are free to move under the influence of anelectric field. So, ...

P.320 Example 12-1

What is the type of bonding present ... ?

P.332 Q44

(c) H o

atom [MgCl 2(s )] =... H o atom for the process =...

(a) Name the types of bonding or intermolecularforces that are ...

2. Dipole-dipole interaction is present betweenpolar molecules; dipole-induced dipole

interaction is present between a polarmolecule and a non-polar molecule, whileinstantaneous ...

**Delete Q 20 part (c).

... and are free to move within layers . So, ...

What are the type s of attractive forces present ... ?

(c) H f o

[MgCl 2(s )] =... H o for the process =...


19/22


A-6 Chapter 1

*Check Point 1-1

(b) X is 88228 X .

Y is 89228 Y .

Z is 90228 Z .

A-6 Chapter 2

*Check Point 2-2

(c) 1.6 g

*Check Point 2-3

(c) V 2 =15.06 atm

*Check Point 2-5

(b) (ii) Partial pressure of He =103 371 Nm 2

Partial pressure of N 2 =22 742 Nm2

Partial pressure of Ar =20 674 Nm 2

Total pressure =146 787 Nm 2

A-6 Chapter 3

*Check Point 3-1

(a) Empirical formula: CH 2

Molecular formula: C 3H6

Structural formula:

(b) X is 88228 Ra .

Y is 89228 Ac .

Z is 90228 Th .

(d) 1.6 g

(c) V 2 =15.06 cm3

(b) (ii) Partial pressure of He = 103 337 Nm 2

Partial pressure of N 2 = 22 752 Nm2

Partial pressure of Ar = 20 697 Nm 2

Total pressure = 146 786 Nm 2

Compound

Empirical Molecular Structural f or mul a f or mul a for mul a

(i) Propene CH 2 C3H6

(ii) Nitric acid HNO 3 HNO 3

(iii) Ethanol C 2H6O C 2H5OH orC2H6O

(iv) Glucose CH 2O C 6H12 O6

H C C

H

H

H H

HC

O

O H O N

H C C O H

H H

H H

CH OH2

OH

OH

H

H

HO

H

HHO

OH

C

H

H H

CH 3C

(a)


20/22

81A nsw ers for C hec k P oints

*Check point 3-8

(b) 55.6%

A-7 Chapter 5

*Check Point 5-2

(a) Gold: [Xe] 6 s 14f 145d 10

(b) 96.7%

[Xe]

6s 4 f 5d

[Rn]

7 s 6d 5 f

[Ne]

3 s 3 p

[Ar]

4s 3d

(b) Uranium: [Rn] 7 s 26d 15f 3

(a) (i) Si: [Ne] 3 s 23p 2

(ii) Cu: [Ar] 4 s 13d 10

A-7 Chapter 6

*Check Point 6-2

(a) (i) 76 kJ mol 1; ...

(ii) 484 kJ mol 1; ...

(iii) 417 kJ mol 1; ...

(iv) 2 081 kJ mol 1; ...

(v) 8 545 kJ mol 1; ...

*Check Point 6-4

(b) 697.2 kJ mol 1

(c) 21.84 kJ mol 1

A-8 Chapter 11

*Check Point 11-13

(a) ...

(b) ...

(c) ...

(d) ... covalent bond and metallic bond

(e) ...

(a) (i) 76 kJ; ...

(ii) 184 kJ; ...

(iii) 484 kJ; ...

(iv) 1 031 kJ; ...

(v) 8 340 kJ; ...

(b) 21.84 kJ mol 1

(c) 697.2 kJ mol 1

*Check point 11- 3

(a) (i) ...

(ii) ...

(iii) ...

(iv) ... covalent bond , metallic bond and

hydrogen bond(v) ...


21/22


21. (e) 90230 Th ...

A-9 Chapter Exer ci se 1

21. 90230 Th ...

A-9 Chapter Exer ci se 2 19. 641 cm 3

20. 320 cm 3

A-10 Chapter Exercise 3

8. 30.96%

10. Volumetric analysis

19. 320 cm 3

20. 641 cm 3

8. 30.96%

9. (a) 2CO( g ) + O 2(g ) 2CO 2(g )

(b) Na 2CO 3(s ) + 2HCl( aq ) 2NaCl( aq ) +H2O( l ) + CO 2(g )

(c) 5I (aq ) + IO 3(aq ) + 6H

+(aq ) 3I 2(aq )+ 3H 2O( l )

(d) 5Fe 2+ (aq ) + MnO 4(aq ) + 8H +(aq )

5Fe 3+ (aq ) + Mn2+ (aq ) + 4H 2O( l )

(e) 5SO 32 (aq ) + 2MnO 4

(aq ) + 6H +(aq )

5SO 42 (aq ) + 2Mn 2+ (aq ) + 3H 2O( l )

10. Volumetric analysis

25. (c) 0.237

(d) 80.1 8 %

6. (b) 2.39 dm 3

(c) 2.56 dm 3

(c) (i) oxygen

(ii) hydrogen

(d) (i) 22.4 cm 3

(ii) 44.8 cm 3

**Delete Q3 part(e).

25. (c) 0.237 m

(d) 80.17 %

A-10 Part 1 Further Exercise

6. (b) 2.37 dm 3

(c) 2.54 dm 3

12. (c) (i) 22.4 cm 3

(ii) 44.8 cm 3

(d) (i) oxygen

(ii) hydrogen


3. (e) 1 s 22s 22p 63s 23p 64s 13d 10


22/22

83A nsw ers for Exercises


3. (d) I.E. =5 320 mol 1

11. (a) Germanium; Period 4; group IV

(b) Chlorine; Period 3; group VII

(c) Chlorine; Period 3; group VII

(d) Scandium; Period 4; Group IIIB

12. (b) (i) H 2

(ii) H 2O

(iii) NH 3


13. (c) 1 390 kJ

16. (a) 54.99 kJ g 1


4. (c) +18.45 kJ mol 1


12. K =4

Cl =4

3. (d) I.E. =5 320 kJ mol 1

11. (a) Germanium; Period 4; Group IVA

(b) Chlorine; Period 3; Group VIIA

(c) Chlorine; Period 3; Group VIIA

(d) Manganese; Period 4; Group VIIB

12. (b) (i) H3

(ii) H7O

(iii) NH 8

13. (c) 3.98 kJ

16. (a) For hydrogen: 140.5 kJ g 1

For methane: 54.99 kJ g 1

4. (c) +20.0 kJ mol 1

12. K + =4

Cl =4

Documents

New Way Chem 1