Atoms , Molecules & Stoichiometry ( STPM + Matriculation )

October 18, 2017 | Author: AlexTanYun-Kai | Category: Isotope, Proton, Deuterium, Ion, Hydrogen
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Collections of first chapter of Chemistry for Malaysia Syllabus 1. MPM Specimen Papers 2. STPM 2013 Sem 1 3. STPM 2...

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Al PM ex T C a he n m is try Atoms, Molecules and Stoichiometry

Contents

1. MPM Specimen Paper

2. STPM 2013 Semester 1 3. STPM 2014 Semester 1

ST

4. Malaysia Matriculation Paper 2013

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Objective questions Q1 ( MPM Specimen Paper )

Al PM ex T C a he n m is try

\

Answer : D

Explanation

Particle

Number of protons

Number of electrons

Number of neutrons

7

7

9

8

10

10

9

9

10

35

34

44

2-

ST

+

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Al PM ex T C a he n m is try

Question 2 ( MPM Specimen Paper )

Answer : D

Explanation 

RAM =

(

) (

)

= 14.00

RMM of X2 is 28 .



Isotopes of X have same number of protons but different in number of

ST



neutrons .

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Question 3 ( STPM 2013 Sem 1 ) Tetraethyllead(IV) , (CH3CH2)4Pb , is added to petroleum as an antiknock agent . (CH3CH2)4Pb is prepared from ethane , CH3CH3 , according to the following reactions : CH3CH3 + Cl2 → CH3CH2Cl + HCl

Al PM ex T C a he n m is try

4CH3CH2Cl + 4NaPb → (CH3CH2)4Pb + 3Pb + 4NaCl Which statement about the reactions is not true ?

A. One mole of Cl2 produces one mole of CH3CH2Cl .

B. Four moles of Cl2 produce one mole of (CH3CH2)4Pb .

C. One mole of HCl is formed for each mole of (CH3CH2)4Pb produced .

D. Four moles of CH3CH3 is required for each mole of (CH3CH2)4Pb produced . Answer : C

Explanation 

Overall equation :

4CH3CH3 + 4Cl2 + 4NaPb → (CH3CH2)4Pb + 3Pb + 4NaCl + 4HCl

1 mole of Cl2 ≡ 1 mole of CH3CH2Cl



4 mole of Cl2 ≡ 1 mole of (CH3CH2)4Pb



1 mole of (CH3CH2)4Pb ≡ 4 mole of HCl



4 mole of CH3CH3 ≡ 1 mole of (CH3CH2)4Pb

ST



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Question 4 ( STPM 2014 Sem 1 ) The proton number of ion Y is 16 , its nucleon number is 33 and it has the same number of electrons and neutrons . What is the charge of ion Y ? A. -2 B. -1

Al PM ex T C a he n m is try

C. +1 D. +2

Answer : B

Explanation Nucleon number

Number of protons

Number of electrons

Number of neutrons

33

16

17

17

Since Y has one electron more than proton , thus it carries a charge of -1 .

ST



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Question 5 ( STPM 2014 Sem 1 ) An element X with a relative atomic mass of 28.1 consists of isotopes 30

X . If the percentage abundance of

29

X and

30

28

X , 29X and

X are the same , what is the

percentage abundance for 28X ? A. 0.32

Al PM ex T C a he n m is try

B. 3.33 C. 93.3 D. 96.7

Answer : C

Explanation 

Let the percentage abundance of 29X and 30X be a .



Let the percentage abundance of 28X be ( 100 – a –a ) .



RAM = 28.1



( )

( )

(

)

= 28.1

a = 3.33 %



Percentage abundance of 29X and 30X = 3.33 %



Percentage abundance of 28X = (100 – 3.33 – 3.33 )% = 93.34 %

ST

= 93.3 %

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Structured & Essays Question

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Question 1 ( MPM Specimen Paper )

(a) (i) Name the subatomic particle Y . [1m] 

Answer : neutron

(ii) Draw the paths of the beams of electrons & hydrogen ions in the above diagram . [2m]



Answering skill : Beam of electrons deflect more than that of H+ ions .

Answer :

ST



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(iii) If a beam of deuterium ions is passed through the electric field , explain the difference in deflection angle between the beam of hydrogen ions and that of deuterium ions .[2m] Answer : The deflection angle for deuterium ions is smaller than that of hydrogen ions .



This is because deuterium ion has a larger m/e value than hydrogen ions .

Al PM ex T C a he n m is try



* Alternative answer : 

The deflection angle for deuterium ion is half of that for hydrogen ion .



This is because the deuterium ion is twice as heavy as hydrogen ions .

* Extra notes 

The heavier the ions ( larger m/e value ) , the smaller the deflection angle for that ion .



Isotopes for hydrogen

Isotopes

Nucleon number

Number of protons

Number of electrons

Number of neutrons

1

1

1

0

2

1

1

1

3

1

1

2

( protium )

( deuterium )

ST

( tritium )

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(b) P+ and Q- ions are isoelectronic with the (i) State the nucleon number of the 

isotope .

isotope .

Answer : 40

(ii) Identify P+ & Q- ions . Answer : ~ P+ ion : K+ ion / potassium ions ~ Q- ion : Cl- ion / chloride ions

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*Extra Note 

Isoelectronic species : species with same number of electrons



For P+ ions :



Number of Number of protons electrons 19 18 For Q ions : Number of electrons 18

ST

Number of protons 17

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Question 2 ( STPM 2013 Sem 1 ) Chlorine gas exists as diatomic molecules , Cl2 . In an experiment , 326.6 g of chlorine was used to react with excess iron to produce iron (III) chloride , FeCl3 . (a) Write a balanced equation for this reaction . [1m] 

Answer : 2 Fe (s) + 3 Cl2 (g) → 2 FeCl3 (s)

Al PM ex T C a he n m is try

(b) Calculate the theoretical yield of FeCl3 . [2m] Answer : 

Number of moles of Cl2 =

mol

= 4.60 moles



Number of moles of FeCl3 produced =

× 4.60 mol

= 3.067 mol



Mass of FeCl3 produced = 3.067 × [ 55.8 + 3( 35.5) ] = 497.8 g ( 4 s.f )

*Extra !!! ( What if we don’t know how to calculate the number of moles of FeCl3

ST

produced ?? )

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(c) If the percentage of the actual yield of FeCl3 obtained in the laboratory is 82.0% , calculate the mass of FeCl3 produced . [2m] Answer : 

Mass of FeCl3 produced = 497.8 × 82.0% = 408.2 g

Al PM ex T C a he n m is try

(d) Explain why there is a difference between the calculated theoretical value yield of FeCl3 with that of the actual yield obtained in the laboratory . [2m] Answer : 

Some of the chlorine gas escapes from the reacting vessel .



Some of FeCl3 will be decomposed and forms FeCl2 .

ST

2 FeCl3 (s)

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2 FeCl2 (s) + Cl2 (g)

Question 3 ( Matriculation 2013 ) Silicon exists as three stable atoms with nucleon number of 28 , 29 and 30 . The proton number of silicon is 14 . (a) Write the symbol for any one of 3 silicon atoms . Indicating its proton and nucleon number in the symbol . [1m]

Al PM ex T C a he n m is try

Answer :



@

@

(b) If the amount of a sample of silicon with nucleon number of 28 , 29 and 30 are 92.2% , 4.68% and 3.12% , calculate the relative atomic mass of silicon . [2m] Answer : 

RAM =

(

) (

) (

)

= 28.1

(c)Sketch and label the mass spectrum of silicon . [2m] Answer :

Answer :

ST



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Question 4 ( Matriculation 2013 ) When 80.0g of ethene gas , C2H4 reacts with 110.0 g of hydrochloric acid , HCl , chloroethane , C2H5Cl is produced . (a) Write a balanced chemical equation for the reaction . [1m] 

Answer : C2H4 (g) + HCl (aq) → C2H5Cl (g)

Al PM ex T C a he n m is try

(b) Determine the limiting reactant in the reaction . [3m] Answer : 

Number of moles of ethene =

(

)

(

)

mol

= 2.86 mol



Number of moles of HCl =

mol

= 3.01 mol



Since the number of moles for ethene is less than that for hydrochloric acid , thus the limiting reactant is ethene gas .

(c) Calculate the mass of C2H5Cl produced . [3m] Answer : 

Number of moles of C2H5Cl = number of moles of ethene gas = 2.86 mol

Mass of C2H5Cl produced = 2.86 × [ 2(12.0) + 5(1.0) + 35.5 ]

ST



= 184 g ( 3s.f )

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(d) Calculate the mass of excess reactant remained at the end of the reaction . [3m] Answer : 

Number of HCl used = number of moles of ethene gas = 2.86 mole



Number of moles of HCl remained = 3.01 - 2.86

Al PM ex T C a he n m is try

= 0.15 mol



Mass of excess HCl remained = 0.15 × ( 35.5 + 1.0 )

ST

= 5.48 g ( 3sf )d

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