CADANGAN JAWAPAN CHEMISTRY 1 SESI 20202021

CADANGAN JAWAPAN CHEMISTRY 1 (SK 015)

SESI 2020/2021

Question Answer

1. a) i.

ii. n= 50
122.1 /

= 0.4095 mol
No of Si atoms = 0.4095 X 6.02 X 1023 atom

= 2.465 X 1023 atoms

b) MnO4- + 8H+ + 5e Mn2+ + 4H2O
Mn2+ + 5Fe3+ + 4H2O
5Fe2+ 5Fe3+ + 5e

MnO4- + 8H+ + 5Fe2+

c) Element K Cr O
Mass (g) 26.58 35.35 38.07
Moles 26.58 35.35 38.07
(mol) 39.1
52 16
Simplest =0.689
ratio 0.689 =0.679 =2.379
0.679 0.679 2.379
0.679 0.679
=1x2
=1x 2 =3.5 x 2

Empirical formula: K2Cr2O7

n= 294.2 /

294.2 /

=1

molecular formula= K2Cr2O7 (Potassium dichromate)

d) 3NO2 + H2O 2HNO3 + NO

i.

2 = 100 2 = 50
46 18

= 2.174 mol =2.778 mol (available)

3 mol NO2 ≡ 1 mol H2O
2.174 mol ≡ 1/3 X 0.725 mol (needed)
nH2O needed < nH2O available
Thus, NO2 is a Limiting reactant

Marziah binti Mohamad

ii. 3 mol NO2 ≡ 2 mol HNO3
2.174 mol ≡ 2/3 X 2.174 mol
= 1.449 mol

Mass HNO3 = 1.449 mol X 63 gmol-1
= 91.287 g

% yield = 80 100%
91.287

= 87.63%

2 a) Series: Paschen series
Region: Infra red (IR)

∆H = ( 1 2 − 1 2 )

= 2.18 x 10-18(612 − 312)

= -1.816 X 10-19 J

∆H = ℎ
ʎ
6.62256 10−34 3.0 108
= 1.816 10−19

= 1.094 X 10-6 m @ 1094 nm

b) When n=3 and l=2, orbital is 3d

*Draw and labelled all 3dxy, 3dyz, 3dxz, 3dy2-x2 and 3dz2

3 a) (-1) (0) (-1)

O O O (+1)

(+1) (0) (0) (0) (0) Cl

Cl C Cl Cl C Cl Cl C

(0)

I II III

b) Structure II because all atoms have zero formal charge

Marziah binti Mohamad

c) (0)

(0) O

Cl (0)

C Cl

(0)

II
Carbon has 3 sigma bonds, thus hybrid sp2
*label sigma and pi bond around C.

d) Trigonal planar

C ground state: 2p O ground state: 2s 2p
2s O exited state: 2s 2p
O hybrid state: sp3
C exited state:

2s 2p

C hybrid state:

sp3

Cl ground state:

3s 3p

sp2 sp2

O

sp2

sp2

C

sp2 sp2
Cl Cl

3s 3s labelled sigma and pi bond

4 a)

b) 2 = 61 = 2
4
32



= 1.906 mol = 0.5 mol

Marziah binti Mohamad

2 =

= 1.906 0.08206 298.15
5

= 9.326 atm

=

= 0.5 0.08206 298.15

5

= 2.446 atm

PT = PO2 + PHe
= 9.326 atm + 2.446 atm
= 11.77 atm

5 a) 2 changes that can increase the yield of ammonia:
-adding N2 and H2. Addition of N2 and H2 will increase the concentration of N2 and
H2. Thus, the reaction will shift forward and increase the amount of ammonia.
-decrease/reduce the temperature. The reaction is an exothermic reaction.

b) Qc = [SO3]2
[SO2]2[O2]

= (0.15)2 −3
(0.2)28.55 10

= 65.78

Qc< Kc
The reaction will shift backward/ from right to left

6 a) CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
00
[ ]i 0.5 M +x +x
∆ -x xx

[ ]f 0.5-x

1.8 x 10-5 = 2 Ka is too small, thus 0.5-x =0.5
0.5−

X= √0.5 1.8 10−5

= 3.0 x 10-3

X=[H3O+]= 3.0 x10-3 M

pH= -log[H3O+]

=2.52

b) CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

25 mL 25 mL

0.5 M 0.5 M

ni 0.0125 mol 0.0125 mol 0 0
change -0.0125 mol - 0.0125 mol + 0.0125mol -
nf
0 0 0.0125 mol

Marziah binti Mohamad

[CH3COONa] = 0.0125
(105000)

= 0.25 M

ii. solution is CH3COONa (salt)
hydrolysis salt to determine the pH

CH3COO-(aq) + H2O (l) CH3COOH (aq) + OH- (aq)

Na+ (aq) cannot hydrolysed

CH3COONa is a basic salt, pH > 7 since OH- ion presence

Marziah binti Mohamad