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INTRODUCTION TO CHEMISTRY
1) Decide whether each of the following process is primarily a physical or a chemical
change and explain briefly.
. Water boiling to become steam – physical (because change of state water H 2
O boils
to water vapour H 2
O)
. Rusting of a bike chain – chemical (new compound formed) Fe-Fe 2
3
(rust)
. Combustion of wood – chemical (combustion that produces mixture of gases)
include CO 2
or CO
. Melting of ice spring – physical (change of state H 2
0 melts too H 2
. Decay of leaves in winter – chemical (transformation/breakdown to new products)
2) Rewrite the following number according to the number of significant figures
specified in each part. Express your answers in scientific notation. 0.
(i) One -2 or 3 x 10
(ii) Two- 2.5 x 10
32333
(iii) Three - 2.48 x 10
3
(iv) Four -2.478 x 10
3
(v) Five - 2.4780 x 10
3
- Air has a density of 0.0013 g/mall What is the mass of a 6.0 L sample of
air?
Convert L to mL; i.e. 6.0 x 1000 = 6000 mL D = m / V; therefore, m = D x V =
0.0013 x 6000 = 7.8 g. (watch out for units)
4) Rewrite the following number according to the number of significant
figures specified in each part. Express your answers in scientific
notation.
(i) one – 1 x 10
(ii)two – 1.4 x 10
(iii)three – 1.38 x 10
(iv)four – 1.377 x 10
(v)five – 1.3768 x 10
- Can you give some examples of Biological processes and decide if there
are Physical or Chemical changes? Also, briefly explain why
. proteins synthesis/ Build up biological molecules – chemical – makes
proteins from different amino acids (new compounds)
. Breaking down biological molecules – chemical – makes new compounds . condensation and hydrolysis – chemical – makes new substance . solvent – solute dissolved in water – physical - no new substance is formed
it’s still water
. sweating – physical – no new substance is formed
- A body has a mass of 35 g and a volume of 18 L. Calculate its density
D=m = 35 = 1.94g/mL L- mL 18 x 1000 =18.
V 18.
a. Calculate the mass of 0.2 L of a body made from the same material as in
point a.
V=0.2L, (L- mL)= 0.2 x 1000= 200 m= d x v m= 1.94 x 200 = 388g
b. A solid has a density of 3.6 g/ml and a mass of 320g. Calculate its volume.
d= 3.6 m= 320 v=m/d = 320/3.6 =88.8mL
c. Calculate the density of a material that has a mass of 862g and a volume of
7.65L. 7.65L-mL = 7.65 x 1000 =
D= m = 862 =0.11g/mL
V 7650
d. The density of aluminium is 27g/ml. Calculate the mass of a block of
aluminium with a volume of 51 L. 51L x 1000=51000mL
M=d x v= 27 x 51000= 1377000g
Atomic Structure and Mole Concept Worksheet
- Neon exists in three isotopes. Calculate the relative atomic mass of Neon.
Show clearly your working.
Isotope Percent Abundance
20
Ne 90.
21
Ne 0.
22
Ne 8.
RAM = isotope x percent abundance
0.26= 0.0026 = (20 x 0.9092) +(21 x 0.0026) +(22 x 0.0882)
100 = 20.18 amu
- The element X has three naturally occurring isotopes. The masses (amu) and
% abundances of the isotopes are given in the table below. Calculate the
average atomic mass of the element. Show clearly your working. ( 5 marks )
Isotope Percent Abundance Mass (amu)
221
X 74.22 220.
220
X 12.78 220.
218
X 13.00 218.
Avg = M 1
P
1
+ M
2
P
2
+ M
3
P
3
Divide the percent abundance by 100
Recall: 1 mole of any substance contains 6.022 x 10
23
entities (atoms or molecules)
Thus: 1 mole of H 2
O contains 6.022 x 10
23
molecules
Therefore: 25.65 moles contain
25.65 x 6.022 x 10
23
molecules
1.54 x 10
25
molecules
- A compound contains 26.1% carbon, 4.3% hydrogen and 69.6% oxygen by mass. Find
the empirical formula of this compound. (A r
for carbon = 12, A
r
for hydrogen = 1, A
r
for
oxygen = 16)
A compound is made from 72 g of carbon and 12 g of hydrogen. Work out its
empirical formula.
Element name
Mass of element (g)
Mass ÷ Relative Atomic Mass
Divide by the smaller number
Ratio
Carbon
72
72 ÷ 12 = 6
6 ÷ 6 = 1
1
Hydrogen
12
12 ÷ 1 = 12
12 ÷ 6 = 2
2
The empirical formula of this compound is CH 2
.
A common salt is analysed and is found to have 52.9 g of sodium and 81.7 g of
chlorine. What is its empirical formula?
Element name
Mass of element (g)
Mass ÷ Relative Atomic Mass
Divide by the smaller number
Ratio
Sodium
52.9 ÷ 23 = 2.
2.3 ÷ 2.3 = 1
1
Chlorine
81.7 ÷ 35.5 = 2.
2.3 ÷ 2.3 = 1
1
The empirical formula of this compound is NaCl.
Aluminium ore may consist of 156 g of aluminium and 278 g of oxygen. Is its
empirical formula AlO 2
or AlO 3
Element name
Mass of element (g)
Mass ÷ Relative Atomic Mass
Divide by the smaller number
Ratio
Aluminium
156
156 ÷ 27 = 5.
5.8 ÷ 5.8 = 1
1
Oxygen
278
278 ÷ 16 = 17.
17.4 ÷ 5.8 = 3
3
The empirical formula of this compound is AlO 3
A commercial paint thinner has the following composition: carbon 25.2 g;
hydrogen 8.5 g; oxygen 33.7 g. What is its empirical formula?
Element name
Mass of element (g)
Mass ÷ Relative
Atomic Mass
Divide by the smaller
number
Ratio
Carbon
25.2 ÷ 12 = 2.
2.1 ÷ 2.1 = 1
1
Hydrogen
8.5 ÷ 1 = 8.
8.5 ÷ 2.1 = 4.
4
Oxygen
33.7 ÷ 16 = 2.
2.1 ÷ 2.1 = 1
1
The empirical formula of this compound is CH 4
O.
A compound has 24 g of carbon and 64 g of oxygen. What is its empirical formula?
Element name
Mass of element (g)
Mass ÷ Relative Atomic Mass
Divide by the smaller number
Ratio
Carbon
24
24 ÷ 12 = 2
2 ÷ 2 = 1
1
Oxygen
64
64 ÷ 16 = 4
4 ÷ 2 = 2
2
The empirical formula of this compound is CO 2
.
hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a
government chemist testing commercial antacids, you use 0.1 M HCl to simulate the acid
concentration in the stomach. How many litres of “stomach acid” react with a tablet
containing 0.10 g of magnesium hydroxide?
Mg(OH) 2
( s ) + 2HCl( aq ) MgCl 2
( aq ) + 2H 2
O( l )
Recall: n = m/M r
N
Mg(OH)
= 0.1 g/58.33 g/mol; Note: Mr Mg(OH) 2
= 58.33 g/mol
= 1.7 x 10
3
mol
I mole Mg(OH) 2
reacts with 2 moles HCl
Hence HCl = 1.7 x 10
3
X 2 = 3.4 x 10
3
mol
Recall: C = n/V (Note: V is in L or dm
3
V = n/C = 3.4 x 10
3
mol/0.1 mol/L
= 3.4 x 10
2
L
- Calculate the molarity of 2.0 L of solution containing 5.0 mol NaOH
Note: Molarity (M) = Concentration (mol/L)
Hence M = n/V = 5/2 = 2.5 M or mol/L
- Calculate the number of mol of solute in 750.0 mL of 0.250 M HCl.
Recall: C = n/V; n = C x V (Note V should be in L or dm3)
n = 0.250 x 0.75 = 0.1875M HCL
- Nitric acid is a major chemical in the fertiliser and explosive industries. In aqueous
solution, each molecule dissociates, and the H becomes a solvated H
ion. What is the
molarity of H
(aq) in 1.4 M nitric acid?
Recall: HNO 3
(nitric acid) is a strong acid and in solution will dissociate completely.
Hence:
HNO
3
( l ) H
( aq ) + NO 3
(aq)
OR
HNO
3
+ H
2
O H
3
O
( aq ) + NO 3
(aq)
Note [H
] = [H
3
O
]
Note [H
] = [H
3
O
]
1.4 M HNO
3
(aq) acid should have 1.4 M [H
] (aq) or [H 3
O
]
- If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution, what is
the concentration of the HCl?
NaOH + HCL= NaCL + H 2
O
n NaOH
= CxV = 0.1 x 0.054 = 0.0054 = 5.4x
mol
1 mole of NaOH reacts with 1 mole of HCL
5.4x
: x
X= n NaOH
= 5.4x
C=n/v = 5.4x
x 1.25x
= 0.0432M
- If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is
the concentration of the NaOH solution?
NaOH + HCL = NaCl + H 2
O
N
HCL
= CxV = 0.05 x 0.025 = 0.001254mol
Ratio 1 : 1
X : 0.
C=n/v = 0.00125/0.345 = 0.0036mol/l = 36x
mol/l
- If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric
acid solution (H 2
SO
4
), what is the concentration of the H 2
SO
4
solution?
H
2
SO
4
+ KOH = K
2
SO
4
+ H
2
O
H
2
SO
4
+ 2 KOH = K
2
SO
4
+ 2 H
2
O
n koh
= CxV = 0.5x0.05 = 0.025mol
ratio 1 : 2
x. : 0.
x = 0.025/2 = 0.
C=n/v = 0.0125/0.125 = 0.1mol/l
ACID-BASE TITIRATION WORKSHEET
- A titration was performed, and the following data were collected at the endpoint:
Volume of acid (H 2
SO
4
) used = 23 mL; Average volume of base (NaOH) used =
29.3 mL; known concentration of base (NaOH) = 0.3 mol/L. What is the molarity
of H 2
SO
4
? Show your calculations
H
2
SO
4
O + Na 2
SO
4
Recall: C =n/V (V in L or dm
3
Note convert volume from mL to L
29.3 mL
=0.0293 L
Number of moles of NaOH = C x V
n NaOH
=0.0293 L NaOH × 0.3 M NaOH =0.00879 mol NaOH
From balance equation, 2 moles of NaOH require 1 mole of H 2
SO
4
Number of moles of H 2
SO
4
X : 8.79x
2x = 8.79x
= 4.395x
mol
C=n/v = 4.395x
/2.3x
= 0.191M
- A student titrated 25.0 mL of solution of sodium hydroxide solution with standard
sulfuric acid. It took 13.4 mL 0f 0.0555M H 2
SO
4
to neutralise the sodium
hydroxide in the solution. What as the molarity of the sodium hydroxide solution?
The equation of the reaction is:
H
2
SO
4
O + Na 2
SO
4
Few drops oof water are left in burette and then used to titrate a base into an acid
soolution to determine the concentration of the acid
a) Will this small amount of water have any effect on the determined value for
the concentration of the acid? If so, what is effect of the water
As water is added to base, it is slightly diluted. This results in a more neutral
solutioon,leading to the apperanace that the acidic solution is tronger than it
actually is.
- With the aid of suitable diagrams describe the type of chemical bonding found in:
(i) Calcium chloride
(ii) Carbon dioxide
- Draw the Lewis structure for:
(i) Carbon dioxide and Hydrogen Cyanide
(ii) How many valence electrons for each molecule?
Hint: Students should be informed that the valence electrons of each atom on the
molecule should be indicated first before the Lewis Structure
Water (H 2
O)
1(2) 6. = valence electrons
H
2
O
H. :O: .H
:O:
/ \
H H
Lithium chloride (L I
Cl)
1 7 = valence electrons
L
I
Cl
.X
L
i
[:Cl::]
Carbon dioxide (CO 2
Hydrogen cyanide (HCN)
- Which of the following are polar or non-polar compounds? State clearly your
reason.
(I)NH
3
(ii) O-S (iii) CO 2
(iv) SiCl 4
(iv) C 2
H
6
(I) Polar (not planar; difference in electronegativity of H and N
(ii) O-S – Polar; O and S may be in the same group but O is more electronegative
than S.
(iii) CO 2
- non-polar; O=C=O; symmetrical
(iv) SiCl 4
- non-polar; planar molecule; though Si and Cl differ in electronegativity.
(v) C 2
H
6
- non-polar; planar; electronegativity difference between C and H is not
much.
- Which have the higher melting and boiling points
Hint: Identify the intermolecular forces (H-bong > dipole-Dipole > Van der
Waal’s forces
(i) H 2
O and C 2
H
4
; H-bonding > Van der Waal’s
(ii) LiCl and ICl; ionic > dipole-dipole
(ii) NH 3
and N 2
; H-bonding and dipole – dipole > Van der Waal’s
(iv) F 2
and Br 2
; Bothe are no-polar; hence Van der Waal’s; however Br2 is bigger
and thus more Van der Waal’s interactions
3-ethyl-4-methlyhexane
2,3,3-trimethylbut-1-ene
- In an experiment, 1.00 g of propanone (CH
COCH
) was completely burned
in air. The heat evolved raised the temperature of 150 g of water from 291.8K to
337.3K. Use this data to calculate the enthalpy of combustion of propanone (the
specific heat capacity of water is 4.18 J g
K
q = mc∆T m = 150. c = 4.18 ∆T = 45.
q = 150 x 4.18 x 45.5 = 28530 J
∆H = q / mol
moles of propane = mass / M r
∆H = –28.53 / 0.01724 = -1650 kJ mol
- In an experiment, 1.00 g of hexane (C 6
H
14
) was completely burned in air. The
heat evolved raised the temperature of 200 g of water from 293.5 K to 345.1 K. Use
this data to calculate the enthalpy of combustion of hexane (the specific heat
capacity of water is 4.18 J g
K
q = mc∆T m = 200 c = 4.18 ∆T = 51.
q = 200 x 4.18 x 51.6 = 43140 J
∆H = q / mol
moles of propane = mass / M r
∆H = –43.14 / 0.01163 = = -3710 kJ mol
- In an experiment, 1.56 g of propan-1-ol (CH 3
CH
2
CH
2
OH) was completely
burned in air. The heat evolved raised the temperature of 0.250 dm
3
of water from
292.1 K to 339.4 K. Use this data to calculate the enthalpy of combustion of propan-
1-ol (the specific heat capacity of water is 4.18 J g
K
q = mc∆T m = 250 c = 4.18 ∆T = 47.
q = 250 x 4.18 x 47.3 = 49430 J
∆H = q / mol
moles of propane = mass / M r
∆H = –49.43 / 0.02600. = -1900 kJ mol
What is the pH of a bleach solution that has a [OH
−
] = 2.8×
−
M?
pOH = -log(OH
) = pOH = -log (2.8x
) =. pOH = 2.55 =. pH +pOH =14 = pH=14-POH
= pH=14-2.55 = pH = 11.
- Calculate the pH of the following solutions:
a) 0.001 moldm
HCl
b) 0.03 moldm
HNO
3
c) 0.002 moldm
KOH
d) 0.003 a moldm
NaOH
- Calculate the molarity (molar concentration) of the following solutions:
a) HCl , pH = 3.
b) NaOH, pH = 11.
c) HNO 3
, pH = 2.
d) KOH, pH = 12.
- Calculate the pH of the following solutions:
a) 0.001 moldm
HCl = [H
]
pH = - log[H
+
] = - log 10
-
b) 0.03 moldm
HNO
3
d) KOH, pH = 12.
pH = - log[H
+
]
[H
+
] = 10
-12.
[H
+
] = 3.16 x 10-
13
-
= 3.16 x 10-
13
[OH
-
]
[OH
-
] = 10
-
/3.16 x 10-
13
= 3.2 x 10-
2
mol dm
-
The combustion of glucose (C 6
H
12
O
6
) is shown in the balanced equation below:
C
6
H
12
O
6
(s) + 6O 2
(g) → 6CO 2
(g) + 6H 2
O(l)
The standard enthalpy change of combustion of glucose is, ΔHº -2840 kJ/mol
(i) What does ΔHº of a chemical reaction represent?
(ii) Define the term standard enthalpy change of combustion.
(iii) The combustion is glucose releases energy. Explain.
(b) Sketch the enthalpy profile diagram for the combustion of octane, labelling
clearly on the diagram the reactants, products, activation energy, E a, and the
enthalpy change, Δ H.
(c)
(i) What is the standard enthalpy formation of oxygen gas?
(ii) Explain your answer.
(i) Standard state enthalpy” refers to a reaction in which reactants and products are
at temperature 298 K (
0
C) and pressure (one atmosphere).
(ii) Is the enthalpy change when one mole of the substance burns completely under
standard conditions.
(iii) Heat is released in the reaction –EXOTHERMIC REACTION.
(b)
(c)
i) zero
ii) oxygen is pure element in its standard state
In a laboratory experiment, 1.45 g of propanone were burned completely in oxygen.
The heat from this combustion was used to raise the temperature of 100 g of water
from 293.1 K to 351.2 K.
(i) Calculate the number of moles of propanone in the 1.45 g.
Number of moles of acetone = n =m/M =1.45/58 = 0.025 moles
(ii) Calculate the heat energy required to raise the temperature of 100 g of water
from 293.1 K to 351.2 K. (The specific heat capacity of water is 4.18 J K
g
heat released = mc∆T (∆T = 351.2 – 293.1 = 58.1) = 100 × 4.18 × 58.1 = 24300 J (or
24.3kJ)
Q
In an experiment, 0.600 g of propane (C 3
H
8
) was completely burned in air. The heat
evolved raised the temperature of 100 g of water by 64.9C. Use this data to
calculate the enthalpy of combustion of propane (the specific heat capacity of water
is 4.18 J g
1
K
1
STEP 1: CALCULATE HEAT EVOLVED
STEP 2: CALCULATE NUMBER OF MOLES OF PROPANE (C
3
H
8
STEP 3: CALCULATE ΔH
(10) (a) The standard enthalpy change of combustion of butane is ΔHº –2877 kJ
mol
(i) What does ΔHº of a chemical reaction represent? ( 2 marks )
It is the enthalpy change of reaction whose reactants and products are in standard
start of 298k (
0
) temperature and 1 atmospheric pressure
(ii) Explain the term standard enthalpy change of combustion ( 2 marks ).
It is the enthalpy change when 1 mole of substance is completely burned in its
standard state under standard condition 25
0
c temperature and 1 atmospheric
pressure
( 5) Butane gas is perfect for camping and caravanning as it can be used for portable
heating. The hydrocarbon mixes with oxygen from the air and burns to form
carbon dioxide and water vapour.
(a) Write a balanced equation for this reaction
C₄H ₁₀+ O
2
= CO
2
+H
2
0(not balanced )
2C₄H ₁₀+ 13O
2
= 8CO
2
+10H
2
0(balanced)
(b) What type of chemical reaction is this? ( 1 mark )
For (c) and (d) below show all your calculations combustion
(a) How many moles of water can be produced by burning 87 g of butane? ( 3
marks )
Calculate the number of moles of octanes: m/M r
(of butane)
87g/58.12g/mol= 1.50moles butane
From stoichiometry of the balanced equation
2C₄H ₁₀ : 10H
2
1.5 : x
1.5x10=
2x/2 =15/2 = 7.5moles
