|
  6b.
Reacting masses, concentration of solution and volumetric titration calculations
NOT
using the mole concept BUT see also
Basic
acid-alkali titration questions using moles
far more appropriate!
Advanced Level acid-alkali volumetric titration calculations
For quantitative chemistry, it is
important to know the concentration of solutions and be able to do calculations
based on experimental results for chemical analysis.
Concentration units
The
concentration of a solution is often measured in grams per decimetre cubed
so the units are expressed
as gdm-3, or g/dm3, which was
g/litre.
concentration = mass / volume
(c = m / V)
concentration in gdm-3,
mass in g, volume in dm3
Its really important remember
that 1 dm3 = 1000 cm3 or
1000 ml
and cm3/1000 =
dm3
rearrangements of the formula
using the triangle to help gives
mass = concentration x volume
(m = c x V)
and volume = mass /
concentration (V = m / c)
In the example questions I have
used the shorthand formulae c = m / V, m = c x V and V = m
/ c
and RFM as an
abbreviation of relative formula mass or molecular mass, and other
shorthands:-
sf means numbers rounded to
2/3/4 significant figures and dp means round to 2/3/4 decimal places
A standard solution is one of
accurately known concentration.
Three simple examples to
illustrate using the formula (formula triangle on right)
(a) What is the concentration of a
salt solution if you dissolve 10g of sodium chloride in 250 cm3 of
water?
250 cm3 is equal to
250/1000 = 0.25dm3
therefore the concentration
c = m /
V =
10/0.25 = 40 g/dm3
(40g/litre in old units, still in
common use!)
(b) What mass of the salt is
required to make 200 cm3 of concentration 15g/dm3?
V = 200/1000 = 0.2 dm3
m = c x V = 15 x 0.2 =
3.0 g
(c) If you were given 8.0 g of salt,
what volume of water, in dm3 and cm3, should you dissolve
it in, to give a salt solution of concentration of 5.0g/dm3?
V = m / c = 8.0 / 5.0 = 1.6 dm3
V = 1.6 x 1000 = 1600 cm3
TOP OF PAGE
QUESTIONS
involving mass, volume and concentration
-
Example Question 6b (Q1)
A potassium sulfate solution A has a concentration of 6.5g/dm3
-
(a) What mass of potassium
sulfate is present in 25cm3 of solution A?
-
(b) What volume of
solution A, in cm3, contains 3.0g of potassium sulphate?
-
(c) What mass of the salt
is required to make 250 cm3 of a solution B so it
contains 3.75 g/dm3 of potassium sulfate?
-
(d) If 12.5g of potassium
sulphate was dissolved in 1500 cm3 of water to make up solution C,
what is the concentration of the salt?
-
-
-
Answers to Question 6b. Q1
A potassium sulfate solution has a concentration of 6.5g/dm3
-
(a) What mass of potassium
sulfate is present in 25cm3 of the solution?
-
(b) What volume of
solution, in cm3, contains 3.0g of potassium sulphate?
-
(c) What mass of the salt
is required to make 250 cm3 of a solution B so it
contains 3.75 g/dm3 of potassium sulfate?
-
(d) If 12.5g of
potassium sulphate was dissolved in 1500 cm3 of water to
make up solution C, what is the concentration of the salt?
-
-
-
Example Question 6b
(Q2) Nitric
acid - sodium hydroxide titration
-
The reaction equation for
the neutralisation of nitric acid by sodium hydroxide to form sodium
nitrate is
-
HNO3(aq) +
NaOH(aq) ===> NaNO3(aq) + H2O(l)
-
(a) Calculate the relative
formula masses of nitric acid, sodium hydroxide and sodium nitrate
-
25.0 cm3 of a
nitric acid solution was pipetted into a conical flask and titrated
with a sodium hydroxide solution of concentration 5.0g/dm3.
If it took 16.8 cm3 of the alkali to neutralise the acid,
calculate the following
-
(b) The mass of sodium
hydroxide that reacted with the nitric acid.
-
(c) Calculate the mass
of nitric acid that reacted with the sodium hydroxide.
-
(d) What was the
concentration of the nitric acid in g/dm3?
-
(e) If the resulting
neutral solution was carefully evaporated to dryness, what mass of
sodium nitrate salt crystals would be left as a residue?
-
-
-
Answers to Question 6b. Q2 Nitric
acid - sodium hydroxide titration
-
(a)
| nitric
acid |
+ |
sodium
hydroxide |
===> |
sodium
nitrate |
+ |
water |
| HNO3(aq) |
+ |
NaOH(aq) |
===> |
NaNO3(aq) |
+ |
2H2O(l) |
| RFM 63 |
|
40 |
|
85 |
|
- |
-
25.0 cm3 of a
nitric acid solution was pipetted into a conical flask and titrated
with a sodium hydroxide solution of concentration 5.0g/dm3.
If it took 16.8 cm3 of the alkali to neutralise the acid,
calculate the following
-
(b) The mass of sodium
hydroxide that reacted with the nitric acid.
-
(c) Calculate the mass
of nitric acid that reacted with the sodium hydroxide.
-
|
HNO3 |
: |
NaOH |
|
63 |
: |
40 |
|
x g |
: |
0.084 g |
- Solving the ratio: x = 0.084 x
63/40 = 0.1323 g HNO3
-
(d) What was the
concentration of the nitric acid in g/dm3?
-
(e) If the resulting
neutral solution was carefully evaporated to dryness, what mass of
sodium nitrate salt crystals would be left as a residue?
-
|
NaOH |
: |
NaNO3 |
|
40 |
: |
85 |
|
0.084 g |
: |
x g |
- Solving the ratio: x = 0.084
x 85/40 = 0.1785 g NaNO3
- You can also use HNO3
instead of NaOH i.e. 63 : 85, it doesn't matter which you use.
-
-
-
Example Question 6b
(Q3)
Limewater analysis
-
The symbol equation for
the neutralisation reaction between calcium hydroxide
solution (limewater) and hydrochloric acid to form calcium chloride
and water is
-
Ca(OH)2(aq)
+ 2HCl(aq) ===> CaCl2(aq) + 2H2O(l)
-
50 cm3 of a
limewater solution was completely neutralised by 9.7 cm3
of a hydrochloric acid solution. If the concentration of the
hydrochloric acid was 3.65 g HCl/dm3 calculate the
concentration of calcium hydroxide in the water by the following
method.
-
(a) calculate the relative
formula masses of the reactants.
-
(b) What mass of
hydrochloric acid (as HCl) reacted with the limewater?
-
(c) What mass of calcium
hydroxide reacted with the hydrochloric acid?
-
(d) What is the
concentration of calcium hydroxide in g/dm3?
-
-
-
-
-
Example Question 6b
(Q4)
Sulfuric acid - potassium hydroxide reaction to make
potassium sulfate
-
The neutralisation
reaction between sulphuric acid and potassium hydroxide is
-
H2SO4(aq)
+ 2KOH(aq) ===> K2SO4(aq) + 2H2O(l)
-
A solution of
potassium hydroxide contained 100g/dm3.
-
A solution of
sulfuric acid contained 120g/dm3.
-
What volume in cm3,
of the sulfuric acid solution, is required to neutralise 250 cm3
of the potassium hydroxide solution?
-
Atomic masses: H =
1, S = 32, O = 16, K = 39
-
Answers to
Question 6b. Q4 Sulfuric acid - potassium
hydroxide reaction to make potassium sulfate
-
H2SO4(aq)
+ 2KOH(aq) ===> K2SO4(aq) + 2H2O(l)
-
Potassium hydroxide
contained 100g/dm3 and sulfuric acid contained
120g/dm3.
-
What volume in cm3,
of the sulfuric acid solution, is required to neutralise 250 cm3
of the potassium hydroxide solution?
-
To calculate mass of
KOH
-
To calculate mass of
H2SO4 needed
-
|
H2SO4 |
: |
2KOH |
|
98 |
: |
2 x 56 = 112 |
|
x g |
: |
25 g |
- Solving the ratio: x = 25 x
98/112 = 21.875 g H2SO4
-
To
calculate the volume of sulphuric acid needed.
-
-
Example
Question 6b (Q5)
0.5g of sodium chloride was dissolved in water
and you are given a silver nitrate solution of concentration 30g/dm3.
When you mix the two solutions you get a white precipitate of silver chloride.
Given the equation and the relative formula masses
(RFMs) calculated from
the atomic masses
Na = 23, Cl = 35.5, Ag = 107.8, O = 16, N =
14
| sodium chloride |
+ |
silver nitrate |
===> |
silver chloride |
+ |
sodium nitrate |
| NaCl(aq) |
+ |
AgNO3(aq) |
===> |
AgCl(s) |
+ |
NaNO3(aq) |
| (a) RFM 58.5 |
|
169.8 |
|
(143.3) |
|
(85) (not needed) |
Using an accurate burette, precisely what volume
(in cm3) of the silver nitrate solution must be added to the sodium
chloride solution in order to precipitate the maximum amount of silver chloride
without wasting a drop of quite a costly solution! Three steps to the
calculation:- (a) calculate the relative formula masses of the reactants, (b)
the mass of silver nitrate reacting and (c) the volume of silver nitrate needed.
(b) From the equation calculate the mass of
silver nitrate that reacts with 0.5 g of sodium chloride.
According to the symbol equation one RFM or 'molecule'
of sodium chloride reacts with one RFM or 'molecule' of silver nitrate.
|
NaCl |
: |
AgNO3 |
|
58.5 |
: |
169.8 |
|
0.5g |
: |
x g |
Solving the ratio gives x = 0.5 x 169.8 /
58.5 = 1.451g of silver nitrate
(b) From your answer to (a) calculate the volume
of silver nitrate needed.
V = m / c = 1.451 / 30 = 0.0484 dm3
V = 1000 x 0.0484 = 48.4 cm3
Example Question 6b (Q6)
The reaction
between hydrochloric acid and sodium hydroxide is
| hydrochloric acid |
+ |
sodium hydroxide |
===> |
sodium chloride |
+ |
water |
| HCl(aq) |
+ |
NaOH(aq) |
===> |
NaCl(aq) |
+ |
H2O(l) |
| RFM = 36.5 |
|
RFM = 40 |
|
|
|
|
According to the symbol equation one RFM or 'molecule'
of hydrochloric acid reacts with one RFM or 'molecule' of sodium hydroxide.
(The atomic masses involved are H = 1, Cl = 35.5,
Na = 23 and O = 16, check the RFM for yourself)
Therefore from the relative formula masses (RFMs)
36.5g of HCl reacts with 40g of sodium hydroxide.
You are given a standard solution of hydrochloric acid of
7.3 g/dm3 (of HCl). 25.0 cm3 of sodium hydroxide solution
was pipetted into a conical flask. On titration with the acid solution using a
burette and suitable indicator, it was found that 14.6 cm3 of the
acid solution was required to completely neutralise the alkaline sodium
hydroxide. Calculate the concentration of the sodium hydroxide solution in g/dm3
via the stages outlined below
(a) The mass of hydrochloric acid reacting.
V = 14.6 cm3/1000 = 0.0146 dm3
m = c x V = 7.3 x 0.0146 = 0.1066 g HCl
(b) Calculate the mass of NaOH that reacts with
0.1066 g of HCl
|
HCl |
: |
NaOH |
|
36.5 |
: |
40 |
|
0.1066g |
: |
x g |
Solving the ratio x = 0.1066 x 40/36.5 =
0.1168 g
(c) Calculate the concentration of NaOH
V = 25 cm3/1000 = 0.025 dm3
c = 0.1168/0.025 = 4.67 g/dm3
(2dp, 3sf)
Now it is possible to titrate the acid solution
with the alkali solution (or vice versa) to obtain an unknown concentration of
one of the solutions. One concentration must be known and the two volumes (acid
and alkali) which react together exactly i.e. no excess of either reactant
solution. How to do this is described in
Section 12. acid-alkali titrations
Example 6b
(Q7)
Citric acid is the most common acid in citrus
fruits and can be estimated by titration with sodium hydroxide.
| citric acid |
+ |
sodium hydroxide |
===> |
sodium citrate |
+ |
water |
 (aq) |
+ |
3NaOH(aq) |
===> |
 (aq) |
+ |
H2O(l) |
| RFM = 192
for C6H8O7 |
|
RFM = 40
but reacting mass is
3 x 40 = 120 |
|
|
|
|
You are a given of sodium hydroxide with a
concentration of 12.0 g/dm3.
10 cm3 of squeezed lemon juice was
pipetted into a flask. Using a burette and suitable indicator the lemon juice
required a titration volume of 35.5 cm3 of the sodium hydroxide
solution to completely neutralise it.
Assuming all the acidity is due to citric acid,
calculate the concentration of citric acid in the lemon juice.
(a) What mass of sodium hydroxide reacted with
the lemon juice?
V = 35.5 cm3/1000 = 0.0355 dm3
m = c x V = 12.0 x 0.0355 = 0.426g NaOH
(b) What mass of citric acid reacted with the
0.426 g of sodium hydroxide?
|
C6H8O7 |
: |
3NaOH |
|
192 |
: |
3 x 40 = 120 |
|
x g |
: |
0.426 g |
Solving the ratio x = 0.426 x 192/120 =
0.6816 g C6H8O7
(c) Calculate the concentration of citric acid in
the lemon juice?
V = 10 cm3/1000 = 0.01 dm3
c = m / V = 0.6816/0.01 = 68.2 g/dm3
C6H8O7
-
Example
Question 6b (Q8)
-
This is a much more elaborate reacting mass calculation involving
solution concentrations and extended ideas from the results.
-
In this exemplar Q I've
used simple formulae a lot for short-hand.
-
A solution of
hydrochloric contained 7.3 g HCl/dm3.
-
A solution of a
metal hydroxide of formula MOH was prepared by dissolving 4.0g of
MOH in 250 cm3 of water.
-
M is an unknown metal but
it is known that the ionic formula of the hydroxide is M+OH-.
25cm3 samples of the MOH solution were pipetted into a
conical flask and titrated with the hydrochloric solution using a
burette and a few drops of phenolphthalein indicator.
-
All the MOH is
neutralised as soon as the pink indicator colour disappears (i.e.
the indicator becomes colourless).
-
On average 19.7 cm3 of
the HCl acid solution was required to completely neutralise 25.0 cm3
of the MOH solution.
-
[Atomic masses: H = 1, Cl = 35.5, O = 16, M =
?]
-
(a) Give
the equation for the reaction between the metal hydroxide and the
hydrochloric acid.
-
MOH(aq)
+ HCl(aq) ==> MCl(aq) + H2O(l)
-
You may or may
not be required to give the state symbols in (), or you may be just
asked to complete the equation given part of it.
-
(b)
Calculate the mass of HCl used in each titration.
-
(c)
Calculate the mass of MOH that reacts with the mass of HCl
calculated in (b).
-
(d)
Calculate the formula mass of HCl.
-
(e)
Calculate the mass in g of MOH that reacts with 36.5g of HCl and
hence the formula mass of MOH.
-
0.1438 g HCl
reacts with 0.40 g MOH
-
therefore 36.5g HCl reacts with z g of MOH
-
solving the
ratio for z, z = 36.5 x 0.40 / 0.1438 = 101.5 g MOH
-
Since the
formula mass of HCl is 36.5 and from the equation, 1 MOH reacts with
1 HCl the experimental formula mass of MOH is found to be 101.5
-
(f) What
is the atomic mass of the metal?
-
(g) From
the formula information on the metal hydroxide deduce the following
giving reasons:
-
Example Question 6b (Q9)
Empirical formula calculation - another more elaborate reacting mass calculation involving
solution concentrations to arrive at a formula mass.
-
I've used
HCl and MOH as shorthand in the question and answers.
-
The idea is to use the
analysis data to work out the atomic mass of metal M, that forms a
hydroxide of formula MOH.
-
A solution of
hydrochloric contained 3.65 g/dm3.
-
A solution of a metal
hydroxide of formula MOH was prepared by dissolving 5.0g of MOH in 1
dm3 of water (M is an unknown metal).
-
25 cm3
of the MOH solution required 22.3 cm3 of the HCl acid
solution to neutralise it in a titration procedure using a pipette
(MOH) and burette (HCl).
-
The equation for
the neutralisation reaction is: MOH + HCl ==> MCl + H2O
-
Atomic masses: H
= 1, Cl = 35.5, O = 16, M = ?
-
(a) Calculate
the mass of MOH neutralised in each titration.
-
(b) Calculate
the mass of HCl reacting in each titration.
-
(d) Calculate
the formula mass of HCl
-
(c) Calculate
the mass of MOH that reacts with 36.5 g HCl and hence the formula
mass of MOH.
-
If 0.125 g MOH
reacts with 0.0814 g HCl
-
z g MOH reacts
with 36.5 g HCl
-
solving the
ratio, z = 36.5 x 0.125 / 0.0814 = 56.1 g MOH
-
Therefore the
experimental formula mass of MOH is 56.1 (~56) because from
the equation 1 HCl reacts with 1 MOH.
-
(d) If the metal
is in the Group 1 of Alkali Metals, what is the atomic mass of M and
what metal is M?
-
If the formula
mass of MOH is 56, atomic mass of M = 56 - 1 -16 = 39
-
The atomic mass
of potassium is 39, so M is potassium.
-
(So 'fictitious'
MOH is really KOH, potassium hydroxide. You are likely to very
familiar with another in the same group, sodium hydroxide NaOH)
-
-
-
Example 6b. (Q10) A Purity calculation based on reacting masses
and solution concentrations.
-
This question is based the
titration of aspirin (an acid), with standard sodium hydroxide solution.
-
0.300g of aspirin was titrated
with sodium hydroxide solution of concentration 4.00g/dm3.
-
If the aspirin required 16.45 cm3
of the NaOH(aq) to neutralise it, calculate the percent purity of the
aspirin.
-
The simplified equation for the
reaction is ...
-
C6H4(OCOCH3)COOH
+ NaOH ==> C6H4(OCOCH3)COONa + H2O
-
Mr(aspirin) = 180,
Mr(NaOH) = 40 (atomic masses: C = 12, H = 1, O = 16,
Na = 23)
-
Therefore the reacting mass
ratio is 180g aspirin reacts with 40g of sodium hydroxide.
-
The titration was 16.45 cm3, so,
converting the cm3 to dm3,
-
the mass of NaOH used in the
titration = 4.00 x 16.45/1000 = 0.0658g,
-
so we can scale this up to get
the mass of aspirin titrated,
-
therefore the mass of aspirin
titrated = 0.0658 x 180 / 40 = 0.296g
-
therefore the % purity
= 100 x 0.296 / 0.300 = 98.7%
-
-
-
Example 6b. (Q11)
also based on reacting masses and solution concentrations.
-
Deposition from hard
water samples
-
On analysis, a sample of hard water was found to contain 0.056 mg of
calcium hydrogen carbonate per cm3 (0.056 mg/ml).
-
If the
water is boiled, calcium hydrogencarbonate Ca(HCO3)2,
decomposes to give a precipitate of calcium carbonate CaCO3,
water and carbon dioxide.
-
(a) Give the
symbol equation of the decomposition complete with state symbols.
-
(b) Calculate
the mass of calcium carbonate in grammes deposited if 2 litres (2 dm3,
2000 cm3 or ml) is boiled in a kettle.
-
[ atomic
masses: Ca = 40, H = 1, C = 12, O = 16 ]
-
the relevant
ratio is based on: Ca(HCO3)2 ==> CaCO3
-
The formula
masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40
+ 12 + 16x3) respectively
-
there the
reacting mass ratio is 162 units of Ca(HCO3)2
==> 100 units of CaCO3
-
the mass of
Ca(HCO3)2 in 2000 cm3 (ml) = 2000 x
0.056 = 112 mg
-
therefore
solving the ratio 162 : 100 and 112 : z mg CaCO3
-
where z =
unknown mass of calcium carbonate
-
z = 112 x
100/162 = 69.1 mg CaCO3
-
since 1g = 1000
mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate
-
(c) Comment
on the result, its consequences and why is it often referred to as
'limescale'?
-
This precipitate
of calcium carbonate will cause a white/grey deposit to be formed on
the side of the kettle, especially on the heating element.
-
Although
0.0691 g doesn't seem much, it will build up appreciably after many
cups of tea!
-
The precipitate is calcium carbonate, which occurs
naturally as the rock limestone, which dissolved in rain
water containing carbon dioxide, to give the calcium hydrogen
carbonate in the first place.
-
Since the deposit of 'limestone'
builds up in layers it is called 'limescale'.
See also 12.
Volumetric titration methods & calculations using
molarity & mole concept
AND see also 11.
Molarity calculations
and 14.3
dilution of solutions calculations
TOP OF PAGE
APPENDIX
- How to make up a standard solution (mass/volume)
The method and procedure of how to make up a
standard solution of a soluble solid e.g. a salt, is fully described.
For
making up a standard solution of known molarity see the page
How to do acid-alkali titrations
and calculations, diagrams of apparatus,
details of procedures
Suppose you want to make up 250cm3 of a salt
solution of concentration 20g/dm3 (20g/litre, 20g/1000cm3,
20g/1000ml).
c = m / v, m = c x v, m = 20 x 250 / 1000 = 5g
so 5g of the salt is needed to be made up into an
aqueous solution of exactly 250.0 cm3.
 An
accurate one pan electronic balanced is set to zero (preferably with an accuracy
of two decimal places). A beaker is placed on the balance and the reading noted
(92.54g).
 Very
carefully, with a spatula (not shown), salt crystals are added to the beaker
until it weighs exactly 5.00 grams more (97.54g). This can be a very fiddly
procedure if you want exactly 5.00g of salt.
 Pure
water is then added to the beaker to completely dissolve the salt and use of a
stirring rod helps to speed up the process. The amount of water you add to the
beaker should be much less than 250cm3 to allow for the transfer and
rinsing of the solution into the standard volumetric flask.
 Eventually
a clear solution of the salt should be seen, there should be no residual salt
crystals at the bottom of the beaker or on the sides of the beaker. You can use
the wash bottle to rinse round the inside if there any crystals on the inside
surface of the beaker.
 An
accurately calibrated 250cm3 volumetric flask should be washed out
and cleaned several times with pure water. Then, the whole of the solution in
the beaker is transferred into the flask with the help of a funnel to avoid the
risk of spillage. To make sure every drop of the salt solution ends up in the
flask, a wash bottle of pure water is used to rinse out the beaker several
times, the stirring rod and in the end the funnel too.
 Then,
very carefully, the flask is topped up with pure water so the meniscus rests
exactly on the 250.0cm3 calibration mark, a teat pipette is useful
for the last portions of water. The stopper is placed on and the flask carefully
shaken quite a few times to ensure the salt solution is completely mixed up.
Finally, check the meniscus is still on the calibration mark, in case another
few drops are needed. Either way, the last drops of water should be added most
carefully with a teat pipette.
For
making up a standard solution of known molarity see the page
How to do acid-alkali titrations
and calculations, diagrams of apparatus,
details of procedures
TOP OF PAGE
Above is typical periodic table used in GCSE science-chemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
OTHER CALCULATION PAGES
-
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
-
Calculating relative
formula/molecular mass of a compound or element molecule
-
Law of Conservation of Mass and simple reacting mass calculations
-
Composition by percentage mass of elements
in a compound
-
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
-
Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination
-
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
-
Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
-
Moles and the molar volume of a gas, Avogadro's Law
-
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants-products)
-
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
-
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
-
Electrolysis products calculations (negative cathode and positive anode products)
-
Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
-
Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
-
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
-
Radioactivity & half-life calculations including
dating materials
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