6b.
Reacting masses, concentration of solution and volumetric titration calculations
NOT
using the mole concept BUT see also
Basic
acidalkali titration questions using moles
far more appropriate!
Advanced Level acidalkali 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/dm^{3}, which was
g/litre.
concentration = mass / volume
(c = m / V)
concentration in gdm^{3},
mass in g, volume in dm^{3}
Its really important remember
that 1 dm^{3} = 1000 cm^{3} or
1000 ml
and cm^{3}/1000 =
dm^{3}
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 cm^{3} of
water?
250 cm^{3} is equal to
250/1000 = 0.25dm^{3}
therefore the concentration
c = m /
V =
10/0.25 = 40 g/dm^{3}
(40g/litre in old units, still in
common use!)
(b) What mass of the salt is
required to make 200 cm^{3} of concentration 15g/dm^{3}?
V = 200/1000 = 0.2 dm^{3}
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 dm^{3} and cm^{3}, should you dissolve
it in, to give a salt solution of concentration of 5.0g/dm^{3}?
V = m / c = 8.0 / 5.0 = 1.6 dm^{3}
V = 1.6 x 1000 = 1600 cm^{3}
TOP OF PAGE
QUESTIONS
involving mass, volume and concentration

Example Question 6b (Q1)
A potassium sulfate solution A has a concentration of 6.5g/dm^{3}

(a) What mass of potassium
sulfate is present in 25cm^{3} of solution A?

(b) What volume of
solution A, in cm^{3}, contains 3.0g of potassium sulphate?

(c) What mass of the salt
is required to make 250 cm^{3} of a solution B so it
contains 3.75 g/dm^{3} of potassium sulfate?

(d) If 12.5g of potassium
sulphate was dissolved in 1500 cm^{3} 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/dm^{3}

(a) What mass of potassium
sulfate is present in 25cm^{3} of the solution?

(b) What volume of
solution, in cm^{3}, contains 3.0g of potassium sulphate?

(c) What mass of the salt
is required to make 250 cm^{3} of a solution B so it
contains 3.75 g/dm^{3} of potassium sulfate?

(d) If 12.5g of
potassium sulphate was dissolved in 1500 cm^{3} 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

HNO_{3}(aq) +
NaOH(aq) ===> NaNO_{3}(aq) + H_{2}O(l)

(a) Calculate the relative
formula masses of nitric acid, sodium hydroxide and sodium nitrate

25.0 cm^{3} of a
nitric acid solution was pipetted into a conical flask and titrated
with a sodium hydroxide solution of concentration 5.0g/dm^{3}.
If it took 16.8 cm^{3} 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/dm^{3}?

(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 
HNO_{3}(aq) 
+ 
NaOH(aq) 
===> 
NaNO_{3}(aq) 
+ 
2H_{2}O(l) 
RFM 63 

40 

85 

 

25.0 cm^{3} of a
nitric acid solution was pipetted into a conical flask and titrated
with a sodium hydroxide solution of concentration 5.0g/dm^{3}.
If it took 16.8 cm^{3} 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.

HNO_{3} 
: 
NaOH 
63 
: 
40 
x g 
: 
0.084 g 
 Solving the ratio: x = 0.084 x
63/40 = 0.1323 g HNO_{3}

(d) What was the
concentration of the nitric acid in g/dm^{3}?

(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 
: 
NaNO_{3} 
40 
: 
85 
0.084 g 
: 
x g 
 Solving the ratio: x = 0.084
x 85/40 = 0.1785 g NaNO_{3}
 You can also use HNO_{3}
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) ===> CaCl_{2}(aq) + 2H_{2}O(l)

50 cm^{3} of a
limewater solution was completely neutralised by 9.7 cm^{3}
of a hydrochloric acid solution. If the concentration of the
hydrochloric acid was 3.65 g HCl/dm^{3} 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/dm^{3}?





Example Question 6b
(Q4)
Sulfuric acid  potassium hydroxide reaction to make
potassium sulfate

The neutralisation
reaction between sulphuric acid and potassium hydroxide is

H_{2}SO_{4}(aq)
+ 2KOH(aq) ===> K_{2}SO_{4}(aq) + 2H_{2}O(l)

A solution of
potassium hydroxide contained 100g/dm^{3}.

A solution of
sulfuric acid contained 120g/dm^{3}.

What volume in cm^{3},
of the sulfuric acid solution, is required to neutralise 250 cm^{3}
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

H_{2}SO_{4}(aq)
+ 2KOH(aq) ===> K_{2}SO_{4}(aq) + 2H_{2}O(l)

Potassium hydroxide
contained 100g/dm^{3} and sulfuric acid contained
120g/dm^{3}.

What volume in cm^{3},
of the sulfuric acid solution, is required to neutralise 250 cm^{3}
of the potassium hydroxide solution?

To calculate mass of
KOH

To calculate mass of
H_{2}SO_{4} needed

H_{2}SO_{4} 
: 
2KOH 
98 
: 
2 x 56 = 112 
x g 
: 
25 g 
 Solving the ratio: x = 25 x
98/112 = 21.875 g H_{2}SO_{4}

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/dm^{3}.
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) 
+ 
AgNO_{3}(aq) 
===> 
AgCl(s) 
+ 
NaNO_{3}(aq) 
(a) RFM 58.5 

169.8 

(143.3) 

(85) (not needed) 
Using an accurate burette, precisely what volume
(in cm^{3}) 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 
: 
AgNO_{3} 
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 dm^{3}
V = 1000 x 0.0484 = 48.4 cm^{3}
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) 
+ 
H_{2}O(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/dm^{3} (of HCl). 25.0 cm^{3} 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 cm^{3} of the
acid solution was required to completely neutralise the alkaline sodium
hydroxide. Calculate the concentration of the sodium hydroxide solution in g/dm^{3}
via the stages outlined below
(a) The mass of hydrochloric acid reacting.
V = 14.6 cm^{3}/1000 = 0.0146 dm^{3}
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/dm^{3}
(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. acidalkali 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) 
+ 
H_{2}O(l) 
RFM = 192
for C_{6}H_{8}O_{7} 

RFM = 40
but reacting mass is
3 x 40 = 120 




You are a given of sodium hydroxide with a
concentration of 12.0 g/dm^{3}.
10 cm^{3} 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 cm^{3} 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 cm^{3}/1000 = 0.0355 dm^{3}
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?
C_{6}H_{8}O_{7} 
: 
3NaOH 
192 
: 
3 x 40 = 120 
x g 
: 
0.426 g 
Solving the ratio x = 0.426 x 192/120 =
0.6816 g C_{6}H_{8}O_{7}
(c) Calculate the concentration of citric acid in
the lemon juice?
V = 10 cm^{3}/1000 = 0.01 dm3
c = m / V = 0.6816/0.01 = 68.2 g/dm^{3}
C_{6}H_{8}O_{7}

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 shorthand.

A solution of
hydrochloric contained 7.3 g HCl/dm^{3}.

A solution of a
metal hydroxide of formula MOH was prepared by dissolving 4.0g of
MOH in 250 cm^{3} of water.

M is an unknown metal but
it is known that the ionic formula of the hydroxide is M^{+}OH^{}.
25cm^{3} 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 cm^{3} of
the HCl acid solution was required to completely neutralise 25.0 cm^{3}
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)} + H_{2}O_{(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.

1 dm^{3}
= 1000 cm^{3}, so in 19.7 cm^{3} of the HCl solution
there will be

19.7 x 7.3 /
1000 = 0.1438 g HCl

(c)
Calculate the mass of MOH that reacts with the mass of HCl
calculated in (b).

25cm^{3}
of the 250 cm^{3} MOH solution was used, so the mass of MOH
titrated is

25 x 4 / 250 =
0.40 g MOH

(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/dm^{3}.

A solution of a metal
hydroxide of formula MOH was prepared by dissolving 5.0g of MOH in 1
dm^{3} of water (M is an unknown metal).

25 cm^{3}
of the MOH solution required 22.3 cm^{3} 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 + H_{2}O

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/dm^{3}.

If the aspirin required 16.45 cm^{3}
of the NaOH(aq) to neutralise it, calculate the percent purity of the
aspirin.

The simplified equation for the
reaction is ...

C_{6}H_{4}(OCOCH_{3})COOH
+ NaOH ==> C_{6}H_{4}(OCOCH_{3})COONa + H_{2}O

M_{r}(aspirin) = 180,
M_{r}(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 cm^{3} to dm^{3},

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 cm^{3} (0.056 mg/ml).

If the
water is boiled, calcium hydrogencarbonate Ca(HCO_{3})_{2},
decomposes to give a precipitate of calcium carbonate CaCO_{3},
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 dm^{3},
2000 cm^{3} or ml) is boiled in a kettle.

[ atomic
masses: Ca = 40, H = 1, C = 12, O = 16 ]

the relevant
ratio is based on: Ca(HCO_{3})_{2} ==> CaCO_{3}

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(HCO_{3})_{2}
==> 100 units of CaCO_{3}

the mass of
Ca(HCO_{3})_{2} in 2000 cm^{3} (ml) = 2000 x
0.056 = 112 mg

therefore
solving the ratio 162 : 100 and 112 : z mg CaCO_{3}

where z =
unknown mass of calcium carbonate

z = 112 x
100/162 = 69.1 mg CaCO_{3}

since 1g = 1000
mg, z = 69.1/1000 = 0.0691 g CaCO_{3}, 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 acidalkali titrations
and calculations, diagrams of apparatus,
details of procedures
Suppose you want to make up 250cm^{3} of a salt
solution of concentration 20g/dm^{3} (20g/litre, 20g/1000cm^{3},
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 cm^{3}.
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 250cm^{3} 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 250cm^{3} 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 250.0cm^{3} 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 acidalkali titrations
and calculations, diagrams of apparatus,
details of procedures
TOP OF PAGE
Above is typical periodic table used in GCSE sciencechemistry 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

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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 GayLussac's Law (ratio of gaseous
reactantsproducts)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)

How to do acidalkali
titration calculations, diagrams of apparatus, details of procedures

Electrolysis products calculations (negative cathode and positive anode products)

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(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy

Energy transfers in physical/chemical changes,
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Gas calculations involving PVT relationships,
Boyle's and Charles Laws

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