Double bonds or not A saturated fat has no C=C double bonds (alkene functional groups) and is...
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- Double bonds or not A saturated fat has no C=C double bonds
(alkene functional groups) and is usually a solid fat like
margarine or animal fat. An unsaturated fat has C=C double bonds
and is usually an oil like vegetable oil. 1
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- Example question 2
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- Mark scheme 3
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- Making margarine To make margarine we have to saturate
vegetable oil by bubbling hydrogen gas through the oil. This
process is called hydrogenation 4
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- Is a fat or oil saturated or not? 5
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- Example question 6
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- Mark scheme 7
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- Hydrolysis When an ester is hydrolysed it goes back to an acid
and alcohol We can hydrolyse by adding acid or alkali (NaOH).
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- Example question 9
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- Mark scheme 10
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- Energy changes in chemistry C7.2 11
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- Quiz When a chemical reaction takes place heat may be given out
or taken in. 1.Can you remember the word we use when heat is given
out? 2.Can you remember the word we use when heat is taken in?
12
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- What do I need to know? 1.Recall and use the terms ENDOTHERMIC
and EXOTHERMIC 2.Describe examples of ENDOTHERMIC and EXOTHERMIC
reactions. 3.Use simple energy level diagrams to represent
ENDOTHERMIC and EXOTHERMIC reactions. 13
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- Change in energy 14
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- Definitions 15
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- Energy level diagrams Heat taken in Heat given out 16
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- Energy level diagrams EndothermicExothermic Heat taken in Heat
given out Energy level of products is higher than reactants so heat
taken in. Energy level of products is lower than reactants so heat
given out. 17
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- Example question 18
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- Mark scheme 19
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- Bond enthalpies C7.2 20
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- Quick quiz 1.Reactions where the products are at a lower energy
than the reactants are endothermic (TRUE/FALSE) 2.Activation energy
is the amount of energy given out when a reaction takes place
(TRUE/FALSE) 3.A reaction which is exothermic transfers heat energy
to the surroundings (TRUE/FALSE) 4.How can we tell if a reaction is
exothermic or endothermic? 5.Sketch the energy profile for an
endothermic reaction. 6.When methane (CH 4 ) burns in oxygen (O 2 )
bonds between which atoms need to be broken? 21
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- Answers 1.Reactions where the products are at a lower energy
than the reactants are endothermic (TRUE/FALSE) 2.Activation energy
is the amount of energy given out when a reaction takes place
(TRUE/FALSE) 3.A reaction which is exothermic transfers heat energy
to the surroundings (TRUE/FALSE) 4.How can we tell if a reaction is
exothermic or endothermic? 5.Sketch the energy profile for an
endothermic reaction. 6.When methane (CH 4 ) burns in oxygen (O 2 )
bonds between which atoms need to be broken? FALSE TRUE Measure the
temperature change CH bonds and O=O bonds 22
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- What do I need to know? 1. Recall that energy is needed to
break chemical bonds and energy is given out when chemical bonds
form 2. Identify which bonds are broken and which are made when a
chemical reaction takes place. 3. Use data on the energy needed to
break covalent bonds to estimate the overall energy change for a
reaction. 23
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- Activation energy revisited What is the activation energy of a
reaction? The energy needed to start a reaction. BUT what is that
energy used for and why does the reaction need it if energy is
given out overall? The activation energy is used to break bonds so
that the reaction can take place. 24
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- Burning methane Consider the example of burning methane gas. CH
4 + 2O 2 CO 2 + 2H 2 O This reaction is highly exothermic, it is
the reaction that gives us the Bunsen flame. However mixing air
(oxygen) with methane is not enough. I need to add energy (a
flame). 25
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- What happens when the reaction gets the activation energy? Bond
Forming Bond Breaking Progress of reaction Energy in chemicals O O
O O H C H H H O O O O C H HHH O C O O O H H H H 26
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- Using bond enthalpies By using the energy that it takes to
break/make a particular bond we can work out the overall
enthalpy/energy change for the reaction. Sum (bonds broken) Sum
(bonds made) = Energy change 27
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- BIN MIX Breaking bonds is ENDOTHERMIC energy is TAKEN IN when
bonds are broken Making bonds is EXOTHERMIC energy is GIVEN OUT
when bonds are made. 28
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- Bond enthalpies BondBond enthalpy (kJ)BondBond enthalpy (kJ)
CH435ClCl243 CC348CCl346 HH436HCl452 HO463O=O498 C=O804 C=C614
29
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- Can you work out the energy change for this reaction? 30
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- The answer is -120 kJ 31
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- Example question part 1 32
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- Question part 2 33
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- Question part 3 34
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- Mark scheme 35
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- Challenge question The true value for the energy change is
often slightly different from the value calculated using bond
enthalpies. Why do you think this is? 36
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- Example question The calculated value is 120 kJ 37
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- Mark scheme 38
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- Definitions Write each of these phrases in your book with a
definition in your own words: Exothermic reaction Endothermic
reaction Activation energy Catalyst Bond energy/enthalpy 39
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- How did you do? Exothermic reaction A reaction which gives
energy out to the surroundings. Endothermic reaction A reaction
which takes in energy from the surroundings. Activation energy The
energy required to start a reaction by breaking bonds in the
reactants Catalyst A substance that increases the rate of a
reaction by providing an alternative pathway with lower activation
energy. It is not used up in the process of the reaction Bond
energy/enthalpy The energy required to break a certain type of
bond. The negative value is the energy given out when that bond is
made. 40
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- Popular exam question 1.Explain why a reaction is either
exothermic or endothermic?
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- Popular exam question 1.Explain why a reaction is either
exothermic or endothermic? In a chemical reactions some bonds are
broken and some bonds are made. Breaking bonds takes in energy.
Making bonds gives out energy. If the energy given out making bonds
is higher than the energy needed to break them the reaction is
exothermic. If the energy needed to break bonds is higher than the
energy given out making them the reaction is endothermic. 42
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- Chemical Equilibria C7.3 Reversible Reactions & Dynamic
Equilibria 43
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- What do I need to know? 1.State that some chemical reactions
are reversible 2.Describe how reversible reactions reach a state of
equilibrium 3.Explain this using dynamic equilibrium model. 44
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- Reversible or not reversible Until now, we were careful to say
that most chemical reactions were not reversible They could not go
back to the reactants once the products are formed. 45
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- Example 46
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- Reversible Some chemical reactions, however, will go backwards
and forwards depending on the conditions. CoCl 2 6H 2 O(s) CoCl 2
(s) + 6H 2 O(l) pink blue 47
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- How do we write them down? This is the symbol for used for
reversible reactions. CoCl 2 6H 2 O(s) CoCl 2 (s) + 6H 2 O(l)
48
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- What is equilibrium? 49
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- Dynamic Equilibrium. 50
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- Example question 51
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- Mark scheme 52
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- Dynamic Equilibria C7.3 Controlling equilibria 53
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- What do I need to know? 1. Recall that reversible reactions
reach a state of dynamic equilibrium. 2. Describe how dynamic
equilibria can be affected by adding or removing products and
reactants. 3. Explain the difference between a strong and weak acid
in terms of equilibria 54
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- Position of the equilibrium Equilibrium can lie to the left or
right. This is in favour of products or in favour of reactants
Meaning that once equilibrium has been reached there could be more
products or more reactants in the reaction vessel. 55
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- Le Chateliers principle If you remove product as it is made
then equilibrium will move to the right to counteract the change If
you add more reactant then equilibrium will move to the right to
counteract the change. In industry we recycle reactants back in and
remove product as it is made to push the equilibrium in favour of
more product. 56
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- Complete When a system is at__________ to make more product you
can_________ product or add more __________ for example by
recycling them back in. To return to reactants you ______ product
or remove_________. [equilibrium, add, reactant, remove, product]
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- Strong and weak acids 58
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- Mark scheme 60
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- Example question 61
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- Mark scheme 62
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- Practicing definitions Write each of these phrases in your book
with a definition in your own words: Reversible reaction Dynamic
equilibrium Position of equilibrium Strong acid Weak acid 63
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- How did you do? Reversible reaction A reaction that can proceed
in the forward or reverse directions (represented by two arrows in
an equation). Dynamic equilibrium The point where the rate of the
forward reaction = rate of the reverse reaction. Position of
equilibrium The point where there is no further change in the
concentration of either reactants or products. The position can lie
to the left (favouring reactants) or right (favouring products).
Strong acid An acid that is completely dissociated in water Weak
acid An acid that is only partly dissociated in water because the
reaction is in dynamic equilibrium and favours the reactants (LHS).
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- Popular exam question 1.Ethanoic acid (CH 3 COOH) is a weak
acid but hydrochloric acid is a strong acid. Use ideas about ion
formation and dynamic equilibrium to explain this difference.
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- Popular exam question Ethanoic acid (CH 3 COOH) is a weak acid
but hydrochloric acid is a strong acid. Use ideas about ion
formation and dynamic equilibrium to explain this difference.
Hydrochloric acid ionises completely So hydrogen ion concentration
is high Ethanoic acid only partly dissociates because the reaction
is reversible Equilibrium is mainly to the left So hydrogen ion
concentration is low. 66
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- Analysis C7.4 Analytical Procedures 67
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- What do I need to know? 1. Recall the difference between
qualitative and quantitative methods of analysis. 2. Describe how
analysis must be carried out on a sample that represents the bulk
of the material under test 3. Explain why we need standard
procedures for the collection, storage and preparation of samples
for analysis 68
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- Qualitative vs. Quantitative 69
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- Which sample should I test? 70
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- Chemical industry 71
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- Chromatography C7.4 paper chromatography 72
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- Chromatography 73
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- Solvents 1.The mobile phase is the solvent the part that moves
2. In paper chromatography it is water or ethanol 74
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- Paper/column 1.The stationary phase is the paper in paper
chromatography or the column in gas chromatography. 2.In thin layer
chromatography it is silica gel on a glass plate 3.The stationary
phase does not move. 75
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- How does the technique work? In chromatography, substances are
separated by movement of a mobile phase through a stationary phase.
Each component in a mixture will prefer either the mobile phase OR
the stationary phase. The component will be in dynamic equilibrium
between the stationary phase and the mobile phase. 76
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- Substance A This is substance A Substance A prefers the
stationary phase and doesnt move far up the paper/column. The
equilibrium lies in favour of the stationary phase. 77
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- Substance B This is substance B Substance B prefers the mobile
phase and moves a long way up the paper/column The equilibrium lies
in favour of the mobile phase 78
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- Using a reference In chromatography we can sometimes use a
known substance to measure other substances against. This will
travel a known distance compared to the solvent and is known as a
standard reference. 79
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- Advantages of TLC TLC has a number of advantages over paper
chromatography. It is a more uniform surface chromotograms are
neater and easier to interpret Solvent can be used which is useful
if a substance is insoluble in water. 80
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- Past Paper Questions 81
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- Past paper question 82
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- Mark scheme 83
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- Describing how chromatography works exam definition stationary
phase is paper and mobile phase is solvent / mobile phase moves up
through stationary phase (1) for each compound there is a dynamic
equilibrium between the two phases (1) how far each compound moves
depends on its distribution between the two phases (1) 84
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- Using an R f value In order to be more precise we can use
measurements on the TLC plate to compare the distance travelled by
our substance (the solute) with the distance travelled by the
solvent. The Rf value is constant for a particular compound. The
distance travelled however could be different on different
chromatograms. The Rf value is always less than 1. 85
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- Rf value 86
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- Example question This question relates to the chromatogram
shown in the earlier question. Refer back 87
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- Mark scheme 88
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- Example question 89
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- Mark scheme 90
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- Past paper question 91
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- Mark scheme 95
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- Gas-liquid chromatography C7.4 GLC 96
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- What do I need to know? 1.recall in outline the procedure for
separating a mixture by gas chromatography (gc); 2.understand the
term retention time as applied to gc; 3.interpret print-outs from
gc analyses. 97
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- Gas chromatography The mobile phase is an unreactive gas known
as the carrier gas this is usually nitrogen The stationary phase is
held inside a long column and is lots of pieces of inert solid
coated in high bp liquid. The column is coiled in an oven The
sample to be analysed is injected into the carrier gas stream at
the start of the column. 98
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- GC 99
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- GC analysis Each component of the sample mixture has a
different affinity for the stationary phase compared with the
mobile phase Therefore each component travels through the column in
a different time. Compounds favouring the mobile phase (usually
more volatile) emerge first. A detector monitors the compounds
coming out of the column and a recorder plots the signal as a
chromatogram 100
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- GLC Chromatograph 101
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- Interpretation The time in the column is called the retention
time Retention times are characteristic so can identify a compound
Area under peak or relative heights can be used to work out
relative amounts of substances 102
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- The key points revise this! the mobile phase carries the sample
(1) components are differently attracted to the stationary and
mobile phases (1) the components that are more strongly attracted
to the stationary phase move more slowly (1) the amount of each
component in the stationary phase and in the mobile phase is
determined by a dynamic equilibrium (1) 103
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- Past paper question 104
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- Mark scheme 108
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- Titration C7.4 109
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- What do I need to know? 1.Calculate the concentration of a
given volume of solution given the mass of solvent; 2.Calculate the
mass of solute in a given volume of solution with a specified
concentration; 3.Use the balanced equation and relative
formula-masses to interpret the results of a titration; 110
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- Concentration We can measure the concentration of solution in
grams/litre. This is the same as g/dm 3 1dm 3 = 1000cm 3 If I want
to make a solution of 17 g/dm 3 how much will I dissolve in 1dm 3.
17 g If I want to make a solution of 17g/dm 3 but I only want to
make 100cm 3 of it how much will I dissolve? 1.7g 111
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- Making standard solutions For a solution of 17g/dm 3 First I
will measure 17g of solid on an electronic balance 112
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- Making standard solutions Now I must dissolve it in a known 1dm
3 of water. I transfer it to a volumetric flask and fill up with
distilled water to about half the flask. I then swirl to dissolve
Top up with a dropping pipette so that the meniscus is on the line.
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- How much to dissolve? Worked example: I want to make 250cm 3 of
a solution of 100g/dm 3. How much solid do I transfer to my 250cm 3
volumetric flask? 114
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- How much to dissolve? Worked example: I want to make 250cm 3 of
a solution of 100g/dm 3. 1. Work out the ratio of 250cm 3 to 1000cm
3 250/1000 = 0.25 2. I therefore need 0.25 of 100g in 250cm 3 which
is 0.25x100=25g 115
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- General rule 116
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- Practie - how much to dissolve? I want to make 250cm 3 of a
solution of 63.5g/dm 3. How much solid do I transfer to my 250cm 3
volumetric flask? 250/1000 x 63.5 = 15.9 g 117
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- Practice - how much to dissolve? I want to make 100cm 3 of a
solution of 63.5g/dm 3. How much solid do I transfer to my 100cm 3
volumetric flask? 100/1000 x 63.5 = 6.35 g 118
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- Concentration from mass and volume We need to rearrange this:
To give 119
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- What is the concentration of? 1.12g dissolved in 50cm 3 2.50g
dissolved in 100cm 3 3.47g dissolved in 1000cm 3 4.200g dissolved
in 250cm 3 120
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- What is the concentration of? 1.12g dissolved in 50cm 3 =
1000/50 x 12 = 240g/dm 3 2.50g dissolved in 100cm 3 =1000/100 x 50
= 500g/dm3 3.47g dissolved in 1000cm 3 = 1000/1000 x 47 = 47g/dm 3
4.200g dissolved in 250cm 3 1000/250 x 200 = 800g/dm 3 121
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- Solutions from stock solutions Stock solution highest
concentration use to make other solutions Extract a portion of
stock solution as calculated Dilute with distilled water Making a
known volume of a lower concentration 122
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- Making solutions from stock solutions If I have a solution
containing 63g/dm 3, how do I make up 250cm 3 of a solution of
concentration 6.3g/dm 3 ? To make 1dm 3 of 6.3g/dm 3 I would need
100cm 3 To make 250cm 3 of 6.3g/dm 3 I would therefore need 25cm 3
and make it up to 250cm 3 with distilled water 123
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- Working out masses We can use the useful relationship Where Mr
is the molecular mass eg Mr of NaOH is (23 + 16 + 1) = 40 This can
help us to calculate an unknown mass 124
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- Titration calculations In a titration we have added a known
amount of one substance usually an acid (in the burette) to a known
amount of another substance usually an alkali (in the conical
flask). The amount added allows us to determine the concentration
of the unknown. 125
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- Titration equipment 126
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- Using a table It can be helpful to sketch a table to keep track
of information you know ValueAcidAlkali Volume (cm 3 ) Mass (g)
Concentration (g/dm 3 ) Molecular weight (Mr) 127
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- Mark scheme 129
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- Mark scheme 131
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- Uncertainty Uncertainty is a quantification of the doubt about
the measurement result. In a titration the uncertainty is the range
of the results. If results are reliable then it will be within
0.2cm 3 NOTE THAT THIS IS RELIABLE NOT NECESSARILY ACCURATE
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- Mark scheme 134
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- Mark scheme 136
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- C7.5 Green Chemistry The Chemical Industry 137
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- What do I need to know? 1.Recall and use the terms 'bulk' (made
on a large scale) and 'fine' (made on a small scale) in terms of
the chemical industry with examples; 2.Describe how new chemical
products or processes are the result of an extensive programme of
research and development; 3.Explain the need for strict regulations
that control chemical processes, storage and transport. 138
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- Bulk processes A bulk process manufactures large quantities of
relatively simple chemicals often used as feedstocks (ingredients)
for other processes. Examples include ammonia, sulfuric acid,
sodium hydroxide and phosphoric acid. 40 million tonnes of H 2 SO 4
are made in the US every year. 139
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- Fine processes Fine processes manufacture smaller quantities of
much more complex chemicals including pharmaceuticals, dyes and
agrochemicals. Examples include drugs, food additives and
fragrances 35 thousand tonnes of paracetamol are made in the US
every year. 140
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- Example question 141
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- Mark scheme 142
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- Research and Development All chemicals are produced following
an extensive period of research and development. Chemicals made in
the laboratory need to be scaled up to be manufactured on the
plant. 143
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- Research in the lab 144
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- Examples of making a process viable Trying to find suitable
conditions compromise between rate and equilibrium Trying to find a
suitable catalyst increases rate and cost effective as not used up
in the process. 145
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- Catalysts Can you give a definition of a catalyst? A substance
which speeds up the rate of a chemical reaction by providing an
alternative reaction pathway. The catalyst is not used up in the
process Catalysts can control the substance formed eg Ziegler Natta
catalysts. 146
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- Regulation of the chemical industry Governments have strict
regulations to control chemical processes Storage and transport of
chemicals requires licenses and strict protocol. Why? To protect
people and the environment. 147
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- Example question 148
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- Mark scheme 149
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- Process development 150
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- Example question part 1 151
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- Example question part - 2 152
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- Mark scheme 153
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- Example question part 3 154
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- Mark scheme 155
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- Factors affecting the sustainability of a process
Sustainability renewable feedstock atom economy type of waste and
disposal energy inputs and outputs environmental impact health and
safety risks social and economic benefits 156
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- Example question 157
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- Mark scheme 158
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- Atom economy 159
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- Atom economy calculation For example, what is the atom economy
for making hydrogen by reacting coal with steam? Write the balanced
equation: C(s) + 2H 2 O(g) CO 2 (g) + 2H 2 (g) Write out the Mr
values underneath: C(s) + 2H 2 O(g) CO 2 (g) + 2H 2 (g) 12 2 18 44
2 2 Total mass of reactants 12 + 36 = 48g Mass of desired product
(H 2 ) = 4g % atom economy = 448 100 = 8.3% 160
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- Example question 161
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- Example question part 2 162
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- Mark scheme 163