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a-level chemistry > Module 2.1 > Flashcards

Module 2.1 Flashcards

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1
Q

First stage in atom change timeline

A

Fitfth century BCE (before the common eraa)
-The greek atom
-thought atoms would be indivisible .

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2
Q

Second stage in atom change timeline

A

A
Early 1800S - Dalton’s atomic theory :
He stated :
-Atoms are tiny particles that make up elements.
-All aotms of a given elemnt are the same .
-Atoms of one element re different form those of every other element .
-Developed the first stable of atomic msses .

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3
Q

Third stage in atom change timeline

A

(1897-1906) J.J Thompson - electron discovery .
He discovered that cathode rays were a stream of partiles with the following properties .
-Had negative charge .
-Could be deflected by both magnet + electric field .
-Very very small mass .
-Concluded electrons came form the atoms of the electrpdes themselves . (DALTON DISAPPROVED).

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4
Q

Fourth stage in the atom change timeline .

A

(1909-1911) Ernest Rutherfords gold-leaf experiment .
alpha particles were filed at a thin sheet of gold foil , the results were .
-Most paricles were not deflected t all .
-However a small percentage were deflected by the source .
-Some deflected bck towards the source .

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5
Q

Fourth stage in the atom change timeline . (1911 ; new model proposed)

A

-Positively charge of an atom + most of its mass re cocentrated t the center (the nucleus ) .
-Negative electrons orbit the nucleus .
-Most of an atom’s vpolume would be the space between the tiny nucleus + negative electrons .
-So positive charge = negative charge .

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6
Q

Q
Fifth stage int he atom changed timeline
(913) Niels Bohr;s planetary model and HENRY mOSELEY’S WORK ON ATOMIC NUMBERS .

A

-Bohr altered Rutherford’s model to allow electron to only follow certain paths . Or they would spiral into the nucleus periodic properties Bohr;s model helped to explain .
-Spectracal lines seen in emission spectrum .
-The energy of electrons at different distances from the nucleus

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7
Q

Henry moseleys’s work (fifth stage continued)

A

discovered a link between x-raus freqiemcoes amd am elment’s atomic number . Rutherford’s discovery of the proton was thena ble to explain Moseley;s findings that an atom;s atomic number was linked to x-ray frequencies . WWe know atomic number = protons .

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8
Q

Sixth stage in the atom change timeline
-Chadwick discovers the neutron

A

He observed a type of radiation made up of uncharged particles with apporximately the same mass as a portien now called neutron .

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9
Q

Sixth stage on the timeline

A

-1923-26
in 1923 the frnech physicist called Lous de Broglie suggesgted that prticles could hve the nature of both a wave and qne q pqerticle .
-In 1926 the austrain phsycis erwin chrodinger suggest that an electron had wave-like properties ina na tom . He also introduced the idea of atomic orbitals .

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10
Q

Lst stage on timeline

A

Modern day
-It is now thought that protons and neutrons themselves are made up of even smaller particls called quakrs . Our understaning of the aom is likely to progress and sceince andvances further and further .

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11
Q

Reactions of isotopes - wh y do different isotopes of the same elements react in the same way

A

This is because
-Chemical reactions involve electrons , and isotopes have the same number nd arrangement of electrons .
-Neutrons make no difference to chemical reactivityy .

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12
Q

formula of the four acids you need to know check sheet

A

HCL –> Hydrochloric acid
nitric acid
sulfuric acid
ethanoic acid
-Key is that all acids contain the element hydorgen .

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13
Q

-when we dissolve an acid in water , the hydrogen is released s the hydrogen ion H+

A

-The hydrogen ion is simply a proton -CHECK THE SHEET - Scientists say that the acid molecule dissociates - splits
-when this happens, it also produces a negative ion (chloride ion in the case of HCL ).

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14
Q

-In the case of hydrocholoric acid, every acid molecule dissociates + releases the hydrogen ion.
Therefore, we say

A

HCL IS A strong acid .
-sulfuric + ntiric acid are also strong acids .

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15
Q

check the sheet ofr the weak acid equation

A

-in the case of ethanoic acid , only a small percentage of the acid molecules dissociate. Therefore as it only partially dissociates it is a weak acid.

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16
Q

Bases - check sheet

A

metal hydroxides , metal carbonates, and ammonia.

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17
Q

what is a base

A

it can neutalise an acid to porduce a salt .

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18
Q

neutralisation reaction , check sheet

A
  1. when we form the salt , the metal ion in the metal oxide has replaced the hdyrogen ions in the acid .
    -In this case ,t he copper ion from the copper oxide has sreplaced the hydrogen ions int he sulfuric acid to form the salt + copper sulfate.
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19
Q

neutralisaiton reaction (2)

A

-Can also see the oxide has formed formt he oxide from the copper oxide and hydrogen ions have formed water.

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20
Q

check sheet to see when a metal hydroxide reacts with an acid .

A

…
-cabornate + acid —> salt + water + carbondixoide .

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21
Q

alkalis are BASES that DISSOLVE in water

A

-both of these re bases, as they neutralise aicds, producing salt.
-copper oxide is insoluble in acid, so is a base only.

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22
Q

however, sodium hydroxide is soluble in water. When we dissolve sodium hydroxide in water, we make a sodium hydroxide solution.

A

sodium hydoricide solution is an alkali

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23
Q

All group 1 metal hydroxides are soluble in water and form alkalis .

A

need to learn this .

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24
Q

ammonia is also an example of this
check sheet for equation

A

-Ammonia is a gas that is highly soluble in water , producing the alkali ammonia hydroxide .

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25
Q

Key features of all alkalis.
-In solution, they release hydroxide ions (oh-)

A

-When an alkali reacts with an acid, we make salt + water
-check the sheet for an example.
but we can summarise his reaction much more simply.
OH-+H+–>H20
ALkali realses hydroxide ion oH + acid releases h+ opens.
-so we can summarise this neutralising reaction by reaching these ions together.

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26
Q

Check equaiton on sheet

A

-Magnesium is more reactive than zinc .
-In this reacton , magnesium atoms dipalces the zinc atom .
-Looking at the reaction we can see that the oxygen atom has transferred from the zic atom to the magnesium atom .

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27
Q

-when oxygen has gained like this we call this an ocidation reaction .

A

-magnwium atom has been oxidised to form magneisum ocide .

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28
Q

-loss of oxygen is called a reudction .

A

so in this reaction , zinc ocide has been reduced to zinc .
-Scienitsts call reactions like this a REDOX reaction as both , REDUCTION and OXIDATIOn has takrn place .

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29
Q

oxidation + reduction can be oth considered in terms of oxygen .

A

we can also consider oxidation + reudction in term sof hydorgen .
-in this reaction hdyorgen has been removced from the methanol , the removal of hydorgen is ocidtion .
Vice vera is reduction /

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30
Q

-oxidation and reudction can also be looked at in terms of electrons .

A

…

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31
Q

look at the zinc oxide and mave a amgnesium aotm which is uncharged .

A

mg —> mg2++
-this looses two electrons to form magnesium ion MG2+
-At the strt of the reaction ,w e have the zinc ion zn2+.
z22+—>zn
-This gains two electroms to form magnesium atom which is uncahged .
-Oxygen remains ucnhagred during this reaction 02- UNCHANGED .

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32
Q

Ions that do not change during a reaction are called what ?

A

Spectator ions .
-loosing electrons is oxidation which i what happened to magnesiu ,.
zinc gained two eelctrons nd gor educed

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33
Q

write the reaction as half equations

A

Mg —> oxidation Mg2++2e-
Xn2+ +2r—. zn
-In anyr edox reacion the chemical accepting the electro is calledthe oxidising agent (zinc ion ) .
-The chemical that is DONATING electrons is called the reducign agent (magnesium).

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34
Q

RECAP
oxidising agent —> reudced
reducing agent –> oxidised

A

…
check half equations video to get more prctice .

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35
Q

What are oxidation numbers

A

Oxidation numbers tell us about how electrons are lost or gianed in a reaction /

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36
Q

MGO

A

-when magneisum oxide forms ,
–Two eectrons are transferred from the mgnesim atom to the ocygena tom .
-This mans that magneisum oxide conains the magnesium ion MG2+ + the oxide ion O2- .
-As the magneisum ATOM lost TWO ELECROS TO ORM THE MANGESIUM ION . iT HAS AN OCIDATION NUMBER OF +2.M
-as OCYEGN HAS gained TWO ELECTRONS IT HAS A OXIDATION NUMER -2.

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37
Q

LOST ELECTRONS …
GAINED ELECTRONS …

A

Postive oxidation numebrs .
Negative oxidation numbers .

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38
Q
  1. for assinging oidation numbers .
A

In a PURE ELEMENT , atoms have an oxidation number of zero .
-Check sheet for examples

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39
Q

whe elements have chemically reacted with other elemtns , then their oxidation number will NO longer be zero .

A

…

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40
Q

Oxidation numbers of non-metals wehn they have reacted with ther elements .

A

…

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41
Q

-element
-oxidation number
-exceptions
(1)

A

-Fluroine
-Always -1
-NO expcetions 0 as fluroine is the most electronegative elemnt , so no other elemnt can remove an electron from fluroine .

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42
Q

-element
-oxidation number
-exceptions
(2)

A

-Oxygen
- -2
–1 in perioxides (hydrogen peroxide e.g)
-+2 when reacted with F f(as fluroine is electronegtive enoguh to remove elctrons from oxygen ).

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43
Q

-element
-oxidation number
-exceptions
(3)

A

-Chlroine , Bromine , Iodine
- -1
Can be postive when bonded to Fluroine or Oxygen .

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44
Q

-element
-oxidation number
-exceptions

A
  • -1 in metal hdyrideds lke Lithiium hdyrides .
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45
Q

Oxidation numbers of metlas when they have reacted with other elements .

A

….

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46
Q

group 1 metals

A

always +1

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47
Q

group 2 metals

A

always +2

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47
Q

aluminium

A

always +3

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48
Q

transition metals ls

A

Variable
-The oxidation number of is variable , therefore w eneed to claculate by looking at other elements present in the compound .

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49
Q

H20 is the forumla of waer .
h20
-2 HYDROGENS CHARGE IS +2. oxidaiton numbe ris +
OXYGEN GAINS TWO ELECTROSN so oxidaiton nmbe ris -2.

A

-Forumula od wae r
-the two hydoreg bodnend to one oxygen
-KEY; oxidaition numbers of atoms in a compund , must dd up to the totls chagre. on the compound .
-QTER moelcule she a ttoalc hagre of 0 .

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50
Q

h3po4

A

-chceck sheet for example .

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51
Q

Na2cr207

A

check sheet again

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52
Q

checknthe ions

A

.. example

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52
Q
A

thIS ROMAN NUMERALS ELLS US HTE OIXDATION NUMBERS AT THE MANGANENSE .

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52
Q

lOOK AT WXmplw 1

A

Roman numerlas as the ixaiotn number of chlorine .
-*Both of these exmpales , ou can see the chlirn has reacted with oxygem .
–Because oxygen is more electroeegatice then chlrine , it can remais as electrons from the chlirne atom .

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53
Q

Example 2

A

roman numerla rells us oxidaiotn on rhw numbwer o the nitroen .

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53
Q

-we can sue ocidaiton numberd to workout which chemical is oxidised and which chemcial si reduced .

A
  1. assign oxidation numbers to all of the atoms in the equaiton .
    -FIRST do the uncobined elements so these have an oxidation number o zero .
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54
Q

check the example on sheet

A
  • On the RIGHT HAND SIDE, THE MAGNESIU + CHLOROIN HAVE reacted to form magnesium chloride.
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55
Q

-Like group 3 elements , whe magneisumr acts it has oxidation number +

A

-chlorne oxidation number -1.
KEY ; magnesium chloride .
-Chlodirde contains two atoms of chlorine .
-HOWEVER , we laays write the oxidation number for one atom.
-NO atter , how manya toms there are

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56
Q

-We can see tht is ocidation number has icnreaed from 0 to +2 .

A
  • when this rocidasiton numebr increases it hshows oxidation ahs tkane place .
    I CBA FOR THE RES HONESLTY JUS CHECK THE SHEET MAN AND THE VIDEO .
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57
Q

cannot easilypredict the charges on the ions of group 3 metals

A

-one we beed to learn . Aluminium charge 3+ .
-silver forms the ion ag+

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58
Q

zinc is not a transition metal and you’ll find out in nd out

A

zinc forunal ion zn2+

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59
Q

“ide” means reacted with other elemens other eleemnt

A

check sheet for chrges of non-metals you need to know

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60
Q

Magnesium Hydroxide

A

-This means that we need to two hydroxides to balance the charge on the magnesium ion .
-This means that the formula of magneisum hydroxides Mg(Oh)2 .
-This formula contains brackets .KEY; remember little number to the right of the brackets mutlipies everuthng inside of the brakce t.

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61
Q

calcium nitrate exmaple check sheet

A

Ca(No3)2

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62
Q

Isotopes are atoms of the same element .

A

with different number of neutrons + sifferen masses ..

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62
Q

All isotopes reaction in the same way , why ?

A

they all have we same electron configuration .

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63
Q

What dies this tell us about how common each isotope is ?

A

abundance

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64
Q

how do we determine the mass

A

mass spectrometer –> number + abundnce oit .

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65
Q

interpeing a mass spectrometer

A

1.First ghing to notice is tht the spectrum has wo main parts peaks .
-This tells us tha tcopper has two main isotopes .
–Y axis ahs the outer elecrons abundance for the isotope , this is gien as s a arecentge of the totl .

-often these re shown at he top of each peak , like i am showing .
-x aci we have the (m/z ration ) . This is the raitio of the mass of each ion to its charge .

JUST THINK m/2 tatio is simplt he rtio of mass of the ion abso reltaiv emass of the ion .

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66
Q

magnesium mass spectoremter

A

-spectrum has thre epeaks so magnesium has THREE MAIN ISOTOPES
-First isotope has a relative mass or abundance of 78.9 .
second isotpo has a relaitve mass of 25 and bundc eof 10 .
3rd isotope ha a relative mass of 26 and an abundance of eleven

-Magnesium also has a range other isotopes witha very low abudncne .

67
Q

masses of atoms - PROBLEM

A

-atom is extereml small .
-So they’ve got extremely small masses . Which is why they do not reduce mases .

68
Q

masses

A

are relatie to carbon-12 which is an isotope of crbon .
-clalled relative msses as they re relative to crbon 12 s it has the mass of ecrtl 12 .

69
Q

reltive isotopic mass

A

mass of an atom of an isope from 1/12 mass oc carbon-12 .

70
Q

key thing baout relative isotopic mass

A

-there is one reltive isotopic mass for ech isotope of an eleemnt .
-relative isotopic mass is always whren there is a decimal .
-ha no unigx

71
Q

what is relative atomic mass

A

the weighted mean mass of an aom of an eleemnt compared with 1/2th the mass oc carbon-12 .
-the mean is weighted for the abundance of each isotope (how common an sitoope is )

72
Q

formula to calcualte relative atomc mass

A

check sheet

73
Q

molecular formula

A

tells us the elemnt in a molecule + the numer of toms of rch element .

74
Q

emprical formula

A

tells us the simplest whole number ratio of the toms of each elemen ina c ompound .

75
Q

molecular forumala or

A

1.firgure out all we wecan use he ar to
-calculate relative moelcule mas for these compoudns/

76
Q

relative moleculr mass

A

weighted mean mass of a molecule compared to 1/12 of ann atom of carbon-1 MR.

77
Q

avogradas constnt (Na)

A

the numer of particles a substnace in one mole of that sibstance .
mol = amount of sutbance mol
mol=amount of substance mol .

78
Q

if we know the amount of a substance in we can calculate the number of partivles …

A

n.o particles - mol*avogrado’s constnant.

79
Q

milliltre = 1/1000 of one
dm3 = volume
1000th of dm3 =cm3

A

dm3 –> cm
/1000

80
Q

concneetration (mol/dm3)

A

amount of susbance (mol) / volume (dm3)

81
Q

one mole of any gas has a volume of approx 2 dm3 at room temperature and pressure

A

room temperautre 20 degree s.
Pressure - 101000PA

82
Q

amoun of substance (mol)

A

volume dm3 / n3

83
Q

Calculart the volume of ydorgen has procued by 130mg of zinc is relsses cacl ex,pa;e

A

check answer

84
Q

ideagas equation

A

pv= nrt learn on sheet

85
Q

pressure

A

-usually given in kpa kist *by 1000 to give the volumeof gas hs to be in m3 .
-This has tp br o m3 , howver this is is givne in dm3
dm2–>m3
(/1000)
cm–>m3 (/1000000)

86
Q

temperature

A

has to be kelvin .
-convert fro deggrees t kelvin add 273 .
-to convert form kelvin to c -273

87
Q

ideal gas constant

A

has a value of 8.341kjmol

88
Q

for an idealgas the following ssumptions are made .

A

-we assume that the gas moelcules are TINY compared o the spaces between the .
-we also asume there are no forces acting between the gas molecules .

89
Q

for an ideal gas other assumptions are also made

A

-assume that gas moelcules move randomly
-we ssume tht yhr ehrn gshd morlvulrd vollifr hr vollidiond str rlsdiv .

90
Q

firsthi thing to do with the idal gs ewuation

A

convert all our values to the correct units .

91
Q

we can use the ideal gas eaution to workout the

A

MOLAR MASS OF A LIQUID CHEMICAL

91
Q

Molar mass of a liquid chemicl (1)

A

-take a gas syringe with small cap at the end .
-+we use a balance to find the mass of the empt syringe + can .

92
Q

Molar mass of a liquid chemicaal (2)

A

Remove cap + allow a small volume of our chemical into the syringe .

93
Q

Molar mass of a liquid chemical (3)

A

now we place the cap to seal the chemcial in the syrine .

94
Q

Molar mass of a liquid chemical (4)

A

-next we reweight the gas syringe containing the chemical plus the cup .
-At this oint ,w e cn now calculte he mass of the chemical int he sutromge

95
Q

Molar mass of a liquid chemical (5)

A

To do this subtract empty the form mass of t=sytine witht he chemical .

96
Q

Molar mass of a liquid chemical (6)

A

boil the liquid so it turns intoo a ga s.
-do thi by putting syringe into a beaker of boling water.
-Once the chemical ahs biled , we cna read the volume of gas from the scal eon the ags syringe . use this work out the molar mass of the chemicl.

97
Q

ths ony works for chemcils with oling point of les thsn hudred ergess as this is boling water

A

temperature greater would have o use n oil bath .

98
Q

-Hydated calcium sulfate .
-crystas of this compound contaain water molecules bonded into the crystal strucuture .

A

dvienistss cslll this water of cyrstallisation shown by the DOT in the formula .
-what this formula tells us is that the crystals contain two omlecules of wter for every calcium sulfate .
-compound like his re clled HYDRTE copouds because they continw ter of crystallisation

99
Q

check sheet for equation of hydrated calcium surface —-> anhydrous calcium surface +water

A

-If we Heat this rhen w egress the bonds holding the water molecules into the .
This drives the water molecules off as steam . We are now left with an anhydrous compound

100
Q
A
101
Q

Meaning of anhydrous

A

It does not contain water or crystallisation

102
Q

Check sheet for hydrated magnesium surface

A

Different hydrated compounds , but in this case , the number of water molecules is x .

103
Q

In exam you could be asked to workout the value of x check sheet for quesiton hydrated magnesium sufferers has a relative formula mass of 247.4g/mol

A
  1. Mr includes =magnesoim .sulfate+water
    Calculate the Mr of sulfate WITHOUTVwater
104
Q

Step 2

A

Mg =24.3 , Sulfur =32.1 , oxygen (16*4)
Total =120.4g/mol

Mgso4 HUDRATED COMPOUJD =240.4g|mol
246.4-120.4=125.0 mass of water

105
Q

Step three calculate the number of molecules of eater

A

Check sheet - each water molecule hwe w relative molecule made of 18 g:mol

106
Q

Divide the total mess of water by 18 step 4

A

7 molecules of water
=7

107
Q

We can also determine the Value of the water of crystallisation

A

We can do this by looking at the HEATIGN method

108
Q

Sample of hudrated copper sulfate .Hydrated copper sulfate forms blue crystals

A

How to determine the value of water of crystallisation :
-Determkne the value of x
- If we hear it , we can drive off the water of crystallisation leading us anhydrous copper sulfate + fhe mass of water —-> can calculate the value of x .

109
Q

Step one of heating method

A
  • take an empty boiling tube and weigh using balance .
  • place several spatulas of hydrated copper sulfate into boiling tube and weigh the boiling tube
110
Q

Step two of heating method

A

-Calculate the mass of the hudrated copper sulfate we added .
To do this we subtract the mass of the empty biking tube from the mass of the boiling tube containing the hudrated copper sulfate .

111
Q

Step threeof heating method

A

Next we hear the boiling tube over a roaring binden burner flame .
This drives off the water of crystallisation as steam .

112
Q

We weigh the biking tube every couple of minutes until the mass stops decreasing .

A

At this point we ahve driven off the wate r and are left with anhydrous copper sulfate which is white .

113
Q

7of heating method

A

At this point we now weigh the biking tube and contents again .
Now we can calculate the mass of anhydrous copper sulfate .
- to dk this , we subtract the mass of the empty boiling tube from the mass of the boiling tube containing the anhydrous copper sulfate. Now we can calculate the value of the water of crystallisation of hydrated copper sulfate .

114
Q

Calculate for heating

A

-using your results we know the mass of hudrated copper sulfate is 7.748g mass of anhydrous copper is 4.96
The difference in the two values represents THE MASS OF WATER DRIVEN KFF in this case 2.794

115
Q

Going back to the hydrated copper sulfate

A
  • we known mass of the copper sulfate + water
    -Now we calculate the n.9 of moles of these todo to this we divide the masses by the MOLAR masses of each compound
    For chopper is 0.032mol
    And for H20is 0.155
    -finally calculate the ratio between these by dividing by the smallest so copper in this case and get 1:5 ration
116
Q

What are some problems with the heating problem (1)

A

If the value of the water of crystallisation is LESS than expected , then that suggests that not all the water molecules were driven off during heating

117
Q

Problems with the heating method 2

A

If the value of water of crystallisation is GREATER than expected , then that suggests the compound has undergone further decomposition .
Once the water had been driven off the anhydrous compound had then defomposed into a different compound .

118
Q

Example of problem 2

A
  • this could take place for example worj a metal carbonate , which could undergo thermal decomposition to form a metal oxide .
    -in fhis reaction csrbodnioxide would he realised .
    fhis would cause rn e compound to LOSE more Mass rhag jf just the water had been removed
119
Q

What a titrations used for

A

To determine the concentration of a solitoomn

120
Q

STAGES OF A TITRATION

A

-we are looking at a simple titration between an acid and aaklon

121
Q

Key :

A

we take a fixed volume of solutions with a known concentration + we then react it with a solution with an unknown concentration .
-If we accuratelrly measure the value needed to react rhen we can determine the concentration of our solution ,

122
Q

Going to react and acid with an unknown concentration with an small with a known concentration

A
  • Measure an accurate dided volume of our small
  • to do this we use a poppers and we measure out 25cm3
  • pipettes has a very fine line at the top this shows where the volume is exactly 25cm3
123
Q

Step 2

A

Rinse the pope tree with distilled water to remove any unwanted chemicals
May contain sitll droplets of water , tjis woild dilute any solution which we used the pipette to make

124
Q

Step three

A

Rinse then pipette with out alkali and discard the alkali down the sink .
Then place our conical flask right next to the beaker containing the alkali .
The fomkc flask has to thenbr risked carefully with distilled water to remove any remains of unwanted chemicals .

125
Q

Step four

A

Place the top
Of the pipette to to the alkali and draw the alkali into the pope tree using the poppers filler .
- need to do this carefully we need to avoid bubbles as if there is the level of alkali would fall so need to make sure the top of the poppets is below the surface

126
Q

Key

A

Fill the pipette last the 25cm3 mark as the level of liquid drops slightly when lifting pipette out of alkaline
- if we measured exactly it will drop after lifting out and therefore our volume won’t be accurate

127
Q

Step 5

A

Carefully lift the poppers out of the alkali and very slowly release drops of the alkali
-bottom of the meniscus should exactly lie on the 25cm3 important to view at eye level

128
Q

Step 6

A

Move the pipette over the conical flask and release the alkali .
When the alakali has been released , you’ll notice a tiny amount left in the pipette .Yo transfer that Weber touch the very top of the pipette r into the alkali

129
Q

Now we ahve Exsctly 25 come of well on the comical task

A

I’m an acid base trust ion we need to use an indiacot to determine the end point

130
Q

Signs. Strong acid strong base titration…

A

We use phenolphthalein kt methyl orange

131
Q

If we are to resting a weak acid with a strong bade

A

Phenolphthalein wokld be suitable

132
Q

If we are trusting as front acid with a weak base

A

Use methyl orange
key: must only add a few drops of the indicator as indicators are weak acids and this could then give inaccurate results .

133
Q

Step 8

A

Using a hit ether - used to measure the VOLUME of acid that reacts with out alkali

134
Q

Rinse the butter with distilled water .This removes any in water chemicals from the burette

A

Key we take a fixed volume of solution with a known concentration and then we react it with a solution of unknown concentration .
If we svvirwfrly measure jr volume needed to react , then we can determine the concentration of our soliton

135
Q

Going to use a hurry yet to measure the volume of ACID that reacts with our alkali

A

First rinse the burette with distilled water to remove an in water chemicals and rinse the burger with acid tk remove anh reaves of water

136
Q

Next we clamp the burette tel thag kt is level and we use a funnel to slowly fill the burette with acid

A

Want the level of acid to be slightly shove zero
Then we remove the funnel
And this prevents acid from dripping from the funnel into the birtete

137
Q

At this stage , we now want to open the tap + allow the acid to slowly leave the burette r.
-We want the bottom of the medicine Exsctly on the ocm3 mark .

A
  • it can be difficult tk make out the position of meniscus so remeber to read frome yelevel
  • jow we place our conical flask containing the alkali onto a white tile (making colour change of the indicator easier to see accurately )
138
Q

Now we open the two of burette and slowly release acid into the conical flask .

A

At the same time we swirl the conical flask .
-swirling ensured the acid + alkali lmk throughly so they can eww .
- while swirling we need to watch the colour of the kenifeote
- we stop adding the acid when we see the colour changes showing end point
- at this step aww read the level of acid on the burette

139
Q

Reading the burette

A

Scr
Measure ngl the nearest 0.1cm3
The uncetrajomty is about 0.05
Sl between two scales js 0.05 so we record the volume to the nearest 0.05

140
Q

Subtracting the start volume
From the final
Volume gives us the titre aka the volume of the acid that reacted with our alkao

A

Our first rotation is considered to be a rough titration
- we then rinse out ojt comical flask ojt and repeat again
Concord and Claus’s are values with&! 0.1 cm3

141
Q

Check sheet how do we work out molar mass
Imagine I knew the certain mass of soemthing

A

We can determine the mass of our hydrated sodium carbonate mh a balance and wwe can ewrom the amount of our substance in moles using titration.
- from these two fwlies we can determine the molar mass of the value x

142
Q

Method :
Weigh a mass of our hydrated sodium carbonate -we need to use a balance with at least 2 decimal places

A

…

143
Q

Step one

A

Start by placing an empty boat onto the balance and then setting the balance to zero

144
Q

Step two

A

Add 1.5 g of hudrated soodim carbonate to the weighing boat
And then transfer the hydrated sodium carbonate to a beaker

145
Q

Steal three

A

Ewww ugh the emlty weighing boar and calculate the mass of hudrated sodium carbonate we added tk Beale r
We do this as a very small amount of hudrated slodim carbonate wk remain attached to the weighingboat

146
Q

Step four

A

Next we add around 100cm3 of distilled water to the beaker and issolve the hudrated sodium carbonate using a stirring rod .
-now we use a funnel to transfer the solution to a 250 cm3 volume for if flask
.
-Af this stage fheee will be traces of our compound on the beaker and string rod .
- so we rinse them with a smal
Amount of distilled wate + rs and fed this into the volumetric flask so this way he can be certain we ahve all of the compound intl the flask

147
Q

Step five

A

Now we dolly fill the volumetric flask with distilled water sk thag the lebel of the solution reaches 250cm3

148
Q

Step five

A

Then we use a poppers to add drops of distilled water until the bottom of the meniscus is at 250cm3 look at eye level for tjis

149
Q

Step six

A

Then we place a stopper on the flask and invert it several times so that solution is mixed .
-now we ahve a solution containing a known mass of our compound , next we need to determine the number of moles of our compound so to do those use titration .
Our compound hudrated thingy and reach with HCL

150
Q

Check sheet for na2co3

A

Now we have known mass of our hydrated compound , dissolve don a given volume of solution .
We now need to workout the number of moles of our compound to do this w ended to use TIRRATION .
-our compounds is hydrated sodium carbonate and react this wotj

151
Q

Nesr wr

A

Use a pip Pete to draw 25 cm3 of sodium carbonate solution into a conical flask

We need to before that rinse our poppers worj distilled water as this removes any other chemicals tjay might be present in the pippete

152
Q

At this point pipette wk still contain reaves or workless wart

A

Bevause of this we would not sacfy measure 253 as traces of distilled water woild contribute to the volume on the poppets al we need to rinse the pipette with sodium carbonate solution as tbks gets rid of any distilled water and allows us to measure an accurate volume

153
Q

Nkw we draw 25 3 of our solution into the pipette (read soliton at eye level)

A

Now we transfer the solution into a conical flask which we previously rinsed with distilled water

154
Q

Next add methyl orange which is used as we have a strong acid hcl and weak base like sodium carbonate

A

Next we take our burette and rinse it with distilled water
Again just like before we now need to remove any slr distilled water as this woild Conroe bite to the volume in the burette and give us an inaccurate reading so we need to rinse burette with hcl .

155
Q

Finally we need to use a funnel to fill the hirreye tk the zero mark with hcl.

A

Remember to remove the funnel as this dam drip hcl which will give us inaccurate results

156
Q

Now sold sonically dark onto white tile add gradually ad hcl acid whole swirling solution

A

We are looking for ojt methyl orange to turn from yellow to orange showing end point
Finally we read the Fokine of acid we added from the scale on the burette to two decimal points
Could of side we called tori we and the first one is rough trite
Repeat Al this titarion again but this point when we reach end point we need to add acid drop by drop whiskey writlkng

157
Q

Again once we reach the end point we can record our tire

A

This tirire will henlroe accurate than the rough tire since we added the acid more sloely near the end point
Repeat until we have got two confidant values

158
Q

Now sofo the titration and ansul use the results

A

How to analyse the results
- stop the triestilnw run we achieve two concordance tires
-then calculate the mean of the Concordant values .
- useful value to workout the values for the water of crystallisation

159
Q

Check sheet for recto on equation w

A

We worked out 10.15,’3 hcl
Need to fscialte number of moles of hcl check triangle
Wthen turn into dm3
Then divide this by 2 to work out moles of soodum carbonate KEY as this only represent 1-10 for the or the sodium carbonate e as we dissolved 250 came bjt we only used 25 cm3 for our titration so we need to multiply our number of moles of sodium carbonate by 10 telling us we have 0.005075
So we neeed

160
Q

Nkw we can calculate the value of wate or crystallisation

A

Mass =1.552
Almond kf substance 0.005075 mol
Work out the most meee check sheet

161
Q

Work out the mass due to me

A

Take always the mr of both things
Yeah check video too cuz I cba

162
Q

What is atom economy , check sheet for equation

A

What presents he of the mass of the reactant atoms end up as the products we want

163
Q

Key points about the atom economy ewuotln

A
  • atom economy is based ojtey on the chemical equation for the reaction + assumes that the percentage yield is 100% .
    )aka all fly rnr wvgwny mlies react)
164
Q

Second point and third point

A

Large numbers in the chemical equation flint for atom economy
- any reaction that only has one lordingist have an atom economy of 100% we will be told desired product in exam

165
Q

Reactions with high atom ECONOMY are MORE SUSTAIMABLE than those with a low atom economy

A

As this tells us that’s mass of reactatsnt that end up in waste products
However there is also other factors like energy use so that means for example , if there is high atom economy bjt large amlknt go energy use that the reaction may ngl be sustainable .
Also waste products could be used as STARTING PRODUCTS AINCRWADING SUSTANWBLY

166
Q

We rarely get maximum ur old , why ?

A

Not all of the reactamsnd may react especially in reversible reactions
- side reactions take place making other products
- product lost e.g during filtiraitkm

167
Q

Check sheet for equation for percentages yield
U can easier use g/mola or mass but you need to stick to one so both things are rhat

A

…

168
Q

What is excess reagent

A

Anh I reached leftovers rectangles

169
Q

Limiting reagent

A

Chemical used up in the reaction the re

170
Q

The amount in moles of the limiting reagent determines the king of PRODUCE FHAT WE MAKE

A

check video for questions

171
Q

what is the molar volume of a gas ?

A

a-level chemistry (7 decks)

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