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Stoichiometric relationships ibdp Flashcards

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1
Q

Characteristics of solids

A
  • particles r close tgthr
  • particles hv lower energy than in the 2 other states
  • particles can only rotate & vibrate abt fixed positions
  • strong foa btwn particles
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2
Q

Consequences of solids

A
  • definite shape
  • definite volume
  • incompressible
  • high density
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3
Q

Characteristics of Liquids

A
  • particles r slightly further apart than in solids
  • particles hv larger amts of energy than those in the solid state
  • particles can move abt quite freely arnd one another while in close proximity
  • moderate foa btwn particles(bcos some of the strong foa in the solid hv been broken)
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4
Q

Consequences of liquids

A
  • indefinite shape
  • definite volume
  • negligible compressibility
  • moderate to high density
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5
Q

Characteristics of Gases

A
  • particles r far apart from one another
  • particles hv much more energy than the othr 2 states
  • particles can move rapidly, randomly and haphazardly into any space avail
  • vry little foa btwn particles
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6
Q

Consequences of Gases

A
  • indefinite shape(occupy whole container)
  • indefinitie volume(affcted by temp & pressure)
  • highly compressible
  • low density
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7
Q

element

A

consists of only one type of atoms

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8
Q

compounds

A

consists of atoms of 2 or more diff elements bounded tgthr chemically in a fixed ratio

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9
Q

mixture

A

combination of 2 or more substances and can be separated by physical methods

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10
Q

homogenous mixture

A

two or more substances tht r evenly distributed throughout the mixture, resulting in the mixture hving an uniform composition n properties

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11
Q

heterogeneous mixture

A

two or more substances tht r not evenly distributed thruout the mixture, resulting in the mixture hving a non-uniform composition n varying properties

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12
Q

quantitatively a chemical eqn tells u

A
  1. the relative no. of molecules of the reactants n pdts
  2. the relative no. of moles(amts) of rctnts n pdts
  3. the relative masses of rctnts n pdts(multiply w Mr)
  4. the relative volumes of rctnts n pdts(if its gasesous)
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13
Q

number of moles=

A

mass(g)/Mr(g/mol)
no. of particles/ L(or NA)
Vol of gas/ 22.7dm3/mol(STP–>273K(0C) & 100kPa)
vol of gas/24,8dm3/mol(SATP–>298K(25C) & 100kPa)
conc (mol/dm3) x vol(dm3)

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14
Q

avagardo’s constant (L or Na)

A

number of constitutent particles per mole of a given substance, 6.02 x 10^23/mol

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15
Q

Molar mass

A

mass of substance in grams tht contains 1 mole of particles, g/mol

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16
Q

Ar

A

relative atomic mass, defined as the weighted average of the atomic masses of its isotopes and their relative abundance compared to 1/12 of the mass of C-12, NO UNITS!!

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17
Q

Mr

A

relative molecular mass, weighted average mass of a molecule compared to 1/12 of the mass of one atom of C-12, no units!!

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18
Q

density

A

mass per vol, si unit: kg/m3 but chemists usually use g/cm3

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19
Q

empirical formula

A

shows the kinds of atoms & their relative no.s in a substance in the smallest possible whole no. ratios

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20
Q

molecular formula

A

chemical formula tht indicates the actual no. of atoms of ech element in one molecule of the substance

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21
Q

headings for empirical formula table

A

no. of moles
divide by smallest no.
smallest mole no.

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22
Q

how to find molecular formula

A

molecular formula= n x empirical formula, n being relative molecular mass/relative mass of the empirical formula

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23
Q

combustion equation

A

CxHy(g) + (x + y/4)O2 (g) –> xCO2(g) + (y/2)H2O(l)
under satp and stp, water is liq hence the vol of water is negligble compared w the volumes of the other gases, from avogardo’s law, equal vol of gases under the same conditions of temp & pressure, contain equal no. of molecules. Hence the reacting molar ratio is = to the reacting ratio by vol

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24
Q

theoretical yield

A

qty of pdt calculated to form when all of the lr is consumed

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25
Q

actual yield

A

amt of pdt actually obtained–> diff frm theoretical yield bcos of diff in rctn conditions & presence of impurities–> give rise to deviation of actual yield frm theroetical yield

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26
Q

% yield

A

experimental yield/theoretical yield x 100%

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27
Q

conc (g/dm3)=

A

conc/molarity (mol/dm3) x molar mass(g/mol)

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28
Q

parts per million, ppm

A 
one ppm is the no. of milligrams of solute per kg of sln, since 1mg= 10^-3g and 1kg=10^3g
mass ratios(solute to sln) x 1 million(10^6)
used for environmental pollution, eg co2 in air, a particular ion in seawater etc
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29
Q

standard sln

A

a sln w a known conc, usually made up of substance tht has high purity and stability. this is bcos it allows the substance to b accurately weighed, such tht an accurately known conc can b used reliably for the calibration of other standards, pepared using a volumetirc flask whr solvent is added to a high purity sample until the solvent rchs the calibrated mark on the flask
egs: ethanedioic acid, butanedioic acid, sodium hydrogencarbonate & anhydrous sodium carbonate. NaOH is not a gd primary standard bcos it absorbs atmospheric CO2
for redox titrations: sodium ethanedioate, potassium iodate(V) and potassium dichromate
sodium thiosulfate–> not a suitable pri stndrd bcos of its water of crystallisation

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30
Q

titration

A

titrant(in burette)–> standard sln added to react w the analyte of an unknown conc. the end point marks the end of the titration, usually indicated by a colour change, this is ideally the same vol req as at the equivalence point, which denotes the exact amt of titrant needed to react w the fixed amt of analyte

31
Q

back titration

A

indirect titration method usually employed whn a cmpnd is an insoluble solid (eg CaCO3 in toothpaste or eggshell) in which the end-point is difficult to detect or the rctn is too slow + cmpnds which contain impurities tht may interfere w direct titration or contain volatile substances( ammonia, iodine)–> result in inaccuracy due to loss of substance frm titration, also req back titration

32
Q

diluting of concentrated slns

A

When a concentrated sln is diluted by adding water, the number of moles of solute in the sln remains unchanged on dilution
M1 x V1=M2 X V2

33
Q

Assumptions of Kinetic-Molecular Theory

A
  1. Gases r made up of vry small particles, seperated by large distances. Most of the vol occupied by a gas is empty space
  2. the gas particles do not attract ech other at all (or thr r negligible foa btwn gas particles)
  3. the gas particles r in a state of continual and random motion
  4. if collisions occur btwn the gas particles, thy r perfectly elastic ie no loss of ke during collisions
  5. the absolute temp(temp measured in Kelvins) of gas is proportional to the avg ke of the gas particles
    under stp, an ideal gas obeys the abv assumption
34
Q

Avogadro’s law

A

Equal vol of all gases at the same temp & pressure contain equal no. of moles irrespective of the type of gases:

  • at 273K and 100kPa (STP), all gases will hv idenitcal vols of 22.7dm3/mol
  • at 298K and 100kPA(SATP), all gases will hv identical bols of 24.8dm3/mol
35
Q

MOLAR VOLUME

A

The volume occupied by one mole of any gas at a particular temp & pressure

36
Q

Differences between real gas and ideal gas

A
  1. Each particle of real gases hv a definite vol and hence, occupies space, so the the vol of the gas is the vol of the particles & spaces between them whereas the particles of ideal gas hv a negligible vol compared to vol of its container, hence, the vol of the gas is the vol of the space between the particles
  2. attractive forces r present between the particles of real gases whereas thr r negligble attractive forces operating btwn particles of ideal gases
  3. collisions btwn the particles of real gases r non-elastic, & hence, thr is loss of ke frm the particles after collision whereas collisions btwn particles of ideal gases r elastic, & hence, thr is no loss of ke frm the particles upon collision
37
Q

real gases can

A

be compressed into liqs but theory predicts tht ideal gasses cannot

38
Q

only ideal gases obey

A

pV=nRT

39
Q

a (real) gas behaves like an ideal gas at

A

high temps and low pressures–> WHY?
at low pressures the molecules(or atoms) r, on avg far apart. This means tht the intermolecular (or interatomic) FOAs r relatively weak/negligible, so thy move arnd independently(like the molecules in an ideal gas). The size of the molecules is also negligible in comparison w the actual vol of the gas
at high temps. the foa btwn the molecules bcomes insignificant. the avg ke of the molecules is much greater than the intermolecular foa

40
Q

do the particles in an ideal gas possess the same ke?

A

no! thr is a rangeee of KEs in an ideal gas(at constant temp) which is described by the Maxwell Boltzmann distribution. Constant elastic collisions rapidly change indiv KEs, but in a large sample of gas particles, the distribution of KEs remains constant. the average KE of the gas particles remain constant

41
Q

The Maxwell-Boltzmann Speed distribution

A

tells us distribution of the molecular speeds of various gas molecules, although the molecules in a sample of gas hv the same avg ke, thy move at various speeds due to collisional changes–> they exhibit a distribution of speeds, some move fast, others relatively slwly, Collisions change indiv molecular speed but the distribution of speeds remains the same, at the same temp, molecules or atoms of lighter gases move, on avg, faster than heavier gases, at higher temps, a greater fraction of the molecules r moving at higher speeds–> impt for activated chemical processes

42
Q

ideal gas eqn

A

PV=nRT,
P=pressure in pascal or Nm^-2(1atm=1.00x10^5Pa)
V=vol in m3(1m3=1000dm3=1000000cm3)
n=amt of gas(mol)
T=thermodynamic or absolutem temp in Kelvin(K=*C+273)
R=gas constant=8.31J/K/Mol

43
Q

if temp is constant

A

PV=nRT=constant, P is inversely proprtional to V

44
Q

Boyle’s Law

A

if temp is conant, PV=constant therefore, P1V1=P2V2=constant, where subscripts 1&2 refer to two gaseous systems at the same temp & containing the same no. of moles of particles

45
Q

if pressure is constant

A

V=nRT/P=(constant/P)xT, thus V is directly proprtional to T fir a fixed mass of gas under a constant P
thus if pressure is constant, V/T=constant, therefore V1/T1=V2/T2=constnat where subscripts 1&2 refer to 2 gasesous systems at the same o abd constaining the same no. of moles of particles

46
Q

Applications based on ideal gas eqn–> can be used to determine the mr of cmpnds in gaseous state

A

PV=mRT/Mr

47
Q

Applications based on ideal gas eqn–> can be used to determine the density, p of a gas

A

p=PMr/RT

48
Q

what is an ideal gas

A

a gas which obeys the ideal gas eqn and obeys the assumptions under all conditions
(while real gas only obeys when its subjected to low p and high t, at high p or low t it wont!)

49
Q

Reduction

A
  • removal of oxy
  • addition of hydro
  • gain of e
  • ⬇️ in os
  • oxidising agent(oxidant)
50
Q

Oxidation

A
  • addition of oxy
  • removal of hydro
  • loss of e
  • ⬆️ in os
  • reducing agent(reductant)
51
Q

redox rules(only the confusing one)

A

the os of the atoms r obtained by assuming the bonds r ‘ionic’ ie the more elctronegative atom(the atom w gr8er attraction for es pairs in covalent bonds) is given the -ve os and vice versa, electronegativity ⬆️es as u move across a period, ⬇️as a u move down a grp thus flourine is the most electronegative

52
Q

Common OA & RAs–> OAs: MnO₄⁻(aq) manganate (VII)

A

become Mn²⁺(aq) (under acidic conditions) manganese(II) ion
change from purple to pale pink/colourless

become MnO₂(s) (under neutral/alkaline conditions), manganese dioxide
change from purple to brown ppt

53
Q

Common OA & RAs–> OAs: Cr₂O₇²⁻ (aq) dichromate(VI)

A

become Cr³⁺ (aq) chromium(III)

change from orange to green

54
Q

Common OA & RAs–> OAs: IO₃⁻ (aq) iodate(V)

A

I₂ (aq) iodine

change from colourless to brown

55
Q

Common OA & RAs–> OAs: I₂ (aq) iodine

A

I⁻ (aq) iodide

change from brown to colourless

56
Q

Common OA & RAs–> OAs: Fe³⁺ (aq) iron(III)

A

Fe²⁺(aq) iron(II)

change from pale yellow to pale green

57
Q

Common OA & RAs–> OAs: H₂O₂ (aq) hydrogen peroxide

A

H₂O(l) water

change from colourless to colourless

58
Q

Common OA & RAs–> OAs: NO₂⁻ (aq) nitrite

A

NO(g) nitrogen monoxide(colourless)

NO can be easily oxidised to NO₂ (brown gas)

59
Q

Common OA & RAs–> RAs: C₂O₄²⁻ (aq) ethanedioate

A

CO₂ (g) carbon dioxide gas

60
Q

Common OA & RAs–> RAs: S₂O₃²⁻ (aq) thiosulfate

A

S₄O₆²⁻ (aq) tetrathionate

61
Q

Common OA & RAs–> RAs: H₂O₂ (aq) hydrogen peroxidehttps://www.brainscape.com/decks/9875786/cards/quick_new_card

A

O₂ (g) oxygen gas

62
Q

Common OA & RAs–> RAs: NO₂⁻ (aq) nitrite

A

NO₃⁻ (aq) nitrate(V)

63
Q

Common OA & RAs–> RAs: I⁻ (aq) iodide

A

I₂ (aq) iodine

change from colourless to brown

64
Q

Common OA & RAs–> RAs: Fe²⁺(aq) iron(II)

A

Fe³⁺ (aq) iron(III)

change from pale green to pale yellow

65
Q

Reduction rctns:

A 
MnO₄⁻ + 8H⁺ + 5e--> Mn²⁺ + 4H₂O
Cr₂O₇²⁻ + 14H⁺ + 6e--> 2Cr³⁺ + 7H₂O
H₂O₂ + 2H⁺ + 2e--> 2H₂O
I₂  + 2e--> 2I⁻ 
Fe³⁺ + e--> Fe²⁺
66
Q

Oxidation rctns:

A 
C₂O₄²⁻ --> 2CO₂ + 2e
2S₂O₃²⁻--> S₄O₆²⁻ + 2e
H₂O₂--> O₂ + 2H⁺ + 2e
2I⁻--> I₂  + 2e
Fe²⁺--> Fe³⁺ + e
67
Q

Potassium manganate(VII)

A

NOT a primary standard!! it is difficult to obtain the substance perfectly pure & completely free from manganese dioxide + ordinary distilled water–> likely to contain reducing substances(traces of organic matter etc) which will react w the potassium mangante(VII) to form manganese(IV) oxide which will catalyse the auto-decomp of the manganate(VII) sln on standing to form manganese(IV) oxide, thus manganate(VII) is inherently unstable cos of the presence on Manganese(II) ions, this rctn is slow in acid sln but is vry rapid in neutral slns

68
Q

sodium thiosulfate

A

used to estimate iodine by a redoc titriation, is a ra, able to reduce iodine to iodide ions. When its added to an aq sln of iodine, the brown colour of the iodine fades, near the end-point, a few drops of starch sln is afded which will cause the formation of a blue-black starch-iodine complex thus making it easier to see the colour change to colourless at the end point bcos all the iodine has reacted,

69
Q

general rules of balancing redox eqns

A
  1. identify the species undergoing oxi and red by checking oxidation numbers
  2. write an unbalanced half-eqn for oxi and red
  3. balance all elements othr than H and O first
  4. balance oxy by adding H2O
  5. for acidic conditions, balance hydrogen by adding H+
  6. balance the charges by adding electrons to the side w excess +ve charge
  7. balance the overall eqn by adding the two half-eqns such tht the no. of es is the same
70
Q

Biochemical Oxygen Demand(BOD),

A

units of ppm refers to the amt of oxy tht wld b consumed if all the organic matters were oxidized in a sample of water at a particular temp over a period of 5days

71
Q

measuring BOD

A
  1. obtain 2 samples of equal vol of water from the area to be tested
  2. dilute ech specime w a known vol of distilled water which has been thoirghly shaken to ensure oxy saturation
  3. use winkler method to determine the conc of dissolved oxy in the first sample
  4. the 2nd sample is then sealed and placed in darkness (to prevent any photochemical rctns) & tested 5days later
  5. BOD is then calculated by subtracting the conc of dissolved oxy in the second sample from the first sample, which gives the conc of oxygen consumed by the oxidation of organic matter in the sample
72
Q

impt of BOD

A

aquatic life–> can only survive in fresh water bcos of the oxy dissolved in it. hence it is impt to monitior the dissolved oxy conc. if this is less than 5mg/dm3 then most species of fish CANNOT survive. the solutbility of oxy in water is temp dependent, at 273K(OC) the solubitility is 14.6mg/dm3(14.4ppm) and 7.6mg/dm3 or ppm at 293K(20C). Solubility of oxy in water decreases w increasing temp, it was observed tht a high level of dissolved oxy is a good indication of a low level of pollution

73
Q

Winkler method

A

Under alkaline conditions, Manganese(II) ions r rapidly oxidised to manganese(III) by dissolved oxygen, producing a pale brown ppt of manganese(II() hydroxide, Mn(OH)3
4Mn(OH)2(s) + O2 + 2H2O–> Mn(OH)3(s)
A sample of river or stream water is shaken with excess alkaline Mn(II) ions, and the resulting pale brown ppt is then reacted w an excess of KI, which it oxidises to Iodine
2KI + 2Mn(OH)3(s) –> I2 + 2Mn(OH)2(s) + 2KOH
The amt of iodine is then determined by titration w sodium thiosulfare of knwon conc, starch is used to show the end point
2S2O3 2- + I2 –> S4O6 2- + 2I-
The winkler method can also be carried out under acidic conditions:
2Mn(OH)2(s) + 02–> 2MnO(OH)2 (s)
MnO(OH)2 (s) + 4H+ + 2I- –> I2 + Mn2+ + 3H2O
2S2O3 2- + I2 –> S$O6 2- + 2I-

Chemistry (8 decks)

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