LAST MINUTE CHEM Flashcards

Question

What's a chlorate ion?

Answer 1

React chlorine gas with cold dilute aqueous sodium hydroxide in a disproportionation reaction

Answer 2

Cl2 + H2O =(reversible reaction) HCl + HClO HClO + H2O = (reversible reaction) ClO- + H3O+ Chlorate ions kill bacteria

Answer 3

Ozone strong oxidising agent for killing bacteria, but expensive and short half life UV light- Kills bacteria by damaging their DNA but inefective in cloudy water

Answer 4

With HCl and Barium Chloride | White precipitate formed if it's present

Answer 5

Add NaOH and warm mixture, if ammonia is given off then ammonium ions were present

Answer 6

Test for carbonates - no CO2, so no barium carbonate precipitate can form Test for sulphates - no sulphate to react with silver nitrate to form silver sulphate Test for halides

Answer 7

In aqueous solution: Chlorine water added: Goes yellow from Br2 Bromine water added: no reaction Iodine solution added: no reaction In organic solution: Chlorine water added: Goes orange from Br2 Bromine water added: no reaction Iodine solution added: no reaction

Answer 8

In aqueous solution: Chlorine water added: Goes orange/brown as I2 displaced Bromine water added: Goes orange/brown as I2 displaced Iodine solution added: no reaction In organic solution: Chlorine water added: Goes purple as I2 displaced Bromine water added: Goes purple as I2 displaced Iodine solution added: no reaction

Answer 9

If there's a change in concentration, pressure or temperature the equilibrium will move to help counteract the change

Answer 10

It will shift the equilibrium to the side with the fewer moles of gas

Answer 11

The equilibrium shifts in the endothermic (positive enthalpy change) direction So if the forward reaction is exothermic it will shift left If the forward reaction is endothermic it will shift to the right

Answer 12

Kc = ({C}^c x {D}^d ) / ({A}^a x {B}^b) Just put in concentrations to find Kc To find units use that each concentration has the units moldm^-3 and then use indices rules to find the units

Answer 13

ΔH, the heat energy transferred in a reaction at constant pressure Units are kj mol^-1

Answer 14

``` 298 kelvin (25 degrees) 100kPa (1 atm) Standard states (1moldm^-3 in solutions) ```

Answer 15

Δ(f)H, enthalpy change when 1 mole of compound is formed from it's elements in their standard states under standard conditions

Answer 16

Δ(c)H, enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions

Answer 17

Δ(neut)H, enthalpy change when an acid and an alkali react together under standard conditions to form 1 mole of water

Answer 18

The overall affect of both of them occuring | Average bond enthalpy of breaking bonds - average bond enthalpy of forming bonds

Answer 19

Heat losss or gained in joules = mass of water in grams x 4.18 x change in temperature Put all values into equation Convert answer from joules to kJ by dividing by 1000 Work out how many moles of fuel produced this heat Divide answer by the amount of moles to find standard enthalpy change (kJmol^-1) also make it negative

Answer 20

Write out the reactants equalling the products, then bellow write out all the elements in their natural states that you require Draw arrows from the natural state elements to the reactants and products, then look how many moles are made and of which substance to work out the enthalpy changes of formation for reactants and products (if element stay the same won't be an enthalpy change, remember to multiply by the amount of moles) Work out total distance from the reactants to products, bearing in mind the first arrow is going in the wrong direction

Answer 21

Same method before, however the direction of arrows have changed and so has what's on the bottom On the bottom have water and CO2 and both the lines will be going downwards with an amount of oxygen next to them to balance the equation Work out total distance taking into account second line is going to wrong way

Answer 22

Number of particles on y axis Kinetic energy on x axis Draw a line with a gradient of 1 going up into a smoth curve which then goes down into an asmyptote Draw a line near the end to show the activation energy

Answer 23

The change in the amount of reactants or products per unit time Units are mol dm^-3 s^-1

Answer 24

The concentration of each reactant doesn't change significantly over time period of clock reaction Temperature stays constant When the endpoint is seen, that the reaction hasn't proceeded too far

Answer 25

H2O2 (aq) + 2I- (aq) + 2H+ (aq) = 2H2O (l) + I2(g) Small amount of Sodium Thiosulfate solution and starch are added to an excess of H2O2 (Hydrogen peroxide), and Iodide ions in acid solution (H+) Starch is an indicator and turns blue/black in presence of iodine Sodium thiosulfate instantly reacts with any Iodine that forms making it go back to iodide ions So once the set amount of Sodium thiosulfate is used up solution will go blue/black, so can vary iodide or H2O2 concs to give you different times

Answer 26

Straight diagonal line going down from y axis

Answer 27

Curved line with diagonal gradient from y axis, reaching an asymptote near x axis

Answer 28

Horizontal line from Y axis, as changing concentration doesn't affect rate

Answer 29

Positive gradient for y=x, through the origin Rate is proportional to concentration

Answer 30

Positive curve going through the origin, Rate is proportional to concentration^2

Answer 31

``` Rate (mol dm^-3 s^-1) = k x (A)^m x (B)^n k= rate constant m= order of reaction of A n = order of reaction of B (A)= concentration of A (B)= concentration of B ```

Answer 32

The time taken for half the reactant to be used up

Answer 33

Constant, will always be the same amount of time for reactant to half Meaning it's half life can be read off it's Conc-Time graph

Answer 34

``` k = ln2 / half life Units = s^-1 ```

Answer 35

Not constant

Answer 36

Is the slowest step in a multi step reaction, which determines the overall rate Probably the first step in the multi-step reaction, and the orders show how many molecules of each there is

Answer 37

ln k = -Ea/Rt + ln A

Answer 38

Plotting ln K on y axis so ln K is substituted for y Plotting 1/T on x axis so 1/T is substituted for x y = mx + c LnK = -Ea/R x + lnA Creating a graph with a gradient of -Ea/R and a y intercept of lnA Calculate the gradient and make it equal to -Ea/R then solve for Ea Find the y intercept and make it equal to ln A and solve for A, or just put Ea into equation (Make sure graph is well proportioned, doesn't need to show Y intercept)

Answer 39

Only aqueous and gaseous reactants and products go into formula (water needs to go in it if the solvent isn't aqueous)

Answer 40

Draw your table with at top the chemical equation with column for each chemical At side have initial concentration, conc change, and equillibrium concentration Fill in table with everything you know, the products initial concentration will always be 0 Whatever the decrease is for the products to equillibrium concentration from initial, will be the increase for the products Remember to take moles into account

Answer 41

It's the sum of all the partial presures

Answer 42

First find mole fraction of the gas in the mixture using Number of moles of gas / total number of all moles of gas in mixture The multiply the mole fraction of gas by the total pressure of the mixture So if they have the same amount of moles in equation will have the same partial pressure

Answer 43

if aA(g) + bB(g) ⇋ cC(g) + dD(g) Then Kp = p(C^c) x p(D^d) / p(A^a) x p(B^b) Use the partial pressures instead of concentrations Units are kPa

Answer 44

Only temperature

Answer 45

HA(acid) + B(base) ⇋ BH+(acid) + A-(base) In forward reaction HA acts as acid as donates a proton to B In reverse reaction A- acts as base to accept proton from BH+ HA and A- are a conjugate pairs, HA being conjugate acid and A- being conjugate base B and BH+ are a conjugate pairs, BH+ being the conjugate acid, and B being conjugate base Will have A1 and B2 on one side, and B1 and A2 on the other numbers link them

Answer 46

pH = -log10 (H+)

Answer 47

(H+) = 10^-pH

Answer 48

H+ concentration = Concentration of acid OH- concentration = Concentration of base

Answer 49

Kw = (H+)(OH-) | If water is pure then ratio between them is 1:1, so Kw is the same as (H+)^2

Answer 50

Don't fully ionise in solution so H+ conc isn't the same as acid concentration Also it's concentration is the same at start and at equillibrium

Answer 51

Ka = ((H+) x (A-)) / (HA) Or Ka = (H+)^2 / (HA) (Ka units are moldm^-3)

Answer 52

pKa = -log10 Ka

Answer 53

Ka = 10^-pKa

Answer 54

A weak acid and it's conjugate base Eg CH3COOH(aq) ⇋ H+(aq) + CH3COO-(aq)

Answer 55

Mix a weak acid with the salt of it's conjugate base Mix an excess of weak acid with a strong alkali

Answer 56

If you add a small amount of acid the H+ ion increases,, meaning the position of equilibrium shifts to the left reducing H+ ion concentration. pH doesn't change much If Alkali is added OH- ion concentration increases, they react with H+ to form water, removing H+ ions from solution. Causing equilibrium to shift right so more CH3COOH dissociates, H+ ion concentration increases, pH doesn't change much

Answer 57

H2CO3 ⇋ H+ + HCO3- | H2CO3 ⇋ H2O + CO2

Answer 58

Assume that initial concentration of salt is the same the equilibrium value of A- HA initial concentration is the same as it's equilibrium concentration Use and rearrange the Ka formula if necessary So trying to solve for H+ as that is what tells us what can tell us what the pH is

Answer 59

The equivalence point | The point where the acid is neutralised and (OH-)=(H+)

Answer 60

Colour at low pH is red Colour at high pH is yellow Approx pH of colour change is between 3 and 4.5 So useful for strong acid/ strong base and strong acid/ weak base

Answer 61

Colour at low pH is colourless Colour at high pH is pink Approx pH of colour change is 8 to 10 So useful for strong acid/ strong base and weak acid/ strong base

Answer 62

They have differently coloured conjugate pairs, so as H+ or OH- concs change the equilibrium positions also do phenolphthallein-H (colourless) ⇋ phenolphthallein-(pink) +H+ Methyl orange-H (red) ⇋ Methyl orange- (yellow) + H+

Answer 63

The enthalpy change when one mole of an ionic lattice is formed from it's gaseous ions under standard conditions (always exothermic)

Answer 64

At the very bottom have the ionic lattice Then above it have the elements of the ionic lattice in their standard states, with line going downwards showing energy change of formation Then do the atomisation energy of each of the elements, the solids will go to gases, and any diatomic molecules will be come singular, the lines are going upwards Do the ionisation energies of what was the metal, gives +1 charge to the metal gaseous ion and produces an electron, lines go up Do the electron affinitys of the other element, decreasing it's by 1 and removing an electron until you reach it's gaseous ions, arrow may go up or down, look to see if value is positive or negative Lattice enthalpy goes from ions to salt, so to calculate it work out value of all the other arrows going in the correct direction to the salt from the gaseous ions

Answer 65

Draw an arrow from the salt to it's aqueous ions, this line is the enthalpy change of solution Bellow it have the gaseous ions Draw arrow from the gaseous ions to the salt, this is the lattice enthalpy Then draw arrow from gaseous ions to aqueous ions,this is enthalpy change of hydration Add hydration and lattice enthalpy to find the one of solution

Answer 66

A measure of disorder of particles, the higher it is the more disordered the particles Is effected by state and amount of molecules (in equations) Has to be positive for a reaction to be feesible

Answer 67

Entropy = Entropy of products - Entropy of reactants So entropy has to increase during a reaction for it to be feesible

Answer 68

ΔG = ΔH - TΔS ``` ΔH = enthalpy change (J mol^-1) T= temperature in Kelvin ΔS= entropy change in (J K^-1 mol^-1) ``` If positive, reaction isn't feesible and can make equal 0 to work out when a reaction

Answer 69

Split it into a oxidation half equation (electrons are lost so on product side) and a reduction half equation (electrons are gained so are on reactant side) If trying to find an unknown half equation, can add H2O, H+ and e- to balance

Answer 70

Reversible | Always put e- on the reactant side

Answer 71

E⦵ value, the more negative it is the easy it is to oxidise So the more negative E⦵value reaction goes backwards And the more positive E⦵ goes forwards

Answer 72

E⦵ = E⦵ of more positive side - E⦵ of more negative side So E⦵ will always be positive

Answer 73

E⦵ of a half cell is the voltage measured under standard conditions when the half cell is connected to a standard hydrogen electrode (0.00v)

Answer 74

Write down both half equations with electrons on the standard left size Look at the E⦵ value, the more positive one the reaction goes forwards (left to right) The more negative one goes backwards (right to left) Combine the 2 half equations remembering to multiply if uneven amount of electrons Check to see if the combination of half equations matches what has been asked in the question If it does then it's feasible, if doesn't match it's not

Answer 75

If you increase the concentration of Zn(2+) equillibrium shifts left, reducing the ease of electron loss, so the electrode potential of Zn/Zn(2+) becomes less negative and the whole cell potential will be lower If you increase the concentration of Cu(2+), the equillibrium will shift right, increasing the ease of electron gain, the electrode potential of Cu(2+)Cu becomes more positive and the whole cell potential becomes higher

Answer 76

At the anode the platinum catalyst splits the H2 into protons and electrons (H2 = 2e- +2H(+)) The polymer electrolyte membrane only allows the H+ across and forces the e- to travel around the whole circuit to get to the cathode An electric current is created in the circuit, which is used to power something At the Cathode, O2 combines with the H+ from the anode and the e- from the circuit, to form H2O which is the only waste product

Answer 77

5Fe(3+) Once the reaction has finished will see a colour change from purple to colourless as MnO4(-) is pink and {Mn(H2O)6}(2+) is colourless How much Oxidising agent we need to reduce the species MY OCTPUS 8 HIS 5 EARS "MUNCH" 4 HIPPOS 5 FISH, 5 FERRETS AND 5 EARS

Answer 78

Cr2O7(2-) + 14H+ + 6e- = 2Cr(3+) + 7H2O 3Zn = 3Zn(2+) +6e- So equals Cr2O7(2-) + 14H+ + 3Zn = 2Cr(3+) + 7H2O + 3Zn(2+) Colour change when reaction is finished will be orange to green as Cr2O7(2-) is orange and {Cr(H2O)6}(3+) looks green How much Oxidising agent we need to reduce the species

Answer 79

Use a certain volume Potassium Iodate solution to oxidise an excess of KI IO3(-) + 5I(-) + 6H+ = 3I2 + 3H2O Find out how many moles of iodine have been produced, by titrating the resulting solution with a known concentration of Sodium Thiosulfate (Na2S2O3), when the solution goes pale yellow showing it's near to the end point, add starch solution, which will show that Iodine is still present by being blue/black, will know reaction is finished when blue/black dissapears I2 + 2S2O3(2-) = 2I- + S4O6(2-) Find the moles of Iodine from second equation, then input it into first equation to find moles of oxidising agent ( divide by 3) and then can find concentration

Answer 80

A d block element, that can form at least one stable ion with an incomplete d sub shell So scandium and Zinc aren't transition elements

Answer 81

Chromium- prefers to have one electron in each orbital of the 3d subshell and just one in the 4s subshell, makes it more stable Copper- prefers to have a full 3d subshell, and only one electron in the 4s subshell

Answer 82

Cu2(+) + 2OH- = Cu(OH)2 goes from pale blue solution to a pale blue precipitate OH is there as NH3 is a weak alkali Then in excess NH3,Cu(OH)2 dissolves to form {Cu(NH3)4(H2O)2}2+ which is a dark blue colour So the reaction is 4NH3 + {Cu(H20)6}2+ = (reversible) {Cu(NH3)4(H20)2}(2+) + 4H2O

Answer 83

Fe(2+) + 2OH- = Fe(OH)2 pale green solution to a green precipitate, OH there as NH3 is a weak alkali

Answer 84

Fe3+ + 3OH- = Fe(OH)3 goes from yellow solution to an orange precipitate, OH there as NH3 is a weak acid

Answer 85

Mn(2+) + 2OH- = Mn(OH)2 goes from pale pink solution to a light brown precipitate

Answer 86

Cr(3+) + 3OH- = Cr(OH)3 goes from a purple solution to a grey-green precipitate Then in excess NH3, Cr(OH)3 dissolves to form {Cr(NH3)6}(3 +)which is a purple solution So the equation is {Cr(H2O)6} + 6NH3 =(reversible) {Cr(NH3)6}(3+) + 6H20

Answer 87

A metal ion surrounded by coordinately (dative bond, both electrons are from the same atom) bonded ligands

Answer 88

Occurs when an ion can exist in 2 non-superimposable mirror images (don't match each other when put on top of each other) So occurs when in octahedral complexes, when 3 biodentate ligands are attached to the central ion Draw by reflecting in a mirror line

Answer 89

When there's only 2 different groups involved The Cis isomers have the same groups on the same side (so are close together as they can be) Trans have the same groups on opposite side

Answer 90

A complex of platinum(II), with 2 chloride ions and 2 ammonia molecules in a square planar shape Treats cancer by chloride ligands being displaced, binds to cancerous cell DNA (at Nitrogens) stops it reproducing, can't fix damage, dies

Answer 91

Work out the half life, and show it;s constant or not constant

Answer 92

Not square

LAST MINUTE CHEM Flashcards

(117 cards)