Chem sadness

Question 1

describe :

How greenhouse gasses maintain the global temperature, and than how they contribute to global warming

Answer 1

These glasses are able to absorb some of the thermal radiation reflected from the earth’s albedo and reradiate it in all directions towards other molecules and to the earth.

Question 2

Explain:

Ocean acidification

Answer 2

Caused by oceans absorbing high levels of CO2 through:
CO2 + H2O H2CO3 [carbonic acid]
H2CO3 + H2O H3O[+] + HCO3[-]
HCO3[-] + H2O H3O[+] + CO3[2-]

Question 3

Explain:

How carbonates react in acidic conditions

Answer 3

Shells are made of calcium carbonate and are vulnerable to dissolution through sulfuric acid:
Full : CaCO3 + H2SO4 -> CO2 + H2O + CaSO4
Ionic : CaCO3 + 1H[+] -> CO2 + H2O = Ca[2+]

Question 4

Explain:

The formation of nitrogen oxides

Answer 4

NO and NO2 are formed in high temperature furnaces and pressure engines through:
N2 + O2 -> 2NO
2NO + O2 -> 2NO2

Question 5

Describe:

The role of nitrogen oxides in formation of ozone in troposphere and how it is bad

Answer 5

Nitrogen oxides form ozone and contribute to photochemical smog:
NO2 -(UV)-> NO + O
O + O2 -> O3
Causes irritation and disrupts photosynthesis of plants.
Damages certain polymers

Question 6

Describe:

Shapes of molecular ozone, methane, and whatever looks like a triangle

Answer 6

Ozone = V-shaped
Methane = tetrahedral 
Triangle = Trigonal planar

Question 7

Describe:

What is needed for photochemical smog

Answer 7

High concentration of pollutants, sunlight, still air and a temperature inversion.

Question 8

Explain:

How catalytic converters do stuff.

Answer 8

The high surface area material can convert photochemical constituents into less harmful materials as catalysts to the reactions:
2NO + 2CO -> 2CO2 + N2

Question 9

Explain:

Conversions between Mol.L / g.L / %W/V / ppm and ppb.

Answer 9

Mol.L xM -> g.L x1000 -> ppm (mg.L) x1000 -> ppb (ug. L)

g.L x1/10 = %W/V (g 100m.L)

Question 10

Explain:

Convention formula used in concentration calculation

Answer 10

C=n/v

c1V1 = c2V2

Question 11

Describe:

Chromatography

Answer 11

TLC, GC, HPLC, IC use a stationary phase and a mobile phase.

The rate of movements is determined by the difference in strength of interactions between molecules and the phases.

Question 12

Describe:

Rf values and retention time and its use

Answer 12

Data can be used to determine the purity of substances, the the retardation factor which is the distance travelled divided by the solvent fronts distance.
Retention time is the time taken from the injection of sample to detection of sample

Question 13

Describe

Equilibrium principles in relation to ion exchange

Answer 13

Cations or anions (sodium) are removed from a resin by replacing them with ions of another type:
Resin(-)Na(+) + NH4(+) ← → Resin(-)NH4(+) + Na(+)

Question 14

Explain:

The subshell notations for atoms and exceptions

Answer 14

1s2 < 2s2 < 2p6 < 3s2 < 3p6 < 4s2 < 3d10 < 4p6 < 5s2 < 4d10 < 5p6
Copper:
1s2 2s2 2p6 3s2 3p6 4s1 3d10
Chromium:
1s2 2s2 2p6 3s2 3p6 4s1 3d5

Question 15

Explain:

The effect of absorption or emission of radiation on the electron configurations in atoms or ions.

Answer 15

Electrons in ground state may transfer to higher states upon absorption of a photon, sending it into a higher energy level. Than it may fall to a lower level, releasing a photon

Question 16

Describe:

Wavelengths of radiation emitted and absorbed by an element

Answer 16

These wavelengths are unique to any element and are caused by electron transfer between atomic energy levels. The technique atomic emission spectroscopy is used to do so.

Question 17

Explain:

Principles of atomic absorption spectroscopy

Answer 17

Sensitive technique which uses a hollow cathode lamp coated in the element of interest, a voltage is applied, causing electrons to be excited to higher states, than releasing a specific frequency of radiation upon falling. This beam is split, one goes through a flame containing material of interest and the other is sent to a detector to establish an incident reading of light intensity.

Question 18

Describe:

Collision theory

Answer 18

Directly affects the rate of reaction
Concentration, temperature, pressure (only for gas), surface area
Increase the number of collisions per unit of time.
Presence of catalyst and temperature increase the number of reactants that possess activation energy.

Question 19

Explain:

Graphs representing change in concentration of reactants and products

Answer 19

In closed systems, over time reversible reactions at a fixed temperature result in equilibrium.

Question 20

Explain:

The writing of kc expressions and what it is good for

Answer 20

Kc value gives the position of equilibrium in chemical system, related to concentrations. It can only be changed by the temperature.
Calculated by:
[Product]^#moles [Product]^#moles / [Reactant]^#moles [Reactant]^#moles

Question 21

Describe:

Changes that can be made to position of equilibrium

Answer 21

Given by Le Chatelier’s principle, factors will cause a change that will stress the reaction and thus the equilibrium will shift to counter the change:
Pressure - by mole ratio
Concentration - reactants/products
Temperature - by deltaH (+ is endo)

Question 22

Describe:

Components of production flow charts

Answer 22

Raw materials- unprocessed used to produce better mats
Waste product- Unusable product of reaction
By-product- usable product of reaction

Question 23

Explain:

Manufacture conditions in relation to environment and cost.

Answer 23

Conditions represent a compromise for high yield to low costs, this goes for temperature and pressure. The use of catalysts will benefit the environment and higher reaction rates are achieved for little cost.

Question 24

Describe:
How to predict melting, boiling point of organic compounds and solubility in water and nonpolar solvents of organic molecules

Answer 24

Points are dictated by the strength of the intermolecular forces. Which is than dictated by the size and polarity of molecules. More polarity has high boiling points.

Solubility is dictated by size and polarity of molecules, smaller is more soluble

Question 25

Explain:
The reactions that can occur with specific alcohols to define them and explain the structural formula that results from reactant to product:

Answer 25

Oxidising agents such as acidified dichromate will react with primary and secondary alcohols but not tertiary alcohols.

Question 26

Describe:

Alcohol group

Answer 26

Hydroxyl groups, primary/secondary/tertiary depends on number of R groups connected to the carbon connected to the hydroxyl group
Oxidation of primary -> aldehyde -> carboxylic acid
Oxidation of secondary -> keytone
Experiences H-bonding

Question 27

Describe:

Ketone and aldehyde group

Answer 27

They have a carbonyl group (=O),
They are formed through the oxidation of primary and secondary alcohols respectively.
Experience dipole-dipole forces with eachother (boiling point) and H-bonding with water(soluble)

Question 28

Explain:

The reactions that can occur with ketones and aldehydes to define them.

Answer 28

Tollens reagent is used to test for aldehydes and will produce elemental silver in oxidation of aldehyde.

Question 29

Explain:

The formula of an aldehyde or ketone after oxidation in acidic or alkaline conditions.

Answer 29

Aldehydes oxidized in basic conditions result in a carboxylate anion(-), in acidic conditions it results in carboxylate salt

Question 30

Describe:

Molecular formula for glucose and the polysaccharides based on its monomers

Answer 30

Is polyhydroxy aldehyde or ketone.
Cyclic for on glucose has a methanol group to the carbon on the right side of the oxygen connector, all other carbons have an OH and a H group attached to them.

Question 31

Explain:

How polysaccharides are formed

Answer 31

Either polyhydroxy aldehydes or polyhydroxy ketones, or substances that form these compounds on hydrolysis go through condensation polymerisation.

Question 32

Describe:

Carboxylic acids

Answer 32

They can be produced by oxidation of aldehydes or primary alcohols and are weak acids and contain a carboxyl group.

Question 33

Explain:

Reactions of carboxylic acids with bases.

Answer 33

Hydroxides - RCOOH + (salt)OH → (salt)RCOO + H2O
Carbonates - RCOOH + (salt)CO3 → (salt)RCOO + CO2 + H2O
Hydrogen carbonates - RCOOH + (salt)HCO3 → (salt)RCOO + CO2 + H2O

Question 34

Explain:

Why sodium carboxylate salts are more soluble in water than their parent acids.

Answer 34

As carboxylate salts, lost their salt in solution and becomes a carboxylate anion, thus experiencing ion-dipole forces of attraction with water which is stronger than H-bonding.

Question 35

Describe:

Amines

Answer 35

An NH2 group which can be NH or N, the N has a lone pair of electrons which should be drawn.
They are either primary, secondary or tertiary which is dependant on the R groups directly attached to the N group. and they may act as a base

Question 36

Explain:

Why the protonated form of an amine is more soluble in water than the boring one and how it may be drawn

Answer 36

The lone pair may accept a H+ group from an acid. When drawn protonated, the + is on the N group.
It is more soluble as protonated has a whole positive charge, thus experiences ion-dipole forces with water, in comparison to weaker H-bonding.

Question 37

Describe:

Esters

Answer 37

They can be formed through the condensation reaction of carboxylic acid and alcohols, resulting in water release.

Polyesters are formed through combination of Diols and Dicarboxylic acids.

Question 38

Describe:

Naming of esters

Answer 38

[Alcohol group] Alkyl + [Carboxylate group] Oate

Alkyl Oate

Question 39

Explain:

The use of heating under reflux and use of concentrated sulfuric acid in laboratory preparation of esters.

Answer 39

Increases proportion of reactants that possess activation energy for esterification, resulting in more successful collisions per unit of time while ensuring that neither products nor reactants escape the reaction vessel.
Acid is a catalyst
Water flows from base to the top.

Question 40

Describe;

Hydrolysis

Answer 40

Chemical compound is broken down in reaction with water.
Esters will break into carboxylic acids and alcohol groups in acidic conditions.
Esters will break into carboxylate anion and alcohol groups in basic conditions.

Question 41

Describe:

Amides

Answer 41

They can be made when carboxylic acids undergo condensation reactions with amines. Looks like an ester with an N instead of a connecting O. They can be broken apart through hydrolysis.
Polyamides are formed with Diamine and Dicarboxylic acids.

Question 42

Describe:

How to separate amides and the products that may be produced

Answer 42

Hydrolysis in acidic or alkaline conditions will separate the groups:
Amide + Water (H+) → Carboxylic acid + Protonated amine
Amind + Water (OH-) → Carboxylate anion + Amine

Question 43

Describe:

Glycerides

Answer 43

Esters of propane-1, 2, 3-triol (glycerol) with various carboxylic acids form edible oils and fats. Under hydrolysis, they are broken into carboxylic acids and glycerol depending on the conditions. Alkaline conditions result in productions of carboxylate ions.

Question 44

Describe:

How the degree of unsaturation can be tested in triglycerides

Answer 44

Through bromine or iodine solution, the brown-purple colour will become clear if there was unsaturation present.
The iodine value is the mass of iodine that reacts with 100g of a triglyceride or fatty acid. Ditto for bromine number.

Question 45

Explain:

How differences in melting point of fats are be.

Answer 45

Determined by the difference in saturation of the fats. More saturation results in higher boiling points as they can move closer together and experience stronger secondary interaction.

Question 46

Explain:

How triglycerides can be made to have higher melting points

Answer 46

Through, pressure, temperature, and a catalyst H gas will convert double bonds into single bonds, increasing saturation of a triglyceride.

Question 47

Explain:
How carboxylate ions can form micelles in solution and how they can dissolve and move non-polar substances through polar water.

Answer 47

Long carbon chain acts as hydrophobic region, carboxyl group is hydrophilic as it has ion dipole interactions. Micelles are formed as the hydrophilic head stays in contact with the water while the hydrophobic tail can interact with non-polar substances where polar water cannot. The micelle is able to flow and interact with polar water.

Question 48

Describe:

proteins

Answer 48

They are polymers of amino acids. Amino acids contain a carboxyl group and an amino group with an R. This occurs when amino acids go through condensation to form peptide links
Amino acids present can be found through identification of N + R + Carboxyl/Carbonyl group

Question 49

Explain:

Why the function of a protein may be affected by pH and temperature

Answer 49

The function is a consequences of its spatial arrangement. Shape is determined by the secondary interactions between the R groups of the amino acid sub-units. Changes in the environment will result altering the interactions between the r groups thus destabilising tertiary structure.