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Piper Dean > Chemistry exam > Flashcards

Chemistry exam Flashcards

1
Q

Cations and anions

A

The loss of electrons results in an ion with a positive charge, positive ions are called cations. The gain of electrons results in an ion with a negative charge, negative ions are called anions.

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2
Q

Atomic structure

A

A: mass number (number of protons and neurons in the nucleus)
Z: atomic number (number of protons in the nucleus)
X: element

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3
Q

Naming ionic compounds

A

1.Name the metal first (f the metal has more than one possible charge the charge should be indicated using roman numerals)
2.Name the stem of the non-metal and add ide.

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4
Q

Diatomic molecules

A

Hydrogen –H2, Nitrogen –N2, Oxygen –O2, Fluorine –F2, Chlorine –Cl2, Bromine –Br2and Iodine –I2

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5
Q

Naming covalent compounds

A
  1. If only one type of atom present just name the element. (Example: S8–sulphur)
    2.Name the first element present.
    3.Name the stem of the second element and add ide.
    4.Use Greek prefixes to indicate the number of atoms present. Do not use mono when naming the first element.
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6
Q

Greek prefixes for naming covalent compounds

A

mono, di, tri, tetra, penta, hexa, hepta, octa, nona, deca

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7
Q

Covalent bonding

A

The sharing of electrons between atoms. Covalent bonds typically form between two or more non-metals.

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8
Q

Ionic compound

A

An ionic compound is formed when electrons are transferred from one element to another element to form a cation and an anion. The attraction between the anion and cation is called an ionic bond. An ionic compound does not have a superscript ionic charge because it is overall neutral, the charges of the cations and anions must be balanced.

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9
Q

Isotopes

A

Atoms of an element that have the same number of protons but a different number of neutrons and therefore a different mass. E.g: ^6Li or Lithium-6

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10
Q

Properties of isotopes

A

Isotopes have the same electron configuration when neutrally charged, therefore their chemical properties are similar.
The physical properties are determined by the mass number of the isotope.

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11
Q

RAM

A

Ar= RIM1 x abundance(%)1/100+ (RIM2 x abundance(%)2/100) + …..

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12
Q

Mass spectronomy

A

An analytical technique that is able to identify specific identities and amounts of atoms or molecules in a sample
Used in particular to determine the isotopic composition of an element
Information from mass spectrometer is made into a graph called a mass spectrum which shows the mass : charge ratio of the ions and their relative abundance.

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13
Q

Absorption spectroscopy

A

When atoms absorb energy through light, heat or electric discharge the electrons can move up one or more energy levels from ground state but not part way between levels. When atoms absorb energy, they only absorb the specific quantities of energy that align with the amount of energy needed for the transitions. When a full spectrum of visible light is passed through a molecule, the light that passes through are the wavelengths that do not align to an absorption transition for that atom-absorption spectrum.

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14
Q

Ground state

A

Ground state is when they electrons are at their lowers energy level.

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15
Q

Emission spectroscopy

A

Electrons in higher energy levels than their ground state are in an excited state.
Electrons in an excited state are unstable and want to move back down to ground state, as they do this the release energy as light. The energy levels for each element are different which means each element emits unique wavelengths of light –emission spectrum.

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16
Q

Atomic absorption spectroscopy

A

Uses absorption spectroscopy in a quantitative way. Used to quantity the amount of an element present in a sample.
Uses the principle of the Beer-Lambert Law: the concentration of the sample element present is proportional to the amount of light that is absorbed.
Light is pulsed at the sample to excited the electrons and when they return to ground state they emit light. The emitted light intensity is detected and the more light that has been absorbed the higher the concentration of the metal.

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17
Q

Pure substance

A

Definite and constant composition –can be element or compound

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18
Q

Mixtures

A

physical combinations of pure substances that have no definite or constant composition

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19
Q

Element

A

made of the same type of atom, cannot be broken into simpler substances

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20
Q

Compound

A

contain atoms of different elements in fixed ratio, can be broken into simpler substances, represented using a formula

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21
Q

Homogenous

A

Uniform composition throughout. If you break matter into smaller pieces all pieces would be the same.

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22
Q

Heterogeneous

A

Non uniform throughout. If you break matter into smaller pieces not all pieces would be the same.

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23
Q

Physical properties

A

Can be observed without changing the substance
Colour
Smell
Melting, boiling and freezing point
Viscosity
Density

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24
Q

Chemical properties

A

Observed through changing the substance
Heat of combustion
Reactivity
pH

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25
Q

Separating mixtures

A

Boiling
Decanting
Filtration

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26
Q

Nanomaterials

A

Materials where the particles are between 1 and 100 nanometers
Very important in the production of nanotechnologies (electronics, healthcare, mechanical)

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27
Q

Four steps in determining IMF

A

Step A: Determine the Lewis Structure for a molecule. A lewis structure is a prediction of the bonding present in a molecule.

Step B: Determine the polarity of the covalent bonds. This involves considering the electronegativity difference of the atoms which are bonded together.

Step C: Determine the shape of the molecule. This involves a consideration of the impact that lone pairs of electrons have on the shape of the molecule.

Step D: Determine the polarity of the molecule. Doe the molecule have a dipole? A slightly negative and a slightly positive END to the molecule.

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28
Q

VSEPR Theory to predict shapes

A

Linear: Central atom has two covalent bonds and no lone pairs
Trigonal Planar: Central atom has three covalent bonds and no lone pairs
Tetrahedral: Central atoms forms four covalent bonds and no lone pairs
Trigonal Pyramidal: Central atom has three covalent bonds and one lone pair
Bent: Central atom has two covalent bonds and two lone pairs

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29
Q

Lewis structure

A
  1. Count valence electrons for each atom
    2.Arrange the atoms in a skeleton structure (H and halogens are usually outside atoms)
    3.Arrange the electrons so that each atom contributes one electron to a single bond between each atom.
    4.Count the electrons around each atom: are the octets complete? If so, your Lewis dot structure is complete
    5.If the octets are incomplete, and more electrons remain to be shared, move one electron per bond per atom to make another bond
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30
Q

Polar bonds

A

In a covalent bond when more electrons end up on one side of a bond and leaves a permanent charge (dipole). Polar bonds don’t cancel eachother out
The larger the electronegativity difference the more polar the bond
electronegativity difference > 0.5

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31
Q

IMF

A

Between the molecules
They are responsible for holding the molecules together
They are not bonds
They affect the properties of the molecules (such as melting and boiling point)
Dispersion forces, dipole-dipole interactions, hydrogen bonding

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32
Q

Dispersion forces

A

Weakest IMF, present in every molecule
Due to the movement of the electrons in the orbitals, the charge distribution will be slightly ‘skewed’ at any one time causing a slight polarity
The slight and temporary negative in one molecule will attract the slight and temporary positive in another molecule

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33
Q

Dipole-dipole interactions

A

A molecule has a permanent positive and negative side, polar bonds that do not ‘cancel’ each other out\
The negative side of one molecule attracts the positive side of the other molecule

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34
Q

Hydrogen bonding

A

A large electronegativity difference between a pair of atoms that includes hydrogen
NOF (hydrogen bonding can happen with nitrogen, oxygen and fluorine WITH hydrogen)
The hydrogen ends up being very positive and the other atom very negative
The positive hydrogens on one molecule are attracted to the non bonding pair of electrons of a negative atom
Strongest IMF

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35
Q

Melting and boiling point

A

When the intermolecular forces are stronger more kinetic energy (heat) is needed to separate the molecules –therefore higher melting point and boiling point

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36
Q

Vapour Pressure

A

High energy particles on the surface of water can vapourise and low energy particles in the air can condense. Vapour pressure is the pressure exerted by gas molecules above a liquid.

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37
Q

Solubility

A

‘Like Dissolves Like’. Polar substances dissolve in polar substances as they both have charges. Non polar substances will not dissolve in polar solvents because the polar molecules are more attracted to themselves.

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38
Q

Surface tension

A

Tension of the surface of a liquid due to the attraction between the particles. High surface tension means a lot of attraction between the particles, therefore high IMF

39
Q

Vapour pressure

A

The pressure exerted by gas molecules above a liquid. When a liquid evaporates it forms gas particles, to form gas particles there needs to be enough energy to overcome IMF. Low vapour pressure therefore indicates high IMF

40
Q

Solubility

A

Polar solute dissolves in a polar solvent. Non-polar solute dissolves in a non-polar solvent.

41
Q

Stationary phase

A

This phase does not move, it is always still. It is made of a solid with a high surface area or a liquid coated onto a solid support. It is the phase to which the components of a sample are absorbed from the mobile phase and then desorbed back into the mobile phase.

42
Q

Mobile phase

A

The phase that flows in chromatography, moving the components of a sample (solute) at different rates over the stationary phase.

43
Q

Paper chromatography

A

The stationary phase is a strip of paper and the mobile phase is often water but could be another solvent such as ethanol.
A dot of the sample to be separated is placed at one end of the paper and sometimes dots of known chemicals (standards) are placed beside it. The position of these dots is known as the origin.
The very end of the paper is placed in the solvent. The solvent is drawn up through the paper.
Chemicals in the unknown and standard samples which have a high affinity for the solvent will travel up the paper further than the chemicals which have a greater affinity for the stationary phase.
Retardation factors (Rf) are determined for each substance (or dot on the chromatogram).

44
Q

Retardation factor (Rf)

A

Rf= distance of sample spot/ distance of mobile phase. They should be expressed as a decimal and cannot be greater or equal to 1. A value of 1 would indicate that the chemical has not been separated from the mobile phase.

Larger Rf value means stronger intermolecular interactions with the mobile phase (migrates further distance)
Smaller Rf value means weaker intermolecular interactions with the mobile phase (migrates less distance)

45
Q

Chromatography

A

An analytical technique that can identify components within a mixture and can determine how much of each component is present. It uses intermolecular forces to separate components of a mixture according to their properties.
Depending on the intermolecular forces present, components of the sample are more attracted to either the mobile or the stationary phase.

46
Q

Affinity

A

Components of a sample are attracted, by intermolecular forces, to the mobile or stationary phase.

47
Q

Adsorption

A

Components of the sample adsorb onto the stationary phase from the mobile phase if they are attracted to, or have an affinity for, it.

48
Q

Desorption

A

Components of the sample desorb off the stationary phase into the mobile phase if they are attracted to, or have an affinity for, it.

49
Q

The system

A

The chemical reaction. This is the energy changes that occur as bonds break and form between atoms within the reaction.

50
Q

The surroundings

A

Everything outside of the chemical reaction.

51
Q

Law of conservation of energy

A

Energy cannot be created or destroyed

52
Q

Exothermic

A

Release heat, negative enthalpy change

53
Q

Endothermic

A

Absorbs heat. Positive enthalpy change

54
Q

Enthalpy

A

(H), the chemical energy of a substance

55
Q

Enthalpy change (delta H)

A

Hproducts - Hreactants
Delta H= Q/n

56
Q

Bond enthalpy

A

Breaking bonds absorbs energy (endothermic)
Forming bonds releases energy (exothermic)

57
Q

Enthalpy change bonds

A

Sum of bonds broken - sum of bonds formed

58
Q

Thermal energy absorbed or released by a system (Q)

A

Q= MC delta T
Delta H= Q/n

59
Q

Calorimetry

A

Experimental method of measuring amount of energy released or absorbed by a chemical reaction

60
Q

Number of moles

A

n=m/M
n= number of particles/6.02 X 10^23
n= V/22.7 (molar volume at STP)

61
Q

Law of conservation of mass

A

Matter (i.e. atoms) is neither created nor destroyed during a chemical reaction.

62
Q

Limiting reagent

A

The reactant that runs out first and limits the reaction, it is entirely consumed in the reaction

63
Q

Excess reagent

A

The reactant that does not run out, there are leftovers because there is nothing left for it to react with

64
Q

Percentage yield

A

(experimental yield/theoretical yield) X 100

65
Q

Theoretical yield

A

the amount of product predicted from stoichiometric calculations

66
Q

Experimental yield

A

The actual amount of product that is produced in a chemical reaction

67
Q

Calculating % yield

A
  1. balanced equation
  2. write knowns/unknowns
  3. calculate moles of reactant
  4. use mole ratio to calculate theoretical moles formed
  5. calculate theoretical mass formed
    Use experimental mass with theoretical mass to calculate percentage yield
68
Q

Percentage composition

A

Organised by element
The percent that each element contributes to the mass of a compound

69
Q

Percentage composition given mass data

A

(Mass of element/mass of compound) X 100

70
Q

Percentage composition given chemical formula

A

(mass of x in 1 mole of compound/ molar mass of compound) X 100
Assume compound is grams

71
Q

Empirical formula

A

The simplest whole number ratio of elements present in a compound

72
Q

Determining empirical formula

A
  1. Identify all elements present in sample
  2. Determine mass of each element
  3. Use mass to determine number of moles of each element present in sample
  4. Determine ratio by dividing all moles by the smallest mole
  5. If necessary reduce/multiply ratio to smallest whole number possible
73
Q

Solute

A

The dissolved particles in a solution

74
Q

Solvent

A

The medium which dissolves the solute

75
Q

Solution

A

The result of a solute dissolving in a solvent

76
Q

Concentration

A

The amount of solute which is dissolved in a given quantity of solvent

77
Q

Concentrated Vs Dilute

A

Concentrated: large mass of of solute dissolved in the solvent
Dilute: small mass of solute dissolved in solvent

78
Q

Calculating molarity

A

C= n/V

79
Q

Calculating grams per litre

A

C= m/V

80
Q

Calculating parts per million

A

C= m (mg) /V

81
Q

Dilution

A

Diluting a solution reduces number of moles in a given volume but total number of moles does not change
C1V1=C2V2
C1 concentration of original solution
V1 volume taken from original solution
C2 concentration of diluted solution
V2 volume of diluted solution

82
Q

Choosing indicators for titration

A

An indicator needs to be chosen so that the end point (colour change) is as close as possible to the pH of the equivalence point. That varies from titration to titration.

A titration curve can be used to determine the equivalence point of a titration. he equivalence point is halfway up the steep section of the curve.

83
Q

Indicator

A

A substance that exhibits a colour change as the concentration of hydrogen (H+) or hydroxide (OH-) ions change in an aqueous solution.

84
Q

Equivalence point

A

The pH at which the number of moles of base (OH-) has completely reacted with the number of moles of acid (H+)

85
Q

End point

A

the pH range at which the indicator changes colour

86
Q

Acids

A

An acid is a substance which is capable of donating a hydrogen ion, H+ (a proton).
Monoprotic: an acid that can donate/release one hydrogen ion per molecule. e.g. HCl, HNO3
Polyprotic: an acid that can donate/release more than one hydrogen ion per molecule. e.g. H2SO4
H2SO4(l) 2H+(aq) + SO42-(aq)

87
Q

Bases/alkalis

A

Base: A base is a substance that produces hydroxide ions (OH-)

Alkali: is a soluble base

88
Q

Three types of base

A

hydroxides e.g. NaOH, Ca(OH)2

Dissociation of hydroxides:

NaOH(s) Na+(aq) + OH-(aq)

oxides e.g. Na2O, MgO

Reaction of oxides with water:

Na2O(s) + H2O(l) 2Na+(aq) + 2OH-(aq)
ammonia NH3

Reaction of ammonia with water:

NaH3(g) + H2O(l) NH4+(aq) + OH-(aq)

89
Q

Strong acid/base

A

A strong acid or base will almost completely dissociate into ions.

90
Q

Weak acid or base

A

Only partially dissociates into ions.

91
Q

Acid and base

A

acid + base —> salt + water

92
Q

Acid + metal

A

Acid + metal —> salt + hydrogen gas

93
Q

Acid + carbonate

A

acid + carbonate salt + carbon dioxide + water

Piper Dean (3 decks)

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