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Piper Dean > Chemistry exam > Flashcards

Chemistry exam Flashcards

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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
Separating mixtures
Boiling Decanting Filtration
26
Nanomaterials
Materials where the particles are between 1 and 100 nanometers Very important in the production of nanotechnologies (electronics, healthcare, mechanical)
27
Four steps in determining IMF
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.
28
VSEPR Theory to predict shapes
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
29
Lewis structure
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
30
Polar bonds
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
31
IMF
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
32
Dispersion forces
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
33
Dipole-dipole interactions
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
34
Hydrogen bonding
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
35
Melting and boiling point
When the intermolecular forces are stronger more kinetic energy (heat) is needed to separate the molecules –therefore higher melting point and boiling point
36
Vapour Pressure
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.
37
Solubility
‘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.
38
Surface tension
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
Vapour pressure
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
Solubility
Polar solute dissolves in a polar solvent. Non-polar solute dissolves in a non-polar solvent.
41
Stationary phase
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
Mobile phase
The phase that flows in chromatography, moving the components of a sample (solute) at different rates over the stationary phase.
43
Paper chromatography
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
Retardation factor (Rf)
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
Chromatography
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
Affinity
Components of a sample are attracted, by intermolecular forces, to the mobile or stationary phase.
47
Adsorption
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
Desorption
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
The system
The chemical reaction. This is the energy changes that occur as bonds break and form between atoms within the reaction.
50
The surroundings
Everything outside of the chemical reaction.
51
Law of conservation of energy
Energy cannot be created or destroyed
52
Exothermic
Release heat, negative enthalpy change
53
Endothermic
Absorbs heat. Positive enthalpy change
54
Enthalpy
(H), the chemical energy of a substance
55
Enthalpy change (delta H)
Hproducts - Hreactants Delta H= Q/n
56
Bond enthalpy
Breaking bonds absorbs energy (endothermic) Forming bonds releases energy (exothermic)
57
Enthalpy change bonds
Sum of bonds broken - sum of bonds formed
58
Thermal energy absorbed or released by a system (Q)
Q= MC delta T Delta H= Q/n
59
Calorimetry
Experimental method of measuring amount of energy released or absorbed by a chemical reaction
60
Number of moles
n=m/M n= number of particles/6.02 X 10^23 n= V/22.7 (molar volume at STP)
61
Law of conservation of mass
Matter (i.e. atoms) is neither created nor destroyed during a chemical reaction.
62
Limiting reagent
The reactant that runs out first and limits the reaction, it is entirely consumed in the reaction
63
Excess reagent
The reactant that does not run out, there are leftovers because there is nothing left for it to react with
64
Percentage yield
(experimental yield/theoretical yield) X 100
65
Theoretical yield
the amount of product predicted from stoichiometric calculations
66
Experimental yield
The actual amount of product that is produced in a chemical reaction
67
Calculating % yield
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
Percentage composition
Organised by element The percent that each element contributes to the mass of a compound
69
Percentage composition given mass data
(Mass of element/mass of compound) X 100
70
Percentage composition given chemical formula
(mass of x in 1 mole of compound/ molar mass of compound) X 100 Assume compound is grams
71
Empirical formula
The simplest whole number ratio of elements present in a compound
72
Determining empirical formula
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
Solute
The dissolved particles in a solution
74
Solvent
The medium which dissolves the solute
75
Solution
The result of a solute dissolving in a solvent
76
Concentration
The amount of solute which is dissolved in a given quantity of solvent
77
Concentrated Vs Dilute
Concentrated: large mass of of solute dissolved in the solvent Dilute: small mass of solute dissolved in solvent
78
Calculating molarity
C= n/V
79
Calculating grams per litre
C= m/V
80
Calculating parts per million
C= m (mg) /V
81
Dilution
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
Choosing indicators for titration
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
Indicator
A substance that exhibits a colour change as the concentration of hydrogen (H+) or hydroxide (OH-) ions change in an aqueous solution.
84
Equivalence point
The pH at which the number of moles of base (OH-) has completely reacted with the number of moles of acid (H+)
85
End point
the pH range at which the indicator changes colour
86
Acids
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
Bases/alkalis
Base: A base is a substance that produces hydroxide ions (OH-) Alkali: is a soluble base
88
Three types of base
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
Strong acid/base
A strong acid or base will almost completely dissociate into ions.
90
Weak acid or base
Only partially dissociates into ions.
91
Acid and base
acid + base ---> salt + water
92
Acid + metal
Acid + metal ---> salt + hydrogen gas
93
Acid + carbonate
acid + carbonate salt + carbon dioxide + water

Piper Dean (3 decks)

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