CCEA A-Level Chemistry

A-Level · CCEAChemistry53 catatan dalam 6 folder, 304 KB

53 catatan belajar, dibuat oleh cookie, dalam 6 bagian yang mengikuti topik kursus.

Notes for CCEA GCE Chemistry (from September 2016), in folders for the six units in the specification's order (AS 1 to AS 3 and A2 1 to A2 3), with one note for each topic or pair of topics and the practical units set out as methods and the chemistry behind them. Delete the A2 folders if you take AS only.

Menambahkannya akan menaruh salinan di catatanmu, dalam folder tersendiri bersama folder di bawahnya. Kamu bisa mengubahnya dan menjadikannya kartu belajar atau deck soal. Mengunduh akan memberimu file zip berisi file markdown yang bisa dibuka di aplikasi catatan apa pun.

Isi di dalamnya

AS 1 Basic concepts in physical and inorganic chemistry
- Formulae, equations and amounts of substance6 KB
- Atomic structure and mass spectra of elements5 KB
- Electron configuration and ionisation energy6 KB
- Ionic, covalent and metallic bonding5 KB
- Electronegativity and intermolecular forces5 KB
- Structure and properties of solids5 KB
- Shapes of molecules and ions5 KB
- Redox4 KB
- Group VII, the halogens6 KB
- Acid-base titrations and standard solutions7 KB
- Qualitative tests4 KB
AS 2 Further physical and inorganic chemistry and an introduction to organic chemistry
- Empirical formulae, gas volumes, yield and atom economy5 KB
- Nomenclature and isomerism in organic compounds7 KB
- Alkanes6 KB
- Alkenes6 KB
- Halogenoalkanes8 KB
- Alcohols5 KB
- Infrared spectroscopy4 KB
- Enthalpy changes and how they are measured6 KB
- Hess's law and bond enthalpies5 KB
- Kinetics: collision theory and the Maxwell-Boltzmann distribution5 KB
- Equilibrium and industrial processes6 KB
- Group II elements and their compounds6 KB
AS 3 Basic practical chemistry
- Inorganic and energetics practicals7 KB
- Organic practicals6 KB
A2 1 Further physical and organic chemistry
- Lattice enthalpy and Born-Haber cycles6 KB
- Entropy and free energy5 KB
- Rate equations and orders5 KB
- Rate-determining steps, mechanisms and the effect of temperature on rate6 KB
- Equilibrium calculations4 KB
- pH, Kw and Ka5 KB
- Buffers, titration curves and salts6 KB
- Isomerism across families and optical isomerism5 KB
- Aldehydes and ketones6 KB
- Carboxylic acids4 KB
- Esters, acyl chlorides, fats and oils6 KB
- Benzene and aromatic chemistry6 KB
A2 2 Analytical chemistry, transition metals, electrochemistry and organic nitrogen chemistry
- Mass spectrometry of compounds5 KB
- Nuclear magnetic resonance spectroscopy6 KB
- Chromatography5 KB
- Volumetric analysis: iodine and thiosulfate, manganate(VII) and back titration6 KB
- Transition metals: electron configurations, complexes and ligands6 KB
- Colours, ligand replacement and hydroxide tests of transition metal ions6 KB
- Vanadium, redox and catalysis by transition metals5 KB
- Electrode potentials, cells and fuel cells8 KB
- Amines5 KB
- Amides, diazonium salts and dyes5 KB
- Amino acids, proteins and enzymes7 KB
- Condensation polymers5 KB
- Chemistry in medicine8 KB
A2 3 Further practical chemistry
- Physical practicals: rates, buffers, titration curves and cells7 KB
- Organic practicals: carbonyl compounds, acids, esters, aspirin and nitration7 KB
- Volumetric analysis, chromatography and transition metal practicals6 KB

Catatan pertama

AS 1 Basic concepts in physical and inorganic chemistry / Formulae, equations and amounts of substance

Writing formulae of ionic compounds

The charge on a simple ion can be predicted from the position of its element in the Periodic Table. Metals in Groups I, II and III (13) lose one, two and three electrons to form $\ce{Na+}$ , $\ce{Mg^2+}$ and $\ce{Al^3+}$ , and non-metals in Groups V, VI and VII gain three, two and one electrons to form $\ce{N^3-}$ , $\ce{O^2-}$ and $\ce{Cl-}$ . Each of these ions has the electron arrangement of a noble gas. An ionic compound is electrically neutral, so the formula is the smallest whole-number ratio of ions in which the charges cancel: magnesium and chlorine give $\ce{MgCl2}$ , and aluminium and oxygen give $\ce{Al2O3}$ , because two $\ce{Al^3+}$ (+6) balance three $\ce{O^2-}$ (−6).

Where an ion contains more than one atom, its formula goes in brackets when more than one is needed, as in $\ce{Ca(NO3)2}$ and $\ce{(NH4)2SO4}$ . The molecular ions to know, with their charges, are in the table.

Name	Formula	Name	Formula
sulfate	$\ce{SO4^2-}$	hydroxide	$\ce{OH-}$
sulfite	$\ce{SO3^2-}$	dichromate	$\ce{Cr2O7^2-}$
thiosulfate	$\ce{S2O3^2-}$	chromate	$\ce{CrO4^2-}$
hydrogensulfate	$\ce{HSO4-}$	permanganate (manganate(VII))	$\ce{MnO4-}$
hydrogencarbonate	$\ce{HCO3-}$	ammonium	$\ce{NH4+}$
carbonate	$\ce{CO3^2-}$	chlorate	$\ce{ClO3-}$
nitrate	$\ce{NO3-}$	hypochlorite	$\ce{ClO-}$
nitrite	$\ce{NO2-}$	phosphate	$\ce{PO4^3-}$

Balanced equations

A balanced equation has the same number of atoms of each element on both sides, and in an ionic equation the total charge is also the same on both sides. State symbols show the physical state: (s), (l), (g), and (aq) for a substance dissolved in water.

A full equation shows every formula, and an ionic equation leaves out spectator ions, which are present unchanged on both sides. The reaction of aqueous barium chloride with aqueous sodium sulfate is written in full as

\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}

and as an ionic equation as

\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}

Sodium and chloride ions take no part, so they do not appear. Equations for reactions not met before can be worked out when the reactants and products are given, by writing the formulae first and only then balancing with coefficients, never by changing a subscript in a formula. For example, the reaction of aluminium with dilute sulfuric acid gives aluminium sulfate and hydrogen. The formulae are $\ce{Al}$ , $\ce{H2SO4}$ , $\ce{Al2(SO4)3}$ and $\ce{H2}$ , and balancing gives

\ce{2Al(s) + 3H2SO4(aq) -> Al2(SO4)3(aq) + 3H2(g)}

Avogadro's constant, the mole and molar mass

The mole is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant, $L = 6.02 \times 10^{23}\ \pu{mol-1}$ . The particles may be atoms, molecules, ions or electrons, so the kind of particle must always be stated: one mole of $\ce{H2}$ molecules contains $\pu{6.02e23}$ molecules and $\pu{1.20e24}$ atoms.

The molar mass, $M$ , is the mass of one mole of a substance, in $\pu{g mol-1}$ . Numerically it equals the relative atomic mass, relative molecular mass or relative formula mass.

n = \frac{m}{M}

where $n$ is the amount in moles and $m$ the mass in grams. The number of particles is $n \times L$ .

For $\pu{3.65 g}$ of hydrogen chloride ( $M = 36.5$ ):

\begin{aligned} n &= \frac{3.65}{36.5} = 0.100\ \text{mol} \\ \text{molecules} &= 0.100 \times 6.02 \times 10^{23} = 6.02 \times 10^{22} \\ \text{atoms} &= 2 \times 6.02 \times 10^{22} = 1.20 \times 10^{23} \end{aligned}

The same quantity is obtained for solutions and gases by other routes. For a solution, $n = c \times V$ , with $c$ in $\pu{mol dm-3}$ and $V$ in $\pu{dm3}$ , and a volume in $\pu{cm3}$ is divided by 1000 to convert it, because $\pu{1 dm3} = \pu{1000 cm3}$ .

Reacting masses

The coefficients in a balanced equation give the ratio of the amounts in moles that react. A reacting-mass calculation has the same steps every time.

Write the balanced equation.
Convert the known mass to moles.
Use the ratio in the equation to find the moles of the substance wanted.
Convert back to a mass or volume.

Where the quantities of two reactants are given, one is usually in excess, and the other, the limiting reagent, decides how much product forms. The limiting reagent is the one that would be used up first.

Worked example with an excess

$\pu{2.43 g}$ of magnesium is added to $\pu{100 cm3}$ of $\pu{1.00 mol dm-3}$ hydrochloric acid.

\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}

\begin{aligned} n(\ce{Mg}) &= \frac{2.43}{24.3} = 0.100\ \text{mol} \\ n(\ce{HCl}) &= 1.00 \times 0.100 = 0.100\ \text{mol} \end{aligned}

All the magnesium would need $2 \times 0.100 = 0.200$ mol of acid, and only 0.100 mol is present, so the acid is limiting and magnesium is in excess. The acid decides the yield:

\begin{aligned} n(\ce{H2}) &= \tfrac{1}{2} \times 0.100 = 0.0500\ \text{mol} \\ m(\ce{H2}) &= 0.0500 \times 2.0 = 0.100\ \text{g} \end{aligned}

The magnesium used is $0.0500 \times 24.3 = \pu{1.215 g}$ , so $\pu{1.215 g}$ is left over.

Hydrated salts and water of crystallisation

An anhydrous substance contains no water. A hydrated crystal has water molecules built into its lattice, called water of crystallisation, in a fixed ratio to the ions, and it is written with a dot, as in $\ce{CuSO4.5H2O}$ . Heating drives the water off, and heating to constant mass (heating, cooling in a desiccator, weighing, and repeating until two weighings agree) shows that no water remains.

Finding the number of molecules of water

$\pu{2.86 g}$ of hydrated sodium carbonate, $\ce{Na2CO3.xH2O}$ , leaves $\pu{1.06 g}$ of anhydrous solid after heating.

\begin{aligned} n(\ce{Na2CO3}) &= \frac{1.06}{106.0} = 0.0100\ \text{mol} \\ m(\ce{H2O}) &= 2.86 - 1.06 = 1.80\ \text{g} \\ n(\ce{H2O}) &= \frac{1.80}{18.0} = 0.100\ \text{mol} \\ x &= \frac{0.100}{0.0100} = 10 \end{aligned}

The formula is $\ce{Na2CO3.10H2O}$ . The same method works from percentage composition: treat the percentages as masses in a 100 g sample, divide each by its molar mass and find the simplest ratio. A hydrate whose mass loss is large compared with its anhydrous mass has a large $x$ .

Percentage of water can also be calculated from the formula: for $\ce{CuSO4.5H2O}$ ( $M = 249.6$ ), five waters weigh $5 \times 18.0 = 90.0$ , and $\frac{90.0}{249.6} \times 100 = 36.1\%$ .

Dan 52 lagi setelah kamu menambahkannya atau mengunduhnya.

Ulasan

Belum ada ulasan tertulis. Tambahkan catatan ini ke catatanmu dan jadilah yang pertama memberi ulasan.