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Summary A Level chemistry amount of substance ANKI flashcards

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ANKI flashcards for AQA A Level chemistry amount of substance module. TXT document can be uploaded onto ANKI to be used in the flashcard format.

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Definition of relative atomic mass (Ar) Weighted average mass of an atom of an
element, taking into account its naturally occuring isotopes, relative to 1/12 of
the mass of a cerbon 12 atom
Definition of relative molecular mass (Mr) Mass of a molecule compared to 1/12
relative atomic mass of a carbon-12 atom (found by adding up the Ar of all atoms in
the molecule)
Avogadro constant definition Number of atoms in 12g of 12C, i.e. number of
entities in a mole<br>6.022 x 10<sup>23</sup>
Moles definition Amount of substance that contains 6.022 x
10<sup>23</sup>&nbsp;particles<br>Applied to electrons, atoms, molecules, ions,
formulas and equations
Number of particles equation Number of particles = no. of moles x (6.022 x
10<sup>23</sup>)
Moles equation Moles = mass (g) / Mr&nbsp;
Percentage composition by mass equation % composition = (total mass of element in
formula / Mr) x100
Reacting masses moles method 1) Work out no. of moles of formula given with
mass<br>2) Use balanced equation to find no. of moles of desired
product/reactant<br>3) Work out mass of this number of moles
Reacting masses ratios method 1) Work out Ar/Mr of species concerned<br>2) Multiply
values by any big numbers in balanced equation<br>3) Mass of reactant / (2) for
reactant = Mass of product / (2) for product
Concentration of a solution definition Expressed in terms of number of moles of
solute dissolved in each dm<sup>3</sup>&nbsp;of solution
Concentration equations No. of moles = (volume (cm<sup>3</sup>) / 1000) x
concentration (mol dm<sup>-3</sup>)<br><br>Concentration (mol dm<sup>-3</sup>) =
(moles x 1000) / volume (cm<sup>3</sup>)<br><br>Volume (cm<sup>3</sup>) = (moles x
1000) / concentration (mol dm<sup>-3</sup>)
Ideal gas equation PV = nRT<br>P = gas pressure, Pa<br>V = gas volume,
m<sup>3</sup>&nbsp; (cm<sup>3</sup>&nbsp;x 10<sup>-6</sup>) (dm<sup>3</sup>&nbsp;x
10<sup>-3</sup>)<br>n = number of moles<br>R = gas constant (8.31
JK<sup>-1</sup>mol<sup>-1</sup>)<br>T = temperature, K (273 + temp in celcius)
What is empirical formula Formula that gives simplest ratio of atoms of each
element in the compound
How to find empirical formula 1) write out mass or % of each element<br>2) divide
each mass or % by Ar of element<br>3) divide these by the smallest answer. If
answers near 1/2s, multiply by 2. If near 1/3s, multiply by 3.
What is molecular formula actual number of atoms of each element in a compound
How to find molecular formula 1) find Mr of empirical formula<br>2) divide Mr by
formula mass of empirical formula<br>3) multiply answer by empirical formula
reactions for ionic equation&nbsp; acid + metal = salt + hydrogen<br>acid +
carbonate = salt + water + carbon dioxide<br>acid + base = salt + water<br>acid +
alkali = salt + water
Ionic equation method 1) Separate ions in aqueos ionic compounds<br>2) Leave
solid compounds<br>3) Leave pure metals + covalent compounds<br>4) Cancel spectator
ions<br>5) Balance total charge on each side
Wgat is the limiting reagent The chemical which is used up first in any chemical
reaction. Once it is used up, the reaction stops
What is atom economy Gives us an indication of proportion of mass of reactants
that ends up as desired product + how much is wasted. In industry, synthetic routes
try to use reactions that give high percentage atom economy.
Atom economy equation Mass of desired product / Total mass of reactants X 100
What is percentage yield Compares how much we actually obtain in a practical
method / process compared to how much we could make in theory
Percentage yield equation no. of moles of product obtained / theoretical max
no. moles X 100<br>or<br>mass of product obtained / theoretical max mass X 100
Determination of % purity of a metal carbonate method 1) Work out mass of metal

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