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physical chem > amount of substance > Flashcards

amount of substance Flashcards

1
Q

equation for mass with molar mass

A

molar mass times no.moles

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2
Q

molar mass rearrangements

A

molar mass = moles/ mass
moles = mass/ molar mass

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3
Q

solutions advantages

A

safer than pure substances
-easier to store

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4
Q

solution

A

when one substances dissolved into another

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5
Q

solute

A

substances that dissolves into solvent

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6
Q

concentration equation

A

moles of solute/ dm3 of solvent

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7
Q

1m3 in dm3

A

1m3 = 1000dm3

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8
Q

1dm3 in cm3

A

1dm3 is 1000cm3

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9
Q

moles in terms of concentration

A

conc times volume

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10
Q

mol dm3 to g dm3

A

concentration in mol dm3 times molar mass

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11
Q

equation for conc of 2 solutions

A

C1V1 = C2V2

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12
Q

conc of c2 equation for 2 solutions

A

C2 = c1v1/v2

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13
Q

steps of running titrations

A

add acid in conical flask
-volume
-add indicator
-known conc NaOH to HCl
-stop when solution turns pink

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14
Q

neutralisation reaction equation

A

acid + base = salt + water

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15
Q

steps for doing titration calculations

A

-write balanced equation
-moles reacted for known conc solution
-use mole ratio to find amount of other solution reacted
-use amount with initial volume to calculate unknown conc

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16
Q

back titration calculations steps

A

-write full equations for both reactions
-identify key substance involved in both reactions as reactant
-calculate amount of substance used up in second reaction
-amount of key substance used up in first reaction
-amount of other reactant in first reaction

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17
Q

purity as a percentage equation

A

mass of solute/ total mass of substance times 100

18
Q

assumptions about gases made

A

-made up of identical particles
-particles move randomly
-when particles collide they lose no energy
-particles collide with the container walls their in
-occupy no space
no intermolecular forces between particles

19
Q

ideal gas equation

A

pV = nRT

20
Q

ideal gas equation components

A

p= pressure in Pa
-V = volume in m3
-n= moles
-R= gas constant
T= temp in K

21
Q

1Pa into 1 atmosphere

A

101325= 1 atmosphere

22
Q

actual yield

A

amount of desired product created in a reaction

23
Q

theoretical yield

A

amount of desired product that could be produced if all reactants fully reacted

24
Q

explanations for lower yield

A

-not every single molecule takes part in reaction
-not every single molecule of desired product is collected

25
Q

calculating theoretical yield

A

-balanced equation of reaction
-moles of limiting product
-use that to find out moles of desired product
-use moles of desired product to work out mass
-

26
Q

percentage yield

A

actual yield as percentage of theoretical yield

27
Q

percentage yield equation

A

actual yield/ theoretical yield times 100

28
Q

percentage atom economy equation

A

Mr of desired products/ Mr of all reactants times 100

29
Q

molar mass of salts

A

Mr added up all together including molar in front of solid and water molecules

30
Q

decomposition reaction

A

lattice of solid and water breaking into only solid and water

31
Q

dissacosiation equations

A

aq along with separate reactants as products

32
Q

acid + metal

A

salt plus hydrogen

33
Q

acid + metal oxide

A

salt + water

34
Q

acid + metal carbonate

A

salt + water + carbon dioxide

35
Q

percentage water content formula

A

molar mass of water/ molar mass of solid + molar mass of water
times 100 to give percentage

36
Q

solving for x moles of water

A

molar mass = molar mass of solid plus x(18)
-rearrange and solve for x with given molar mass

37
Q

solving for x when given water percentage

A

percentage = molar mass of water/ molar mass of solid + molar mass of water
times 100

38
Q

finding x using experiment

A

heat liquid until only crystal salts left
-measure weight of salt
-mass at start and end measured
-

39
Q

how to make sure all water removed when finding x using experiment

A

mass will stop decreasing

40
Q

how to find mass of water in hydrated salt

A

total mass of hydrated salt and crucible minus anhydrous salt and crucible mass

41
Q

solving for x using calculations equations

A

-find moles of solid by dividing mass by molar mass
-use moles of anhydrous solid to work out moles of hydrous salt
-molar mass of hydrous = mass/ moles
-molar mass of salt = molar mass of solid/ molar mass of water
x=molar mass of hydrous salt - anhydrous molar mass/ 18

42
Q
A

physical chem (14 decks)

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