Amount of Substance 2.1.3

Question 1

When do we use the term Mole

Answer 1

-When counting atoms we use the term mole (abbreviation mol)

Question 2

Define Mole

Answer 2

A mole is the amount of substance that has the same number of particles as there are carbon atoms in exactly 12.0g of carbon 12

Question 3

Define Avogadro’s Constant

Answer 3

The number of atoms in 12.0g of 12C , 6.022 x1023

Question 4

What is the Avogadro’s Constant number

Answer 4

6.022 x10 to the power of 23

Question 5

What is molar mass

Answer 5

-The mass of one mole of a substance (g)

Question 6

What are the units for Molar Mass

Answer 6

units= g per mole to the power of -1

Question 7

Define Mole

Answer 7

The number of particles in exactly 12.0g of 12C

Question 8

What does n represent

Answer 8

We use the letter n (lower case) to represent :

the amount of substance in moles. i.e. number of moles and the units ‘mol’

Question 9

Formulae with Moles:
For mass:

Answer 9

n = m/M
moles= mass/ molar mass
n= moles
m= mass
M= Relative Formula mass or Mr

Question 10

Formulae with Moles:
For volume of gases:

Answer 10

n = v/Vm
moles=volume/ molar volume
n= moles
v= Volume
Vm= 24000cm3 or 24dm3

Question 11

Formulae with Moles:
For concentration of solutions:

Answer 11

n = cV
moles= Concentration x volume
n= moles
c= concentration
V= volume

Question 12

Why is the molar volume always 24000cm3 or 24 dm3

Answer 12

Because gas has weak/no inter-molecular forces between particles and particles are far apart moving in all directions, the same number of particles of any gas will take up the same volume (space) at the same temperature and pressure

Question 13

Volume of gas is affected by…

Answer 13

Temperature and pressure

Question 14

What is the molar volume for gas at room temperature and pressure in dm3

Answer 14

Molar volume for gas at room temp. and pressure is 24dm3 mol-1

Question 15

What do you do to get from cm3 into dm3

Answer 15

-cm3 —-> dm3 you divide by 1000

Question 16

What is the unit for Concentration

Answer 16

Concentration of moles units: mol dm-3

Question 17

What formula do you use to find the number of particles of a substance

Answer 17

 Number of particles = n x NA

n= moles
Na= Avogadro’s Constant

Question 18

What is a hygroscopic substance

Answer 18

They absorb water from the atmosphere e.g. NaCl known as Sodium Chloride

Question 19

Give an example of a hygroscopic substances

Answer 19

NaCl

Question 20

What is an anhydrous compound

Answer 20

-An anhydrous compound is one which has no water molecules inside of it

Question 21

How can you make a substance anhydrous
Give 2 examples

Answer 21

-To make an anhydrous substance you can either :
-heat it to evaporate off the water
- you can dry it in a drying cabinet

Question 22

If a compound is hydrated, what are the two ways to write its formula

Answer 22

1) Empirical formula
OR
2) Dot formula

Question 23

What is an empirical formula

Answer 23

A formula that shows the simplest whole number ratio of atoms of each element present in a compound

Question 24

Give an example of how we can use dot formula

Answer 24

Copper Sulfate Pentahydrate becomes: CuSO4 * 5H2O

Question 25

Define Water of Crystallisation

Answer 25

Water molecules that are bonded into a crystalline structure of a compound

Question 26

Some substances can trap water molecules inside of them, The water contained is called the...

Answer 26

The water of crystallisation

Question 27

The Empirical formula is the simplest ratio of the...

Answer 27

different atoms in a substance .

Question 28

What is the Empirical formula for Ethane

Answer 28

Ethane = CH3.

Question 29

How can you work out the Empirical formula of a compound

Answer 29

You can work out the empirical formula of a compound if you know the mass of each element in it.

Question 30

Define Molecular Formula

Answer 30

A formula that shows the number and type of atoms of each element present in a molecule

Question 31

The molecular formula gives the ...

Answer 31

...actual number of atoms of each element in one molecule of the compound E.g. Ethane has a molecular formula C2H6 but its empirical formula is CH3

Question 32

One Mole of C atoms have a mass of...

Answer 32

12.0 g

Question 33

One Mole of H atoms have a mass of ...

Answer 33

1.0 g

Question 34

One Mole of Mg atoms have a mass of ...

Answer 34

24.3 g

Question 35

One Mole of Fe atoms have a mass of ...

Answer 35

55.8 g

Question 36

What do the following mean? - 1 mol of H -1 mol of H2

Answer 36

1 mol of H= 1 mol of hydrogen atoms 1 mol of H2= 1 mol of hydrogen molecules

Question 37

Calculate the amount of substance in moles, in 96.0 g of carbon, C.

Answer 37

n=m/M 96.0/12.0 =8.0 mol

Question 38

Calculate the mass in g, of 0.050 mol of NO2

Answer 38

n=m/M m=n x M =0.050 x 46.0 = 2.3 g

Question 39

Calculate the molar mass when 2.65 g contains 0.025 mol of a substance

Answer 39

n= m/M M= m/n =2.65/0.025 = 106.0 g mol-1

Question 40

Do Summary questions on page 21 of Kerboodle Textbook Year 1 AS chemistry

Answer 40

Answers are under the section that say 3.1 https://www.kerboodle.com/api/courses/14989/interactives/102929.html

Question 41

What is relative formula mass

Answer 41

The weighted mean mass of the formula unit of a compound compared with 1/12th of the mass of an atom of C-12

Question 42

In an experiment 1.203 g of calcium combines with 2.13 g of chlorine to form a compound. Work out the empirical formula of the compound

Answer 42

1) Convert mass into moles of atoms using n=m/M 2) Find the smallest whole number ratio by dividing by the smallest whole number ratio 3) Write the empirical formula 1) Moles of Ca= 1.203/40.1 = 0.030 mol Moles of Cl= 2.13/35.5 = 0.060 mol 2) n(Ca) : n(Cl) 0.030/0.030 = 1 0.060/0.030 = 2 Ratio of n(Ca) : n(Cl) 1 : 2 3) CaCl2

Question 43

Do Worked Example: Determination of a molecular formula on page 23 in kerboodle year 1 AS chemistry text book

Answer 43

It a worked example it tells you what to do and how it got the answer- make a flashcard after working through and understanding example

Question 44

When blue crystals of hydrated copper (II) sulphate are heated what happens to the bonds and what happens to the Copper (II) sulphate as a result

Answer 44

The bonds holding the water within the crystal are broken and the water is driven off, leaving behind white anhydrous copper (II) sulphate.

Question 45

Why is it difficult to heat and remove the last traces of water from the anhydrous copper (II) sulphate

Answer 45

Because we don't want to break the compound down only the water within the crystalline structure

Question 46

How could you carry out an experiment to determine the water of crystallization in hydrated crystals

Answer 46

This experiment uses hydrated copper (II) sulphate but you can use it for any hydrated salt Step 1: Weigh an empty and dry crucible Step 2: Add the hydrated salt into the weighed crucible. Weigh the crucible and the hydrated salt, make sure to write down these results Step 3: Using a pipe-clay triangle, support the crucible containing the hydrated salt on a tripod. Heat the crucible and its contents gently for about 1 minute. Then heat it strongly for a further 3 minutes, lifting the lid every minute usin tongs to let some of that water evaporate and see the color change of the salt from when it was hydrous to anhydrous Step 4: Leave the crucible to cool. Then weigh the crucible and anhydrous salt

Question 47

Use the results below to Calculate the water present in the crystalline structure of the Copper (II) Sulphate Mass of crucible/ g ------->18.742 Mass of crucible + hydrated salt / g------> 28.726 Mass of crucible+ anhydrous salt / g------> 25.126

Answer 47

Step1: Calculate the amount in mol of anhydrous copper (ii) sulphate Mass of CuSO4 formed= 25.126-18.742= 6.384 g Moles= mass/Mr moles= 6.384/159.6 Anhydrous CuSO4 = 0.0400 mol Step 2: Calculate the mass and amount in water mass of H2O formed: = 28.726-25.126= 3.600 g moles (H2O)= 3.600/18.0 = 0.200 mol Step 3: Find the smallest whole number ratio n(CuSO4) : n(H20) = 0.0400 : 0.200 1 : 5 Step 4: Write down the value of x and the formula for hydrated copper sulphate x= 5 so the formula is CuSO4. 5H20

Question 48

Do example of water crystallization on bottom right hand side of page 24 in year 1 AS Chemistry Kerboodle textbook

Question 48

What is the first assumption you have to be make in the experiment to determine the water of crystallization of hydrated salts and how can we overcome this problem

Answer 48

Assumption1= All of the water has been lost ( you only see the surface of the crystals therefore we don't know if water was left inside, it is not easy especially if the hydrated and anhydrous forms are similar colors) Solution 1= To heat to constant mass/ concordant results, when crystals are reheated until no change in mass it suggests all water has been removed

Question 49

What is the second assumption you have to be make in the experiment to determine the water of crystallization of hydrated salts

Answer 49

Assumption 2= No further decomposition, ( many salts decompose further when heated ) This can be very difficult to judge if there is no color change e.g. if heated very strongly Copper(II) Sulphate decomposes to form black Copper(II) Oxide.

Question 50

Do summary questions on page 25 of Kerboodle AS Chemistry textbook

Answer 50

Page 273 under the section 3.2

Question 51

1 dm3= how many cm3

Answer 51

1 dm3 = 1000cm3

Question 52

1000cm3= how many ml

Answer 52

1000cm3= 1000ml

Question 53

What is the simplest ratio of C6H6---->

Answer 53

C6H6 ----> CH

Question 54

What is the simplest ratio of C2H4

Answer 54

C2H4 ---->CH2

Question 55

What is the simplest ratio of C12H26

Answer 55

C12H26 ---->C6H13

Question 56

What is the simplest ratio of C2H5OH

Answer 56

C2H6O

Question 57

What is the simplest ratio of A compound that contains 3.2g of S and 3.2g of O

Answer 57

SO2

Question 58

What is the empirical formula of a compound that contains 27.3% carbon and 72.7% oxygen by mass?

Answer 58

Step 1 :Find, and write down the relative atomic masses of the elements involved. Ar (C) = 12 and Ar (O) = 16 Step 2 : Assume that you have 100g of the compound (this makes the maths easy) Step 3 Work out how many moles of each element this must be using Moles= mass/Mr Moles = 27.3/12 Moles in Carbon = 2.275 Moles=72.7/16 Moles in Oxygen= 4.55 Step 4 Divide both numbers of moles by the smallest number (2.275 in this case). Moles in Carbon = 2.275/2.275 = 1 Moles in Oxygen = 4.55/2.275 = 2 In the empirical formula, there are 1 carbon atom and 2 oxygen atoms, so it is CO2. iF YOU DONT UNDERSTAND LOOK ON SECOND PAGE OF CHEM BOOK IN FIRST PAGE

Question 59

How can you work out the answer to this question What is the empirical formula of a compound that contains 27.3% carbon and 72.7% oxygen by mass?

Answer 59

Use the table in the chemistry book= IT'S REALLLYYYY HELPFUL YOU NEED TO MEMORISE THAT WAY OF DOING IT

Question 60

A compound has the empirical formula CH2 and we are told that it’s molecular mass is 28. Find out its molecular formula.

Answer 60

CH2 = 14 Molecular Formula = 28 ÷ 14 = 2 = C2H4

Question 61

What is the Ideal Gas Law Equation and units for each letter

Answer 61

pV=nRT p is pressure and must be in Pa V is volume and must be in m3 n is the number of moles R is the gas constant = 8.314 J K-1 mol-1 T is temperature and must be in Kelvin 0oC = 273 K

Question 62

In the Ideal Gas Law Equation what does p stand for and its units

Answer 62

p is pressure and must be in Pa

Question 63

In the Ideal Gas Law Equation what does V stand for and its units

Answer 63

V is volume and must be in m3

Question 64

In the Ideal Gas Law Equation what does n stand for

Answer 64

n is the number of moles

Question 65

In the Ideal Gas Law Equation what does R stand for and its units

Answer 65

R is the gas constant = 8.314 J K-1 mol-1

Question 66

In the Ideal Gas Law Equation what does T stand for and its units

Answer 66

T is temperature and must be in Kelvin 0oC = 273 K

Question 67

Rearrange the Ideal Gas Law Equation to find Pressure

Answer 67

P= nRT/ V

Question 68

Rearrange the Ideal Gas Law Equation to find Ideal Gas Constant

Answer 68

R= pV/ Tn

Question 69

Rearrange the Ideal Gas Law Equation to find Volume

Answer 69

V = nRT/ P

Question 70

Rearrange the Ideal Gas Law Equation to find Temperature

Answer 70

T= pV/nR

Question 71

Rearrange the Ideal Gas Law Equation to find Moles

Answer 71

n= pV/ RT

Question 72

Find the volume of 2 moles of gas at 30*C, and a pressure of 100 000 Pa

Answer 72

Answer= 0.0504m3 Working out: V= nRT/p V= (2 x 8.31 x 303)/100000 V = 0.05036 m3

Question 73

Find the temperature of 2 moles of 20cm3 of a gas at a pressure of 10kPa.

Answer 73

Answer= 0.012 K Working out: pV=nRT T=pV/ nR T= 10000 x 2x10 to the power of -5 / 2 x 8.31 0.012 K

Question 74

Find the pressure of 0.5 moles of gas if the volume is 11000cm3 and the temperature is 25oC

Answer 74

Answer= 112562.7 pa Working out: P=nRT/V P = (0.5 x 8.31 x 298)/0.011 P= 112562.7 pa

Question 75

Find the number of moles of H2 in 48 000cm3, at a pressure of 100 kPa and a temperature of 25oC

Answer 75

Answer= 1.94 moles Working out: n= pV/RT n= (100000 x 0.048) / (8.31 x 298) n = 1.94

Question 76

Calculate the volume occupied by one mole of gas at 25oC and 100kPa.

Answer 76

0.024m3

Question 77

Calculate the mass of a sample of Carbon Dioxide which occupies 20dm3 at 27oC and 100kPa.

Answer 77

35.3g

Question 78

Calculate the relative molecular mass of a gas if a 500cm3 sample at 20oC and 1 pa has a mass of 0.66g.

Answer 78

31.7

Question 79

At 25oC and 100kPa a gas occupies a volume of 20dm3. Calculate the new temperature of the gas if: a) The volume is decreased to 10dm3 at constant pressure. b) The pressure decreased to 50kPa at a constant volume.

Answer 79

a= 149K b= 149K

Question 80

At room temp. and pressure, how many moles in: 320dm3 of oxygen gas ?

Answer 80

n=v/Vm n = 320/24 = 13.3 mol

Question 81

At room temp. and pressure, how many moles in: 48 000cm3 of CO2(g)

Answer 81

n=v/Vm V = 48000/24000 = 2.00 mol

Question 82

At room temp. and pressure, how many moles in: 80cm3 of chlorine gas?

Answer 82

n=v/Vm V = 80/24000 = 3.33 x10-3 mol

Question 83

Sodium oxide was neutralized using hydrochloric acid. Na2O + 2HCl------> 2NaCl + H2O What mass of sodium chloride is produced from 3.46g of sodium oxide?

Answer 83

Answer= 6.53g Working out: 3.46g Na2O This is a mass so use n=m/M M(Na2O) = (23.0x2) +16.0 = 62g n = 3.46/62 n = 0.0558 mol Na2O 1Na2O : 2NaCl 0.0558 : (2x 0.0058) =0.1116 mol NaCl This is a mass so use n=m/M (m= n x M) M(NaCl) = (23.0 + 35.5) = 58.5g m = 0.1116 x 58.5 m = 6.53g

Question 84

2Cu + O2 2CuO What mass of copper oxide is produced from 2.94g of copper metal?

Answer 84

3.68g if not understood than look at flashcard number 84 and see how others are worked out

Question 85

2Fe + 3Cl2----> 2FeCl3 What mass of iron (III) chloride is produced from 1kg of Fe?

Answer 85

2908.6g if not understood than look at flashcard number 84 and see how others are worked out

Question 86

ZnCO3----> ZnO + CO2 What mass of zinc carbonate is needed to produce 10g of zinc oxide?

Answer 86

15.4g if not understood than look at flashcard number 84 and see how others are worked out

Question 87

If only 80% of the ZnCO3 decomposes calculate what mass of zinc carbonate is needed to produce 100g of ZnO.

Answer 87

192.6g if not understood than look at flashcard number 84 and see how others are worked out

Question 88

sodium carbonate was neutralized using hydrochloric acid. Na2CO3 + 2HCl ---->2NaCl + H2O +CO2 What volume of carbon dioxide is produced from 12.2g of sodium carbonate?

Answer 88

STEP 1 – triangle 12.2g Na2CO3 (Mass so use n=m/M) M(Na2CO3) = (23x2) +12.0 + (3x16) = 106g n = 12.2/106 = n = 0.115 mol Na2O STEP 2 – ratio 1Na2CO3 : 1CO2 0.115 : (1x 0.115) =0.115 mol CO2 STEP 3 – triangle (Gas Vol so use V= n x Vm) Vm = 24dm3 (24000cm3) V = 0.115 x 24 V = 2.76dm3 (2760cm3)

Question 89

2Na + 2HCl ----> 2NaCl + H2 What Volume of hydrogen gas is produced from 2.43g of sodium metal?

Answer 89

1.268 dm3 or 1268 cm3 If not understood look at flashcard 89 for worked example

Question 90

2Fe + 3Cl2 -----> 2FeCl3 What volume of chlorine gas is required to react with 45.2g of Fe?

Answer 90

29.16 dm3 If not understood look at flashcard 89 for worked example

Question 91

ZnCO3 -----> ZnO + CO2 What mass of zinc carbonate is needed to produce 200dm3 of carbon dioxide gas?

Answer 91

1045g If not understood look at flashcard 89 for worked example

Question 92

2Mg(NO3)2----> 2MgO + 4NO2(g) + O2(g) What is the total volume of gas produced from 10.1g Mg(NO3)2

Answer 92

4.1dm3 If not understood look at flashcard 89 for worked example

Question 93

Sodium hydroxide was neutralized using hydrochloric acid. NaOH + HCl----> NaCl + H2O What is the concentration of the HCl if 20.2cm3 of HCl is needed to neutralise 25cm3 of 0.1 moldm-3 NaOH solution?

Answer 93

STEP 1 – triangle 25cm3 of 0.1 moldm-3 NaOH solution (Vol and concentration so use n = c x V) n = 0.1 x 25/1000 n = 0.0025 mol NaOH STEP 2 –Ratio 1NaOH : 1HCl 0.0025 : (1x 0.0025) =0.0025 mol HCl STEP 3 – triangle (concentration so use c = n/V) V= 20.2 cm3 = V = 0.0202 dm3 c = 0.0025/ 0.0202 c = 0.124 mol dm-3

Question 94

2Na2CO3 + 2HCl ---> 2NaCl + H2O + CO2 30.1cm3 of 0.25 moldm-3 Na2CO3 solution was needed to neutralise 25cm3 HCl solution. Calculate the concentration of the HCl?

Answer 94

Look at flashcard 94 and follow working out

Question 95

NaOH + HNO3---->NaNO3 + H2O What volume of 0.08 moldm-3 HNO3 is required to completely neutralise 15cm3 of 0.02 moldm-3 NaOH solution?

Answer 95

Look at flashcard 94 and follow working out

Question 96

2HCl + Ca(OH)2 ----> CaCl2 + 2H2 O What volume of 0.2 moldm-3 HCl is required to completely neutralise 25cm3 of 0.1 moldm-3 Ca(OH)2 solution?

Answer 96

Look at flashcard 94 and follow working out

Question 97

What is the limiting reagent

Answer 97

The reactant that is not in excess which will be used up first and stop the reaction

Question 98

6.54g of zinc was added to 25cm3 of a 1.00moldm-3 solution of copper (II) sulfate – what is the limiting reagent?

Answer 98

Write out the balanced equation Look at the ratio of moles of each reagent Work out the number of moles of each reagent Find the one which is the limiting reagent and base your calculation on this. Zn + CuSO4 -> ZnSO4 + Cu 1 mol reacts with 1 mol n of Zn = 6.54/65.4 = 0.100 n of CuSO4 =1 x 0.025 = 0.025 The limiting reagent is therefore the copper (II) sulfate

Question 99

25.0cm 3 0.0100 moldm-3 solution of sodium hydroxide is reacted with 25.0cm3 of a 0.0120 moldm-3 solution of sulfuric acid. Which reagent is in excess?

Answer 99

Write out the balanced equation Look at the ratio of moles of each reagent Work out the number of moles of each reagent Find the one which is the limiting reagent and base your calculation on this. Equation: 2NaOH + H2SO4---> Na2SO4 + 2H2O Ratio: 2 : 1 n of NaOH =0.0100 x 0.025 = 2.50 x 10 -4 - requires ½ x 2.5 x 10 -4 = 1.25 x 10 -4 moles of alkali n of H2SO4 = 0.0120 x 0.025 = 3.00 x 10 -4 The acid is in excess.

Question 100

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Answer 100

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Question 101

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Answer 101

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Question 102

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Answer 102

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Question 103

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Answer 103

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Question 104

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Answer 104

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Question 105

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Answer 105

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Question 106

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Answer 106

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Question 107

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Answer 107

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Question 108

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Answer 108

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Question 109

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Answer 109

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Question 110

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Answer 110

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Question 111

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Answer 111

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Question 112

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Answer 112

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Question 113

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Answer 113

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Question 114

What are the properties of Molecules in an ideal gas

Answer 114

Molecules in an ideal gas: -show random motion -have negligible size -have no intermolecular forces- the molecules do not attract each other so they bounce off each other when they collide

Question 115

Why does pressure arise between gas molecules

Answer 115

Pressure arises because of collisions between the gas molecules and the sides of the container they are in

Question 116

The pressure of gas molecules depends on what

Answer 116

- the number of particles in the container - the size of the container -the absolute temperature

Question 117

How do you go from cm3 to m3

Answer 117

x 10 to the power of -6

Question 118

How do you go from dm3 to m3

Answer 118

x 10 to the power of -3

Question 119

How do you go from *C to K

Answer 119

+273

Question 120

How do you go from kPa to Pa

Answer 120

x 10 to the power of 3

Question 121

What is percentage Yield

Answer 121

A way of comparing the actual yield (amount of product made) with the predicted yield (amount of predicted product to be made)

Question 122

How can you find percentage Yield of a substance

Answer 122

% Yield= Actual Yield (g) / x 100 Predicted Yield (g)

Question 123

What does 100% Yield mean

Answer 123

That there is no loss of product

Question 124

What does 0 % Yield mean

Answer 124

That there is no product made

Question 125

The Percentage Yield will always be less than 100%, but why

Answer 125

Products can be lost in -filtration -evaporation -transferring liquids -heating

Question 126

Why is a high yield important

Answer 126

It's important for: -reducing waste -using less raw material - using less energy -increasing profit

Question 127

How can we calculate Atom Economy

Answer 127

Atom Economy (%)= Sum of Molar mass of desired products / Sum of Molar masses of all products multiplied by 100

Question 128

What is Actual Yield

Answer 128

The amount of product obtained from a reaction

Question 129

Do Task 1- Atom Economy and Percentage Yield AND Task 1 Percentage Yield First Link= Questions on document Second Link= Answers on document

Answer 129

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Question 130

Do Task 2- Calculate the % atom economy of ethylene oxide First Link= Questions on document Second Link= Answers on document

Answer 130

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Question 131

Do Task 3- A little Trickier First Link= Questions on document Second Link= Answers on document

Answer 131

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Question 132

This question is about expressing numbers in standard form. Express the following number in standard form: 1) 0.0023 2) 1032 3) 2, 750, 000 4) 0.000528

Answer 132

1) 2.3 x 10-³ 2) 1.032 x 10 ³ 3) 2.75 x 10⁶ 4) 5.28 x 10-⁴

Question 133

Work out the following numbers in ordinary form: a) 2.01 x 10³ b) 5.2 x 10-² c) 8.41x 10² d) 1.00 x 10-⁴

Answer 133

a) 2010 b) 0.052 c) 841 d) 0.0001

Question 134

Do Question 1c on page 4 of Amount of Substance Booklet

Answer 134

Answers are next to qs

Question 135

Do Question 2 on Page 4 of Amount of Substance Booklet

Answer 135

Answers present by Qs

Question 136

Do Question 3 on Page 4 of Amount of Substance Booklet

Answer 136

Answers present by Qs

Question 137

Do the Re arranging Equations on page 5

Answer 137

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Question 138

Write the formula for Magnesium Hydroxide

Answer 138

Mg²+ OH- Mg(OH)2

Question 139

Write the formula for Aluminium Nitrate

Answer 139

Al³+ NO3 ²- Al2(NO3)3

Question 140

Do Task 1 on bottom of page 6 Writing the formulae in Amount of Substance Booklet

Answer 140

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Question 141

Do Task 1 on bottom of page 6 and top of 7 Relative Formula Mass in Amount of Substance Booklet

Answer 141

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Question 142

Do How many moles task on top of page 7

Answer 142

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Question 143

How many molecules are there in 42.5g of Ammonia, NH3

Answer 143

Relative Atomic Mass= (1 x N)+(3 x H) =(1 x 14.0)+ (3 x 1.0) = 17.0 g mol-1 Note: Give all molar masses to 1d.p. Mass= Mr x Moles Moles= mass/ Mr = 42.5g/ 17.0g mol-1 =2.5 Number of particles= n x Avogadros number 2.5 x 6.02 x 10²³ = 1.505 x 10²⁴ = 1.51 x 10²⁴

Question 144

Do Task 2 Question 1 page 8 of Amount of substance booklet

Answer 144

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Question 145

Do Task 2 Question 2 page 8 of Amount of substance booklet

Answer 145

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Question 146

Do Task 2 Question 3 page 8 of Amount of substance booklet

Answer 146

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Question 147

Do Task 2 Question 4a, and 4b on page 8 of Amount of substance booklet

Answer 147

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Question 148

Do Task 2 Question 4 on bottom of page 8 of Amount of substance booklet

Answer 148

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Question 149

Do all of page 9 in amount of substance booklet

Answer 149

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Question 150

Do Box 1 on page 10 of Amount of substance booklet

Answer 150

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Question 151

Do Box 2 on page 10 of Amount of substance booklet

Answer 151

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Question 152

Do Box 3 on page 10 of Amount of substance booklet

Answer 152

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Question 153

Do Box 4 on page 10 of Amount of substance booklet

Answer 153

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Question 154

Do Box 5 on page 10 of Amount of substance booklet

Answer 154

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Question 155

Do Box 6 on page 10 of Amount of substance booklet

Answer 155

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Question 156

Do Box 7 on page 10 of Amount of substance booklet

Answer 156

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Question 157

Do Task 4 on page 11 in Amount of Substance Booklet

Answer 157

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Question 158

Do Task 4 on Page 13 in Amount of Substance Booklet

Answer 158

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Question 159

What 2 assumptions does the ideal gas equation rely on

Answer 159

- forces between molecules are negligible -gas molecules have negligible size compared to the size of their container

Question 160

Do Task 5 on page 14 of Amount of Substance Booklet

Answer 160

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Question 161

Do Task 6 on page 15 in Amount of Substance Booklet

Answer 161

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Question 162

Do Page 16 in Amount of Substance booklet

Answer 162

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Question 163

Write a method for determining the formula of a hydrated salt

Answer 163

Look on Kerboodle or Physics and maths tutor

Question 164

Do All the questions on page 18 and 19 of classwork booklet

Answer 164

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Question 165

Do pages 20 to 23 in Amount of Substance Booklet

Answer 165

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Question 166

Do Task 14 in Amount of Substance Booklet

Answer 166

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Question 167

Do Page 25 and 26 in Amount of Substance Booklet

Answer 167

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Question 168

Do Task 16 in Amount of Substance Booklet

Answer 168

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Question 169

Do Pages 28 and 29 in Amount of Substance Booklet

Answer 169

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Question 170

Do page 30 in Amount of Substance Booklet

Answer 170

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Question 171

In practical work, you will use glassware graduated in ml and l, how should you record these readings

Answer 171

Cm3 and dm3

Question 172

What is a solute

Answer 172

The dissolved compound

Question 173

A 1 mol dm-3 solution contains how many moles of solute dissolved in how many dm3 of solution

Answer 173

A 1moldm-3 solution contains 1 mol of solute dissolved in each 1dm3 of solution

Question 174

In the equation n =c x V The volume is measured in what

Answer 174

dm3

Question 175

In a practical you usually use cm3 graduations on equipment, how would you convert between solution volumes using the formula n= c x V

Answer 175

n= c x (V(cm3) divided by 1000)

Question 176

Calculate the amount of NaCl, in mol in 30.0cm³ of a 2.00mol dm-³ solution

Answer 176

n (NaCl)= c x V(cm³)/ 1000 = 2.00 x (30.0/1000) =0.0600mol

Question 177

Calculate the volume of a 0.160 mol dm-³ solution that contains 3.25 x 10-³ mol of NaCl

Answer 177

n = c x V(cm³)/ 1000 So, V= 1000 x n/ c = 1000 x 3.25 x 10-³ divided by 0.160 =20.3 cm3

Question 178

What is a standard solution

Answer 178

A solution of known concentration

Question 179

How are standard solutions prepared

Answer 179

-By dissolving an exact mass of the solute in a solvent and making up the solution to an exact volume

Question 180

How do you convert from concentration to mass concentration. Use Na2CO3 as an example. For the solution of Na2CO3, the concentration is 0.250moldm-3.

Answer 180

To work out mass concentration you need to convert between moles and grams mol dm-3 -----------> gdm-3 n= m / M so m = n x M = 0.250 x 106.0 = 26.50 g Mass concentration of Na2CO3= 26.5 gdm-3

Question 180

Calculate the mass of Na2CO3 required to prepare 100 cm³ of a 0.250 mol dm-³ standard solution

Answer 180

Step 1: Work out the amount in moles required n(Na2CO3)= c x V(cm³)/ 1000 =0.250 x 100 / 1000 = 0.0250 mol Step 2: Work out molar mass of Na2CO3 M(Na2CO3) = 23.0 x 2+12.0+16.0 x 3= 106.0 g mol-1 Step 3:Rearrange n =m / M to calculate the mass of Na2CO3 required m= n x M = 0.0250 x 106.0 = 2.65g

Question 180

You can prepare a standard solution by dissolving an exact mass of the solute in a solvent and making up the solution to an exact volume. But how do you know the mass required to prepare a standard solution

Answer 180

You can work out the mass required to prepare a standard solution using your understanding of moles

Question 181

What is the Molar Gas Volume Vm

Answer 181

The molar gas volume Vm is the volume per mole of gas molecules at a stated temperature and pressure

Question 182

At RTP, 1 mole of gas molecules has a volume of approximately what

Answer 182

24.0dm3 = 240000 cm3

Question 183

What is the molar gas volume at RTP

Answer 183

24.0 dm3 mol-1

Question 184

Calculate the amount (mol) of hydrogen, H2 (g), in 480 cm3 at RTP

Answer 184

n= V (in cm3) divided by 24000 = 480 / 24000 = 0.0200 mol of H2

Question 185

Calculate the volume in dm3 of 0.150mol of O2 (g) at RTP

Answer 185

V(dm3)= n x 24.0 so V = 0.150 x 24.0 = 3.60dm3

Question 186

Assumptions for molecules in an ideal gas:

Answer 186

-random motion -elastic collision -negligible size - no intermolecular forces

Question 187

In the ideal gas law temp is always in K, what does k stand for

Answer 187

Kelvin

Question 188

In ideal gas law when Kelvin is at 0* what temp is this in Celcius

Answer 188

-273*c

Question 189

Go Page 30 on kerboodle textbook and make flashcards on method of finding relative molecular mass

Question 190

Do Kerboodle Summary questions on page 31

Answer 190

Answers are on page 273 under 3.3

Question 191

What do chemists use balanced equations to find

Answer 191

-the quantities of reactants required to prepare a required quantity of a product - the quantity of products that should be formed from certain quantities of reactants

Question 192

Calculate the mass of Aluminium oxide Al2O3 formed when 8.10g of Aluminium completely reacts with oxygen

Answer 192

Step1 : Calculate amount in mol of Al that reacts n(Al)= m / M = 8.10/ 27.0 = 0.300 mol Step 2: Write balanced equation and work out Ratio 4Al(s)+ 3 O2(g)-----> 2Al2O3 (s) So, 4 moles of Aluminium react with 3 moles of Oxygen to form 2 moles of Aluminium oxide We need to Look at ratio between Al and Al2O3 n= Al= 0.300 n= Al2O3= 0.150 mol Because theres twice as much Al as there is Al203 Step3: Calculate mass of Al2O3 formed m(Al2O3)= m/ M m= n x M =0.150 x (27.0 x2+ 16.0 x3) = 0.150 x 102.0 = 15.3 g

Question 193

What is the practical for identifying an unknown metal e.g. Group 2 metal and how the results can be analyzed using the set method for reacting quantities

Answer 193

1. Set up the apparatus as shown in figure 1 (look at amount of substance booklet page 30 or Kerboodle textbook year 1 as chemistry page 33) 2. Weigh a sample of the metal and add to the flask 3. Using a measuring cylinder add 25.0 cm3 1.0moldm-3 HCl (aq) (an excess) to the flask and quickly replace the bung 4. Measure the maximum volume of gas in the syringe Make sure to learn this practical**

Question 194

What is the theoretical yield

Answer 194

The maximum possible amount of product that could form

Question 195

Why is it so difficult to achieve the theoretical yield

Answer 195

-the reaction may not have gone to completion -other reactants (side reactants) may have taken place alongside the main reception -purification of the product may result in loss of some product

Question 196

The actual yield obtained from a reaction is usually...

Answer 196

... lower than the theoretical yield

Question 197

How do you find the percentage yield

Answer 197

Percentage yield= actual yield/ theoretical yield all times by 100

Question 198

What is a limiting reagent

Answer 198

The reactant that is not in excess, that is completely used up first and stops the reaction

Question 199

What is atom economy

Answer 199

The atom economy of a chemical reaction is a measure of how well atoms have been utilized

Question 200

Reactions with high atom economy : (2marks)

Answer 200

- produce a large proportion of desired products and few unwanted waste products -are important for sustainability as they make the best use of natural resources

Question 201

What happens if we start to improve atom economy

Answer 201

Improving atom economy: -makes industrial processes more efficient -preserves raw materials - reduces waste