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A Level Chemistry > Amount of Substance 2.1.3 > Flashcards

Amount of Substance 2.1.3 Flashcards

1
Q

When do we use the term Mole

A

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

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2
Q

Define Mole

A

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

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3
Q

Define Avogadro’s Constant

A

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

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4
Q

What is the Avogadro’s Constant number

A

6.022 x10 to the power of 23

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5
Q

What is molar mass

A

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

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6
Q

What are the units for Molar Mass

A

units= g per mole to the power of -1

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7
Q

Define Mole

A

The number of particles in exactly 12.0g of 12C

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8
Q

What does n represent

A

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

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

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9
Q

Formulae with Moles:
For mass:

A

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

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10
Q

Formulae with Moles:
For volume of gases:

A

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

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11
Q

Formulae with Moles:
For concentration of solutions:

A

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

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12
Q

Why is the molar volume always 24000cm3 or 24 dm3

A

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

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13
Q

Volume of gas is affected by…

A

Temperature and pressure

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14
Q

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

A

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

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15
Q

What do you do to get from cm3 into dm3

A

-cm3 —-> dm3 you divide by 1000

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16
Q

What is the unit for Concentration

A

Concentration of moles units: mol dm-3

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17
Q

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

A

 Number of particles = n x NA

n= moles
Na= Avogadro’s Constant

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18
Q

What is a hygroscopic substance

A

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

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19
Q

Give an example of a hygroscopic substances

A

NaCl

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20
Q

What is an anhydrous compound

A

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

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21
Q

How can you make a substance anhydrous
Give 2 examples

A

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

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22
Q

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

A

1) Empirical formula
OR
2) Dot formula

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23
Q

What is an empirical formula

A

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

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24
Q

Give an example of how we can use dot formula

A

Copper Sulfate Pentahydrate becomes: CuSO4 * 5H2O

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25
Q

Define Water of Crystallisation

A

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

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26
Q

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

A

The water of crystallisation

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27
Q

The Empirical formula is the simplest ratio of the…

A

different atoms in a substance .

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28
Q

What is the Empirical formula for Ethane

A

Ethane = CH3.

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29
Q

How can you work out the Empirical formula of a compound

A

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

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30
Q

Define Molecular Formula

A

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

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31
Q

The molecular formula gives the …

A

…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

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32
Q

One Mole of C atoms have a mass of…

A

12.0 g

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33
Q

One Mole of H atoms have a mass of …

A

1.0 g

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34
Q

One Mole of Mg atoms have a mass of …

A

24.3 g

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35
Q

One Mole of Fe atoms have a mass of …

A

55.8 g

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36
Q

What do the following mean?

  • 1 mol of H
    -1 mol of H2
A

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

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37
Q

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

A

n=m/M
96.0/12.0
=8.0 mol

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38
Q

Calculate the mass in g, of 0.050 mol of NO2

A

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

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39
Q

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

A

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

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40
Q

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

A

Answers are under the section that say 3.1

https://www.kerboodle.com/api/courses/14989/interactives/102929.html

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41
Q

What is relative formula mass

A

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

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42
Q

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

A

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

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43
Q

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

A

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

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44
Q

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

A

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

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45
Q

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

A

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

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46
Q

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

A

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

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47
Q

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

A

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

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48
Q

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

A
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48
Q

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

A

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

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49
Q

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

A

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.

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50
Q

Do summary questions on page 25 of Kerboodle AS Chemistry textbook

A

Page 273 under the section 3.2

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51
Q

1 dm3= how many cm3

A

1 dm3 = 1000cm3

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52
Q

1000cm3= how many ml

A

1000cm3= 1000ml

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53
Q

What is the simplest ratio of
C6H6—->

A

C6H6 —-> CH

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54
Q

What is the simplest ratio of
C2H4

A

C2H4 —->CH2

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55
Q

What is the simplest ratio of
C12H26

A

C12H26 —->C6H13

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56
Q

What is the simplest ratio of
C2H5OH

A

C2H6O

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57
Q

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

A

SO2

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58
Q

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

A

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

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59
Q

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?

A

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

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60
Q

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

A

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

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61
Q

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

A

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

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62
Q

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

A

p is pressure and must be in Pa

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63
Q

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

A

V is volume and must be in m3

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64
Q

In the Ideal Gas Law Equation what does
n stand for

A

n is the number of moles

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65
Q

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

A

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

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66
Q

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

A

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

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67
Q

Rearrange the Ideal Gas Law Equation to find Pressure

A

P= nRT/ V

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68
Q

Rearrange the Ideal Gas Law Equation to find Ideal Gas Constant

A

R= pV/ Tn

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69
Q

Rearrange the Ideal Gas Law Equation to find Volume

A

V = nRT/ P

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70
Q

Rearrange the Ideal Gas Law Equation to find Temperature

A

T= pV/nR

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71
Q

Rearrange the Ideal Gas Law Equation to find Moles

A

n= pV/ RT

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72
Q

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

A

Answer= 0.0504m3

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

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73
Q

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

A

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

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74
Q

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

A

Answer= 112562.7 pa

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

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75
Q

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

A

Answer= 1.94 moles

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

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76
Q

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

A

0.024m3

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77
Q

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

A

35.3g

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78
Q

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

A

31.7

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79
Q

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.

A

a= 149K

b= 149K

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80
Q

At room temp. and pressure, how many moles in:

320dm3 of oxygen gas ?

A

n=v/Vm

n = 320/24 = 13.3 mol

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81
Q

At room temp. and pressure, how many moles in:

48 000cm3 of CO2(g)

A

n=v/Vm

V = 48000/24000
= 2.00 mol

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82
Q

At room temp. and pressure, how many moles in:

80cm3 of chlorine gas?

A

n=v/Vm

V = 80/24000
= 3.33 x10-3 mol

83
Q

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

A

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

84
Q

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

A

3.68g

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

85
Q

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

A

2908.6g

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

86
Q

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

A

15.4g

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

87
Q

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

A

192.6g

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

88
Q

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?

A

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)

89
Q

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

A

1.268 dm3 or 1268 cm3

If not understood look at flashcard 89 for worked example

90
Q

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

A

29.16 dm3

If not understood look at flashcard 89 for worked example

91
Q

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

A

1045g

If not understood look at flashcard 89 for worked example

92
Q

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

A

4.1dm3

If not understood look at flashcard 89 for worked example

93
Q

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?

A

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

94
Q

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?

A

Look at flashcard 94 and follow working out

95
Q

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?

A

Look at flashcard 94 and follow working out

96
Q

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?

A

Look at flashcard 94 and follow working out

97
Q

What is the limiting reagent

A

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

98
Q

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

A

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

99
Q

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?

A

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.

100
Q

Do Task A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

101
Q

Do Task B

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

102
Q

Do Task C

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

103
Q

Do Task D

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

104
Q

Do Task E
https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

105
Q

Do Task F

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

106
Q

Do Task G

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

107
Q

Do Task H

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

108
Q

Do Task I

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

109
Q

Do Task J

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

110
Q

Do Task K

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

111
Q

Do Task L

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

112
Q

Do Task M

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

113
Q

Do Task N

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B51097D20-4727-49B2-AC4A-914FB767A35A%7D&file=AoS%20Revision%20Questions.docx&action=default&mobileredirect=true

A

Answers are in Mikes team= class material= lesson 13

114
Q

What are the properties of Molecules in an ideal gas

A

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

115
Q

Why does pressure arise between gas molecules

A

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

116
Q

The pressure of gas molecules depends on what

A
  • the number of particles in the container
  • the size of the container
    -the absolute temperature
117
Q

How do you go from cm3 to m3

A

x 10 to the power of -6

118
Q

How do you go from dm3 to m3

A

x 10 to the power of -3

119
Q

How do you go from *C to K

A

+273

120
Q

How do you go from kPa to Pa

A

x 10 to the power of 3

121
Q

What is percentage Yield

A

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

122
Q

How can you find percentage Yield of a substance

A

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

123
Q

What does 100% Yield mean

A

That there is no loss of product

124
Q

What does 0 % Yield mean

A

That there is no product made

125
Q

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

A

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

126
Q

Why is a high yield important

A

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

127
Q

How can we calculate Atom Economy

A

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

multiplied by 100

128
Q

What is Actual Yield

A

The amount of product obtained from a reaction

129
Q

Do Task 1- Atom Economy and Percentage Yield
AND
Task 1 Percentage Yield

First Link= Questions on document
Second Link= Answers on document

A

Questions =

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B955384CE-D3AB-471E-BA0F-7112B67BF9D8%7D&file=Atom%20economy%20and%20percentage%20yield%20questions.docx&action=default&mobileredirect=true

Answers =

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7BF3BFD1B5-E3DE-4682-B00F-61E54D6CDB1E%7D&file=ANSWERS-13%20%25%20yiel%20%2B%20atom%20economy.doc&action=default&mobileredirect=true

130
Q

Do Task 2- Calculate the % atom economy of ethylene oxide

First Link= Questions on document
Second Link= Answers on document

A

Questions =

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B955384CE-D3AB-471E-BA0F-7112B67BF9D8%7D&file=Atom%20economy%20and%20percentage%20yield%20questions.docx&action=default&mobileredirect=true

Answers =

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7BF3BFD1B5-E3DE-4682-B00F-61E54D6CDB1E%7D&file=ANSWERS-13%20%25%20yiel%20%2B%20atom%20economy.doc&action=default&mobileredirect=true

131
Q

Do Task 3- A little Trickier

First Link= Questions on document
Second Link= Answers on document

A

Questions =

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7B955384CE-D3AB-471E-BA0F-7112B67BF9D8%7D&file=Atom%20economy%20and%20percentage%20yield%20questions.docx&action=default&mobileredirect=true

Answers =

https://edukedstac.sharepoint.com/:w:/r/sites/CX-Y12-C1KE23/_layouts/15/Doc.aspx?sourcedoc=%7BF3BFD1B5-E3DE-4682-B00F-61E54D6CDB1E%7D&file=ANSWERS-13%20%25%20yiel%20%2B%20atom%20economy.doc&action=default&mobileredirect=true

132
Q

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

A

1) 2.3 x 10-³

2) 1.032 x 10 ³

3) 2.75 x 10⁶

4) 5.28 x 10-⁴

133
Q

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

A

a) 2010

b) 0.052

c) 841

d) 0.0001

134
Q

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

A

Answers are next to qs

135
Q

Do Question 2 on Page 4 of Amount of Substance Booklet

A

Answers present by Qs

136
Q

Do Question 3 on Page 4 of Amount of Substance Booklet

A

Answers present by Qs

137
Q

Do the Re arranging Equations on page 5

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

138
Q

Write the formula for Magnesium Hydroxide

A

Mg²+ OH-

    Mg(OH)2
139
Q

Write the formula for Aluminium Nitrate

A

Al³+ NO3 ²-

    Al2(NO3)3
140
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

141
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

142
Q

Do How many moles task on top of page 7

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

143
Q

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

A

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

144
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

145
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

146
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

147
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

148
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

149
Q

Do all of page 9 in amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

150
Q

Do Box 1 on page 10 of Amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

151
Q

Do Box 2 on page 10 of Amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

152
Q

Do Box 3 on page 10 of Amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

153
Q

Do Box 4 on page 10 of Amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

154
Q

Do Box 5 on page 10 of Amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

155
Q

Do Box 6 on page 10 of Amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

156
Q

Do Box 7 on page 10 of Amount of substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

157
Q

Do Task 4 on page 11 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

158
Q

Do Task 4 on Page 13 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

159
Q

What 2 assumptions does the ideal gas equation rely on

A
  • forces between molecules are negligible
    -gas molecules have negligible size compared to the size of their container
160
Q

Do Task 5 on page 14 of Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

161
Q

Do Task 6 on page 15 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

162
Q

Do Page 16 in Amount of Substance booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

163
Q

Write a method for determining the formula of a hydrated salt

A

Look on Kerboodle or Physics and maths tutor

164
Q

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

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

165
Q

Do pages 20 to 23 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

166
Q

Do Task 14 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

167
Q

Do Page 25 and 26 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

168
Q

Do Task 16 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

169
Q

Do Pages 28 and 29 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

170
Q

Do page 30 in Amount of Substance Booklet

A

https://edukedstac.sharepoint.com/:w:/r/sites/CX-2Y-CSKE23/_layouts/15/Doc.aspx?sourcedoc=%7B589F4841-5D11-4C00-8A52-F8788B1AA52F%7D&file=AOS%20Answers%202023.docx&action=default&mobileredirect=true

171
Q

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

A

Cm3 and dm3

172
Q

What is a solute

A

The dissolved compound

173
Q

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

A

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

174
Q

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

A

dm3

175
Q

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

A

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

176
Q

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

A

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

177
Q

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

A

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

178
Q

What is a standard solution

A

A solution of known concentration

179
Q

How are standard solutions prepared

A

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

180
Q

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.

A

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

180
Q

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

A

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

180
Q

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

A

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

181
Q

What is the Molar Gas Volume Vm

A

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

182
Q

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

A

24.0dm3 = 240000 cm3

183
Q

What is the molar gas volume at RTP

A

24.0 dm3 mol-1

184
Q

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

A

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

185
Q

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

A

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

186
Q

Assumptions for molecules in an ideal gas:

A

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

187
Q

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

A

Kelvin

188
Q

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

A

-273*c

189
Q

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

A
190
Q

Do Kerboodle Summary questions on page 31

A

Answers are on page 273 under 3.3

191
Q

What do chemists use balanced equations to find

A

-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

192
Q

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

A

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

193
Q

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

A
  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**

194
Q

What is the theoretical yield

A

The maximum possible amount of product that could form

195
Q

Why is it so difficult to achieve the theoretical yield

A

-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

196
Q

The actual yield obtained from a reaction is usually…

A

… lower than the theoretical yield

197
Q

How do you find the percentage yield

A

Percentage yield= actual yield/ theoretical yield

all times by 100

198
Q

What is a limiting reagent

A

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

199
Q

What is atom economy

A

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

200
Q

Reactions with high atom economy :

(2marks)

A
  • 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

201
Q

What happens if we start to improve atom economy

A

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

202
Q
A

A Level Chemistry (13 decks)

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