Eduqas 2022 chemistry paper unit 1

Question 1

Polar covalent bond?

Answer 1

a bond where the electrons are not shared equally

Question 2

a mass of 2.750g is weighed by difference
calculate the percentage error when a 3 decimal place is used?

Answer 2

uncertainty in 1 weighing = 0.0005
percentage error = 2 x 0.0005 / 2.750 x 100 = 0.036%

Question 3

Thermal decomposition of Calcium Hydroxide?

Answer 3

Ca(OH)2 — CaO + H2O

Question 4

The equilibrium mixture formed contained NOCl, at a concentration of 0.126 mol dm–3, and NO, at a concentration of 5.73 × 10–2 mol dm–3.
The value of Kc for the equilibrium at this temperature was 7.40 × 10–3.
Calculate the concentration of Cl2 in this equilibrium mixture?

Answer 4

Cl2 = (NO)2(Kc) / NOCl^2
7.4 x 10^-3 x (0.126)^2/(0.0573)^2
0.0358

Question 5

How would the amount of chlorine change if the reaction was carried out at a higher pressure?

Answer 5

Less chlorine would be produced ,
the equilibrium would shift to oppose the
increase in pressure
so it would move to the side with fewer gas moles

Question 6

Explain why the first ionisation energy of xenon is lower than that of krypton?

Answer 6

Krypton is in the next period
therefore, xenon has a bigger atomic radius and therefore it is easier to remove outer electrons

Question 7

Why is the first ionisation energy of xenon higher than that of iodine?

Answer 7

Xenon has a greater nuclear charge
similar atomic radius and so harder to remove outer electron

Question 8

Electronegativity def?

Answer 8

the ability of an atom to attract a pair of electrons within a covalent bond

Question 9

Calculate the volume of xenon gas required at a temp of 400 degrees and a pressure of 1 x 10^5 Pa to make 5g of XeF2

Answer 9

n( XeF2) = 5/169 = 0.0296
pV = nRT
v = 0.0296 x 8.31 x 673 / 1.01 x10^5
volume of xenon = 1.66dm3

Question 10

Using your knowledge of VSEPR theory, suggest why electron pairs arrange themselves in this way>

Answer 10

electron pairs arranged as far apart to minimise repulsion
lone pairs repel more than bond pairs so bond angle = greater

Question 11

Use the oxidation states of xenon and oxygen t show why xenon difluoride is an oxidising agent in this reaction?

Answer 11

Oxidation state of Xenon changes from +2 to 0 and oxidation state of oxygen
changes from -2 to 0

Question 12

Equation for xenon difluoride reacting with water to produce hydrogen fluoride and 2 elements?

Answer 12

2XeF2 + 2H2O —– 2Xe+O2 + 4HF

Question 13

How does the mass spec work and the mass spec of chlorine?

Answer 13

Mass spectrometer
Element is vaporised
ionised by having electrons knocked off
positive ions are formed
acceleration by charge plates
made into beam by slits
deflected by electromagnet
deflected according to m/z value
whole system is under vacuum

mass spec of Cl
peak at 35 and 37 -2 isotopes 35cl + 37cl
these peaks are in a 3:1 ratio
peaks due to chlorine molecule at 70,72 or 74
peaks are at a ratio of 9:6:1

Question 14

Given that both water samples contain the maximum mass of these contaminants ,according to the limits in the table
determine which sample contains greater number of moles?

Answer 14

Mass of arsenic = 10 x 10^-6 x 330/1000 = 3.30 x 10^-6
moles of arsenic = 3.30x 10^-6 / 74.9 = 4.41 x 10^-8
mass of copper = 2x10^-3x110/1000 = 2.20 x 10^-4 g

moles of copper = 2.20 x 10^-4 / 63.5 = 3.46 x 10^-4
therefore more moles of copper

Question 15

explain own view on fluoridation of drinking water

Answer 15

Fluoride is added to water to reduce tooth decay
However, it introduces mass medication where people do not get a say as to whether they can have
fluoride in their water which is not acceptable
fluorosis
therefore I disagree

Question 16

Describe solid structure of aluminium?

Answer 16

Regular array of metal ions
Ions have a 3+ charge
sea of delocalised electrons

Question 17

Use this information to explain the unusual melting temperature of gallium compared to aluminium?

Answer 17

Gallium is made of molecules and there are only weak forces between thesemolecules
whereas, aluminium is made of metal ions and delocalised electrons held together by strong
electrostatic forces of attraction
a lot less energy is needed to allow the particles in Gallium to overcome the forces holding them
in the solid structures than is the case in aluminium

Question 19

given that the relative atomic mass of natural gallium is 69.798, determine the percentage abundance of these two isotopes

Answer 19

69.78 = 69 x 100 - Ga71 + 71 x Ga71 / 100
Ga71 = 6979.8 - 6900 / 2 = 39.9%
Ga-69 = 60.1%
abundance of Ga-69 = 60.1%

Question 20

Write an equation for the first ionisation energy of gallium.

Answer 20

Ga(g) — Ga+(g) + e-

Question 21

Calculate the wavelength of radiation, in m, that will ionise an atom of gallium to form a Ga+ ion.

Answer 21

E = h x c/ lamder
lamder = Hc / e
= 6.63 x 10 ^-34 x 3.00 x 10^8 / 9.61 x 10^-19 = 2.07 x 10^-7 m

Question 22

Suggest a reason for removing the funnel after it has been used for filling the burette.

Answer 22

droplet from the funnel could enter burette

Question 23

Write an equation for the reaction between sodium hydroxide and carbon dioxide

Answer 23

2NaOH + CO2 → Na2CO3 + H2O

Question 24

Calculate the number of moles of hydrogen carbonate ions reacting in the second stage of the reaction

Answer 24

v -= 2.50 cm3
n = 1 x 2.5/1000 =2.50 × 10–3

Question 25

Using your answers from parts I. and II., determine the number of moles of carbonate ions and hence the number of moles of hydroxide ions in the 25.0 cm3 sample.

Answer 25

Moles ( CO32-) = 2.50 X 10^-3
Moles OH = 2.25 x 10^-2 - 2.50 x 10^-3 = 0.0200

Question 26

Using the number of moles of carbonate ions determined in part III., calculate the volume of carbon dioxide absorbed into the original solution at 298 K and 1 atm.

Answer 26

Moles CO2 absorbed = moles CO23- made = 2.50 x 10^-3 in 25cm3
moles in 1dm3 = 2.50 x 10^-3 x 40 = 0.100
volume = 0.100 x 24,5 = 2.45dm3

Question 27

Explain what is meant by a strong acid and write an equation to show how hydrochloric acid behaves as a strong acid.

Answer 27

Fully diassociates into aqueous solution
HCl—- H+ + Cl-

Question 28

Calculate the pH of 500 cm3 of 0.50 mol dm–3 hydrochloric acid.

Answer 28

pH = –log [0.50] = 0.30

Question 29

Calculate the volume of water that must be added to this solution to increase its pH to a value of 1.0.

Answer 29

new [H+] = 10–1.0 = 0.10 mol dm–3
new V = n/c = 0.25 / 0.10 = 2.50dm3
volume water added = 250-0.50 = 2dm3