AQA AS Level Chemistry Paper 1 Exam fully
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Terms in this set (49)
Give the full electron Argon
configuration of the 1s2 2s2 2p6 3s2 3p6
element in period 3 with
the highest first ionisation
energy (1)
give an equation including Na+ (g) --> Na2+ (g) + e-
state symbols to represent
the process that occurs
when the second
ionisation energy of
sodium is measured (1)
table 1 shows some Element: Sulfur (1)
successive ionisation
energies for an element in Explanation: big difference between the 6th and 7th
period 3. Identify the ionisation energies (1)
period 3 element. Explain which indicates electron moves from the 2nd energy
your answer (3) level which has lower energy (1)
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,10/22/25, 7:05 AM AQA AS Level Chemistry Paper 1 Exam fully solved & updated 2025-2026(latest version verified for accuracy) | 2024\2025Latest!!…
this question is about the Magnesium has a lattice of Mg2+ ions. (1)
elements in group 2.
Describe the structure There is strong electrostatic attraction between the
and bonding of Mg2+ ions and there are delocalised electrons in the
magnesium (2 marks) lattice (1)
Trend: Atomic radius increases (1)
state the trend in the
atomic radius of the
Reason: As you move down a group the number of
elements down Group 2
energy levels increases (1) (which creates more
from Mg to Ba. Give a
distance between the nucleus and electrons so the
reason for this trend (2)
atomic radius increases)
Give an equation, equation:
including state symbols, Mg(s) + H2O(l) --> MgO(s) + H2(g)
for the reaction of
magnesium with steam. Observation 1: Bright light
State 2 observations for
this reaction. (3) Observation 2: Smoke (because magnesium burns)
The sulfates of the Formula:
element in group 2 from BaSO4 (1)
Mg to Ba have different
solubilities. Use:
State the formula of the X-rays (1)
least soluble of these
sulfates.
Give a use for this sulfate
(2)
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A sample of strontium is 1) abundance of 87Sr = X
made up of only 3 AND abundance of 86Sr = 1-0.83-X
isotopes: 86Sr, 87Sr and =0.17-X
88 Sr.
This sample contains 2) 87.73= (88 x 0.83) + (87 x X) + (86 x (0.17-X))
83.00% by mass of 88Sr 87.73 = 87.66 + X
This sample of strontium
has Ar=87.73 3) 87Sr = 0.07 = 7%
Calculate the percentage
abundance of each of the 4) abundance of 86Sr = 1-0.83-0.07=0.1
other 2 isotopes in this which is 10%
sample (4)
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