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2018 Paper One
A substantial 18-mark question.
Many students failed to consider why the attraction between the nucleus and the electron might be different.
A number of students gave the incorrect formula.
The students who gave a correct algebraic expression with one unknown could work through the calculation and generally gained full marks.
Common errors included:
Failure to use the Avogadro constant
Not converting mass to kg
Incorrect rearrangement of expressions.
The full configuration is required:
In Ca+, the electron in the outer shell is further from the nucleus than the outer electron in K+
Ca+ loses electron from a 4(s) orbital and K+ loses electron from a 3(p) orbital
There is also more electron shielding in Ca+
The first ionization energy generally decreases as we move down a group in the periodic table, but increases as we move from left to right across a period. Therefore, the s-block element with the highest first ionization energy would be the element at the top right corner of the s-block, which is beryllium (Be). Beryllium has a higher nuclear charge and smaller atomic radius than the other s-block elements, which makes it more difficult to remove an electron from its outer shell.
The solubility of hydroxides of Group 2 elements in water generally increases as we move down the group. Therefore, the least soluble hydroxide would be that of the element at the top of Group 2, which is magnesium (Mg).
Ionic equation for the formation of the precipitate:
Calculate the amount of each reagent present and compare them.
Amount of barium chloride = 0.25 x 6/1000 = 0.0015 mol
Amount of sodium sulfate = 0.15 x 8/1000 = 0.0012 mol
From the balanced equation, the reactants react in a molar ratio therefore the limiting reagent is sodium sulfate.
To calculate the total volume of the excess reagent that should be used, we need to determine how much of the excess reagent is required to react with all of the limiting reagent. Since 1 mole of sodium sulfate reacts with 1 mole of barium chloride, we need 0.0015 moles of sodium sulfate to react with all of the barium chloride.
Amount of sodium sulfate required = 0.0015 mol
Concentration of sodium sulfate = 0.15 mol dm-3
Volume of sodium sulfate required = 0.0015/0.15 = 0.01 dm3 or 10 cm3
The student needs to use an additional 10 cm3 of the sodium sulfate solution to react with all of the barium chloride solution and obtain a filtrate containing only one solute.
Same electronic configuration
The relative atomic mass of the sample is given as 87.7, which can be expressed as a weighted average of the atomic masses of the isotopes, using their respective abundances:
(86.0x% + 87.0x% + 88.0(100 - 2x)%) / 100% = 87.7
Simplifying we get:
(86.0x + 87.0x + 88.0(100 - 2x) / 100 = 87.7
Solve for x:
% abundance of 88 isotope:
100 – 2(10) = 80%
This will be the heaviest of the 3 - make sure you indicate the ion not the isotope itself.
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