Question
Which allotrope, oxygen or ozone, has the stronger bond between O atoms, and which absorbs higher frequency UV radiation in the atmosphere?
▶️Answer/Explanation
Markscheme: C
Since double bonds are stronger than single bonds, this means that the O-O bonds in oxygen are stronger than those in ozone.
The UV rays absorbed by oxygen atoms are very, very high in energy, but the Sun does not emit much of these rays. So oxygen atom absorption is not important.
The UV rays absorbed by oxygen molecules are high in energy and there is a medium amount of these rays emitted by the Sun. So oxygen molecule absorption is important, not only because UV rays are absorbed, but because the absorption helps create ozone.
The UV rays absorbed by ozone molecules are of medium energy and there is a lot of these rays emitted by the Sun. So ozone molecule absorption is very important.
Question
What are the numbers of neutrons and electrons in \(^{32}_{16}S^{2-}\)?
▶️Answer/Explanation
>Markscheme: B
The symbol \(^{32}_{16}\rm S^{2-}\) represents the sulfur ion with a charge of -2. The superscript (top number) indicates the mass number, the subscript (bottom number) indicates the atomic number, and the charge is indicated as a superscript on the upper right side.
For a neutral sulfur atom (\(^{32}_{16}\rm S\)), the number of electrons is equal to the atomic number, which is 16.
Since the ion has a charge of -2, it means that there are two more electrons than the number of protons. Therefore, for \(^{32}_{16}\rm S^{2-}\), there are \(16 + 2 = 18\) electrons.
To find the number of neutrons, you can subtract the number of protons (which is equal to the atomic number) from the mass number. For \(^{32}_{16}S\), there are \(32 – 16 = 16\) neutrons.
So, the numbers of neutrons and electrons in \(^{32}_{16}\rm S^{2-}\) are 16 neutrons and 18 electrons.
Question
Which of the following is consistent with Avogadro’s law?
A. \(\frac{P}{T} = \) constant (\(V\), \(n\) constant)
B. \(\frac{V}{T} = \) constant (\(P\), \(n\) constant)
C. \(Vn = \) constant (\(P\), \(T\) constant)
D. \(\frac{V}{n} = \) constant (\(P\), \(T\) constant)
Answer/Explanation
Markscheme
D
Examiners report
A number of respondents felt that this question was off-syllabus. This was discussed at length during GA and it was felt that the question itself is clearly on-syllabus as it relates to AS 1.4.4 which states that candidates should be able to apply Avogadro’s law to calculate reacting volumes of gases. In order to apply Avogardo’s law candidates should be able to understand the underlying principle of the law itself. The question was asked within the confines of P1 and a formal definition was not asked, such as could be asked in P2, which in such a case then would be deemed off-syllabus.
The correct answer, D, with \(V{\text{/}}n = \) constant, was given by 40% of candidates and hence it was felt that the question although tough could be answered with a clear understanding of the nature of Avogadro’s law as cited in AS 1.4.4. Some respondents did comment that the law is only valid if \(P\) and \(T\) are constant, which is a fair comment and it would have been better if for each of the answers A-D that the variables for the other constants had been put in brackets.
In the case of the correct answer D, it was assumed that \(P\) and \(T\) are constant and hence as D is the best answer of those given it was decided to keep this question, as 40% of candidates gave D as the correct answer.
Question
Which sample has the greatest mass?
A. 1 mol of \({\text{S}}{{\text{O}}_{\text{2}}}\)
B. 2 mol of \({{\text{N}}_{\text{2}}}{\text{O}}\)
C. 2 mol of Ar
D. 4 mol of \({\text{N}}{{\text{H}}_{\text{3}}}\)
Answer/Explanation
Markscheme
B
Examiners report
Question
What is the number of ions in 0.20 mol of \({{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\)?
A. \(8.0 \times {10^{ – 1}}\)
B. \(1.2 \times {10^{23}}\)
C. \(4.8 \times {10^{23}}\)
D. \(2.4 \times {10^{24}}\)
Answer/Explanation
Markscheme
C