Differentiate the function \(f(x)=\arcsin \left(x^5\right)\). (1 mark)
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Differentiate the function \(f(x)=\arcsin \left(x^5\right)\). (1 mark)
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\(f^{\prime}(x)=\dfrac{5 x^4}{\sqrt{1-x^{10}}}\)
\(f(x)=\sin ^{-1}\left(x^5\right)\)
\(f^{\prime}(x)=5 x^4 \times \dfrac{1}{\sqrt{1-\left(x^5\right)^2}}=\dfrac{5 x^4}{\sqrt{1-x^{10}}}\)
Using the substitution \(u=x-1\), find \(\displaystyle \int x \sqrt{x-1}\, d x\). (3 marks)
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\(\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c\)
\(\displaystyle \int x \sqrt{x-1}\, d x \)
\(u=x-1\) | \(\Rightarrow \ x=u+1 \) | |
\(\dfrac{d u}{d x}=1\) | \(\Rightarrow \ d u=d x\) |
\(\displaystyle\int(u+1) \sqrt{u+1-1}\, d u\) | \(=\displaystyle{\int}(u+1) \sqrt{u} \, d u\) | |
\(=\displaystyle{\int} u^{\frac{3}{2}}+u^{\frac{1}{2}}\, d u\) | ||
\(=\dfrac{2}{5} u^{\frac{5}{2}}+\dfrac{2}{3} u^{\frac{3}{2}}+c\) | ||
\(=\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c\) |
Consider the vectors \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}\) and \(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\).
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i. \(\displaystyle \binom{7}{0}\)
ii. \(5\)
i. \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}, \ \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\)
\(2 \underset{\sim}{a}-\underset{\sim}{b}=2 \displaystyle \binom{3}{2}-\binom{-1}{4}=\binom{6}{4}-\binom{-1}{4}=\binom{7}{0}\)
ii. \(\underset{\sim}{a} \cdot \underset{\sim}{b}=\displaystyle\binom{3}{2}\binom{-1}{4}=3 \times(-1)+2 \times 4=5\).
Which is the correct balanced formula equation for the reaction of potassium with water?
\(C\)
→ The correct chemical equation is:
\(\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)}\)
→ The state of water is always liquid. An aqueous solution refers to a solution where a substance has been dissolved into water.
\(\Rightarrow C\)
In an experiment, calcium carbonate \(\ce{(CaCO3)}\) is heated strongly to produce calcium oxide \(\ce{(CaO)}\) and carbon dioxide according to the reaction below:
\(\ce{CaCo3(s) -> CaO(s) + CO2(g)}\)
A student starts with 50.0 g of calcium carbonate. After heating, they collect 28.0 g of calcium oxide.
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a. \(22.0\ \text{g}\)
b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.
→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.
→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.
a. According to the law of conservation of mass, the total mass of reactants must equal the total mass of products.
Since mass must be conserved:
\(m\ce{(CO2)}\) | \(=m\ce{(CaCO3)}-m\ce{(CaO)}\) | |
\(= 50.0-28.0\) | ||
\(=22.0\ \text{g}\) |
b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.
→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.
→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.
What numbers are required to correctly balance this equation?
__\(\ce{Fe2O3 +}\) __\(\ce{CO ->}\) __\(\ce{Fe +}\) __\(\ce{CO2}\)
\(A\)
→ Start by balancing the iron \(\ce{Fe}\) atoms:
→ There are 2 Fe atoms in \(\ce{Fe2O3}\), so we place a 2 in front of \(\ce{Fe}\) on the product side.
\(\ce{Fe2O3 + CO -> 2Fe + CO2}\)
→ Next, balance the oxygen \(\ce{O}\) atoms:
→ \(\ce{Fe2O3}\) has 3 oxygen atoms, and we need 3 \(\ce{CO}\) molecules on the reactant side to balance the 3 \(\ce{CO2}\) molecules on the product side.
\(\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}\)
→ Finally, check all atoms to ensure balance.
\(\Rightarrow A\)
Balance the following chemical equations:
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a. \(\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}\)
b. \(\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}\)
c. \(\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}\)
d. \(\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}\)
a. \(\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}\)
b. \(\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}\)
c. \(\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}\)
d. \(\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}\)
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a. \(0.5\ \text{L}\)
b. The exact volumes of solutions must be known as:
→ The concentration of each solution must be precise, as it affects the molar ratios and yields.
→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.
a. \(V=\dfrac{n}{c}=\dfrac{0.6}{1.2}=0.5\ \text{L}\)
b. The exact volumes of solutions must be known as:
→ The concentration of each solution must be precise, as it affects the molar ratios and yields.
→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.
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a. \(1.5\ \text{mol}\)
b. \(87.99\ \text{g}\)
a. \(n = c \times V = 0.75 \times 2 = 1.5\ \text{mol}\)
b. \(MM\ce{(NaCl)} = 22.99 + 35.45 = 58.44\ \text{g/mol}\)
\(m\ce{(NaCl)} = MM \times n = 58.44 \times 1.5 = 87.99\ \text{g}\)
A student prepares a solution of potassium nitrate by dissolving 0.05 kg of \(\ce{KNO3}\) in enough water to make 2000 mL of solution. Which of the following correctly calculates the concentration of the solution in mol L\(^{-1}\)?
\(D\)
→ Mass of \(\ce{KNO3}\) is 50 g.
→ Volume of solution is 2 L.
\(c\ce{(KNO3)} = \dfrac{50}{2} = 25\ \text{g L}^{-1}\)
\(\Rightarrow D\)
Consider the functions \(y=f(x)\) and \(y=g(x)\), and the regions shaded in the diagram below.
Which of the following gives the total area of the shaded regions?
\(D\)
\(\text{Intervals where}\ f(x) \gt g(x)\ \ \Rightarrow\ \text{Positive area values}\)
\(\text{Intervals where}\ g(x) \gt f(x)\ \ \Rightarrow\ \text{Negative area values}\)
\(\Rightarrow D\)
A student prepares a standard solution of sodium chloride. They dissolve 5.85 g of sodium chloride \(\ce{(NaCl)}\) in enough water to make 1.00 L of solution. Determine the concentration of this solution?
\(A\)
\(n\ce{(NaCl)} = \dfrac{5.85}{22.99 + 35.45} = 0.100\ \text{mol}\)
\(\ce{[NaCl]} = \dfrac{n}{V} = \dfrac{0.100}{1} = 0.100\ \text{mol L}^{-1}\)
\(\Rightarrow A\)
Consider the function \(g(x) = 2 \sin^{-1}(3x)\).
Which transformations have been applied to \(f(x) = \sin^{-1}(x)\) to obtain \(g(x)\)?
\(C\)
\(\text{A vertical dilation of factor 2:}\)
\(f(x) = \sin^{-1}(x)\ \ \rightarrow \ \ f_1(x) = 2\sin^{-1}(x)\)
\(\text{A horizontal dilation of factor}\ \dfrac{1}{3}:\)
\(f_1(x) = 2\sin^{-1}(x)\ \ \rightarrow \ \ f_2(x) = 2\sin^{-1}(3x)\)
\(\Rightarrow C\)
The polynomial \(x^{3} + 2x^{2}-5x-6\) has zeros \(-1, -3\) and \(\alpha\).
What is the value of \(\alpha\)?
\(B\)
\(\alpha \beta \gamma = -\dfrac{\text{b}}{\text{a}} = 6\)
\(-1 \times -3 \times \alpha\) | \(=6\) | |
\(\alpha\) | \(=2\) |
\(\Rightarrow B\)
A floor plan for a living area is shown. All measurements are in millimetres.
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a. \(4\ \text{m}\times 0.5\ \text{m}\)
b. \(160\ \text{tiles}\Rightarrow\ 11\ \text{boxes}\Rightarrow\ $1100\)
a. \(4\ \text{m}\times 0.5\ \text{m}\)
b. \(\text{Method 1}\)
\(\text{Total area to be tiles}\) | \(=6\times 3-4\times 0.5\) |
\(=16\ \text{m}^2\) | |
\(\text{Area of each tile}\) | \(=0.2\times0.5\) |
\(=0.1\ \text{m}^2\) | |
\(\text{Tiles needed}\) | \(=\dfrac{16}{0.1}\) |
\(=160\ \text{tiles}\) | |
\(\text{Number of boxes}\) | \(=\dfrac{160}{15}\) |
\(=10.66\dots\) | |
\(=11\ \text{boxes}\) | |
\(\text{Total cost}\) | \(=$100\times 11\) |
\(=$1100\) |
\(\text{Method 2}\)
\(\text{Tiles to fit 6 m width}\ =\dfrac{6}{0.2}=30 \)
\(\text{Tiles to fit 2 m width}\ =\dfrac{2}{0.2}=10 \)
\(\text{Total tiles}\ =30\times 5\ \text{rows}+10\times 1\ \text{rows}\ =160\)
\(\text{Number of boxes}\) | \(=\dfrac{160}{15}\) |
\(=10.66\dots\) | |
\(=11\ \text{boxes}\) | |
\(\text{Total cost}\) | \(=$100\times 11\) |
\(=$1100\) |
A standard solution is best described as:
\(B\)
→ A standard solution is a solution with a precise and known concentration, allowing accurate calculations in quantitative chemical analysis.
→ It is prepared by dissolving an exact amount of a primary standard in a specific volume of solvent.
\(\Rightarrow B\)
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a. \(0.24\ \text{mol L}^{-1}\)
b. \(41.6\ \text{g}\)
a. \(MM\ce{(KCl)}= 39.10 + 35.45 = 74.55\ \text{g mol}^{-1}\)
\(n\ce{(KCl)}= \dfrac{m}{MM} = \dfrac{4.56}{74.55} = 0.061\ \text{mol}\)
\(c\ce{(KCl)}=\dfrac{n}{V}= \dfrac{0.061}{0.25} = 0.24\ \text{mol L}^{-1}\)
b. \(n\ce{(CaCl2)} = c \times V = 0.25 \times 1.5 = 0.375\ \text{mol}\)
\(m\ce{(CaCl2)} = MM \times n = (40.08 + 2(35.45)) \times 0.375 = 41.6\ \text{g}\)
A wheel is shown with the numbers 0 to 19 marked.
A game is played where the wheel is spun until it stops.
When the wheel stops, a pointer points to the winning number. Each number is equally likely to win.
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a. \(8\ ,\ 10\ ,\ 12\ ,\ 14\ ,\ 16\ ,\ 18\)
b. \(\dfrac{7}{10}=0.7\)
a. \(8\ ,\ 10\ ,\ 12\ ,\ 14\ ,\ 16\ ,\ 18\)
b. | \(P(\text{not even }>7)\) | \(=1-P(\text{even }>7)\) |
\(=1-\dfrac{6}{20}\) | ||
\(=\dfrac{14}{20}=\dfrac{7}{10}=0.7\) |
Mark buys one raffle ticket in a raffle with 1000 tickets.
Which of the following best describes the probability that Mark wins?
\(C\)
\(\text{P(win)}=\dfrac{1}{1000}\)
\(\therefore\ \text{Unlikely}\)
\(\Rightarrow C\)
Four people completed the same fitness activity.
The graph shows the heart rate for each person before and after completing the activity.
Which person had the LEAST difference in heart rate?
\(B\)
Option A: | Jo | \(=120-80=40\) |
Option B: | Kim | \(=120-100=20\ \checkmark\) |
Option C: | Lee | \(=120-90=30\) |
Option D: | Mal | \(=150-100=50\) |
\(\Rightarrow B\)
A monic polynomial, \(f(x)\), of degree 3 with real coefficients has \(3\) and \(2+i\) as two of its roots.
Which of the following could be \(f(x)\) ?
\(B\)
\(\text{Since coefficients are real, roots are:}\ \ 3, 2+i, 2-i\)
\[ \sum \text{roots} = 7 = \dfrac{-b}{1}\ \ \Rightarrow \ b=-7\]
\(\text{Product of roots}\ = 3(2+i)(2-i)=15=\dfrac{-d}{1}\ \ \Rightarrow \ d=-15\)
\(\Rightarrow B\)
Consider the statement:
'If a polygon is a square, then it is a rectangle.'
Which of the following is the converse of the statement above?
\(A\)
\(\text{Statement:}\ \P \Rightarrow \Q\)
\(\text{Converse of statement:}\ \Q \Rightarrow \P\)
\(\Rightarrow A\)
Consider the following statement written in the formal language of proof
\(\forall \theta \in\biggl(\dfrac{\pi}{2}, \pi\biggr) \exists\ \phi \in\biggl(\pi, \dfrac{3 \pi}{2}\biggr) ; \ \sin \theta=-\cos \phi\).
Which of the following best represents this statement?
\(C\)
\(\Rightarrow C\)
A network of towns and the distances between them in kilometres is shown.
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a. \(TYWH\)
b. \(\text{Length of shortest path}\ (YWHMG) = 89 \text{km}\)
a. \(TYH=30+38=68, \quad TYWH=30+15+20=65\)
\(\therefore \text{ Shortest Path is}\ TYWH.\)
b. \(Y W C M G=15+25+25+25=90\)
\(YWHMG=15+20+29+25=89\)
\(\Rightarrow \ \text{All other paths are longer.}\)
\(\therefore\text{ Length of shortest path = 89 km}\)
The graph of the function \(f(x) = \ln(1 + x^{2})\) is shown.
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\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 0.25 & 0.5 & 0.75 & 1 \\
\hline
\rule{0pt}{2.5ex} \ln(1+x^2) \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 0.0606 & 0.2231 & 0.4463 & 0.6931 \\
\hline
\end{array}
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a. \(f(x)= \ln(1+x^2)\)
\(f^{′}(x)=\dfrac{2x}{1+x^2}\)
\(f^{″}(x)\) | \(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\) | |
\(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\) | ||
\(=\dfrac{2(1-x^2)}{(1+x^2)^2}\) |
\(\text{In domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)
\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)
\(\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) \)
\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)
b. \(\text{Total shaded area}\ \approx 0.5383\ \text{(4 d.p.)}\)
c. \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)
\(\text{Since the graph is concave up in the given domain, the trapezoidal rule will overestimate the area.}\)
a. \(f(x)= \ln(1+x^2)\)
\(f^{′}(x)=\dfrac{2x}{1+x^2}\)
\(f^{″}(x)\) | \(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\) | |
\(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\) | ||
\(=\dfrac{2(1-x^2)}{(1+x^2)^2}\) |
\(\text{In domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)
\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)
\(\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) \)
\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)
b. \(\text{Total shaded area}\)
\(\approx 2 \times \dfrac{h}{2}[ y_0 + 2(y_1+y_2+y_3) + y_4] \)
\(\approx 2 \times \dfrac{0.25}{2}[ 0 + 2(0.0606+0.2231+0.4463) + 0.6931] \)
\(\approx 0.538275 \)
\(\approx 0.5383\ \text{(4 d.p.)}\)
c. \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)
\(\text{Since the graph is concave up in the given domain, the trapezoidal rule will overestimate the area.}\)
A vertical tower \(T C\) is 40 metres high. The point \(A\) is due east of the base of the tower \(C\). The angle of elevation to the top \(T\) of the tower from \(A\) is 35°. A second point \(B\) is on a different bearing from the tower as shown. The angle of elevation to the top of the tower from \(B\) is 30°. The points \(A\) and \(B\) are 100 metres apart.
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\(194^{\circ}\)
a. \(\text{In}\ \Delta TCA:\)
\(\tan 35°\) | \( =\dfrac{40}{AC}\) | |
\(AC\) | \( =\dfrac{40}{\tan 35°}\) | |
\(=57.125…\) | ||
\(=57.13\ \text{m (2 d.p.)}\) |
b. \(\text{In}\ \Delta TCB:\)
\(\tan 30°\) | \( =\dfrac{40}{BC}\) | |
\(BC\) | \( =\dfrac{40}{\tan 30°}\) | |
\(=69.28\ \text{m}\) |
\( \text{Find} \ \angle BCA \ \text{using cosine rule:}\)
\(\cos \angle B CA\) | \( = \dfrac{57.13^2+69.28^2-100^2}{2 \times 57.13 \times 69.28}\) | |
\(= -0.2446 \)… | ||
\(\angle BCA\) | \( = 104.2° \) |
\(\therefore\ \text{Bearing of}\ B\ \text{from}\ C= 90+104=194^{\circ} \text{(nearest degree)} \)
The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .
Population \(W\) is modelled by the equation \(y=A \times(1.055)^x\).
Population \(K\) is modelled by the equation \(y=B \times(0.97)^x\).
Complete the table using the information provided. (3 marks)
\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}
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\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}
\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}
\(\text{Calculations:}\)
\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)
\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)
\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)
Sarah, a 60 kg female, consumes 3 glasses of wine at a family dinner over 2.5 hours.
Note: there are 1.2 standard drinks in one glass of wine.
The blood alcohol content \((BAC)\) for females can be estimated by
\(B A C_{\text {female}}=\dfrac{10 N-7.5 H}{5.5 M},\)
where \(N\) | = number of standard drinks | |
\(H\) | = number of hours drinking | |
\(M\) | = mass in kilograms |
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The time it takes a person's BAC to reach zero is given by
\(\text {Time}=\dfrac{B A C}{0.015}.\)
Calculate the time it takes for Sarah's BAC to return to zero, assuming she stopped drinking after 2.5 hours. Give your answer to the nearest minute. (2 marks)
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a. \(BAC=0.052\)
b. \(\text{3 h 28 m}\)
a. \(N=3 \times 1.2=3.6, H=2.5, M=60\)
\(\therefore B A C\) | \(=\dfrac{10 \times 3.6-7.5 \times 2.5}{5.5 \times 60}\) | |
\(=0.052 \ \text{(3 d.p.)}\) |
b. | \(T\) | \(=\dfrac{0.052}{0.015}\) |
\(=3.466\) | ||
\(=3 \text{ h 28 m}\) |
A teacher was exploring the relationship between students' marks for an assignment and their marks for a test. The data for five different students are shown on the graph.
The least-squares regression line is also shown.
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a. \(\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)\)
\(y \,\text {-intercept}=6\)
\(m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4\)
\(\text{LSRL}\ \ \Rightarrow \ y=1.4x+6\)
b. \(\text {The student’s mark \((5,12)\) sits below the LSRL.}\)
\(\text {Therefore the student did worse than expected.}\)
a. \(\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)\)
\(y\,\text {-intercept}=6\)
\(m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4\)
\(\text{LSRL}\ \ \Rightarrow \ y=1.4x+6\)
b. \(\text {The student’s mark \((5,12)\) sits below the LSRL.}\)
\(\text {Therefore the student did worse than expected.}\)
The cost of electricity is 30.13 cents per kWh .
Calculate the cost of using a 650 W air conditioner for 6 hours. (2 marks)
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\(\text{Cost} =\$ 1.18\)
\(\text{Usage}=6 \times 650=3900\, \text{W}=3.9\, \text{kW}\)
\(\text{Cost} =3.9 \times 30.13=117.507 \,\text{c}=\$ 1.18 \, \text {(nearest cent)}\)
A network of towns and the distances between them in kilometres is shown.
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a. \(TYWH\)
b. \(\text{Length of shortest path}\ (YWHMG) = 89 \text{km}\)
a. \(TYH=30+38=68, \quad TYWH=30+15+20=65\)
\(\therefore \text{ Shortest Path is}\ TYWH.\)
b. \(Y W C M G=15+25+25+25=90\)
\(YWHMG=15+20+29+25=89\)
\(\Rightarrow \ \text{All other paths are longer.}\)
\(\therefore\text{ Length of shortest path = 89 km}\)
A project involving nine activities is shown in the network diagram.
The duration of each activity is not yet known.
The following table gives the earliest start time (EST) and latest start time (LST) for three of the activities. All times are in hours.
\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Activity} \rule[-1ex]{0pt}{0pt} & EST & LST \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ 0\ \ \ \ \ \ & \ \ \ \ \ \ 2\ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 0 & 1 \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 12 & 12 \\
\hline
\end{array}
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a. \(\text{Critical Path:}\ BEGI\)
b. \(\text{Duration of}\ I =7\ \text{hours}\)
c. \(\text{Max tim}\ =9\ \text{hours}\)
a. \(\text{Activity}\ A\ \text{and}\ C: \ LST \gt EST\)
\(\Rightarrow\ \text{Activity}\ A\ \text{and}\ C\ \text{not on critical path.}\)
\(\text{Critical Path:}\ BEGI\)
b. \(\text{Duration of}\ I = 19-12=7\ \text{hours}\)
c. \(\text{Activity}\ I\ \text{must start at 12 hours (on critical path).}\)
\(\text{Given duration activity}\ C = 3:\)
\(\text{Max time from start of}\ F\ \text{to end of}\ H = 12-3=9\ \text{hours}\)
Anna is sitting in a carriage of a Ferris wheel which is revolving. The height, \(A(t)\), in metres above the ground of the top of her carriage is given by
\(A(t)=c-k\,\cos\Big( \dfrac{\pi t}{24}\Big) \),
where \(t\) is the time in seconds after Anna's carriage first reaches the bottom of its revolution and \(c\) and \(k\) are constants.
The top of each carriage reaches a greatest height of 39 metres and a smallest height of 3 metres.
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\(B(t)=c-k\,\cos\Big( \dfrac{\pi}{24}(t-6)\Big) \),
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a. \(c=\ \text{centre of motion}\ = \dfrac{39+3}{2} = 21\)
\(k=\ \text{amplitude}\ = \dfrac{39-3}{2}=18\)
b. \(T= 48\ \text{seconds}\)
c. \(h_1=4.37\ \text{m,}\ h_2=37.63\ \text{m}\)
a. \(c=\ \text{centre of motion}\ = \dfrac{39+3}{2} = 21\)
\(k=\ \text{amplitude}\ = \dfrac{39-3}{2}=18\)
b. \(A(t)=21-18\,\cos\Big( \dfrac{\pi t}{24}\Big)\ \Rightarrow\ \ n=\dfrac{\pi}{24} \)
\(T=\dfrac{2\pi}{n} = 2\pi \times \dfrac{24}{\pi} = 48\ \text{seconds}\)
c. \(\text{Billie’s carriage is 6 seconds behind Anna’s.}\)
\(\text{Angle between the 2 carriages}\ = \dfrac{\pi \times 6}{24} = \dfrac{\pi}{4} \)
\(\text{By inspection:}\)
\(\text{Heights are the same:}\)
\(h_1=21-18\cos(\dfrac{\pi}{8}) = 4.370… = 4.37\ \text{m (2 d.p.)}\)
\(h_2=21-18\cos(\dfrac{7\pi}{8}) = 37.629… = 37.63\ \text{m (2 d.p.)}\)
In a game, the probability that a particular player scores a goal at each attempt is 0.15.
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a. \(0.7225\)
b. \(n=10\)
a. \(P(G)=0.15, \ \ P(\bar{G})=0.85\)
\(P(\bar{G}\bar{G}) = 0.85^2=0.7225\)
b. \(\text{2 attempts:}\ P(\text{at least 1 goal})=1-P(\bar{G}\bar{G})=1-0.85^{2}\)
\(\text{3 attempts:}\ P(\text{at least 1 goal})=1-0.85^{3}\)
\(\text{n attempts:}\ P(\text{at least 1 goal})=1-0.85^{n}\)
\(\text{Find}\ n\ \text{such that:}\)
\(1-0.85^{n}\) | \(\gt 0.8\) | |
\(0.85^{n}\) | \(\lt 0.2\) | |
\(n \times\ln(0.85)\) | \(\lt \ln(0.2)\) | |
\(n\) | \(\gt \dfrac{\ln(0.2)}{\ln(0.85)}\) | |
\(\gt 9.9…\) |
\(\therefore\ \text{Least}\ n=10\)
$\(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\) $ represents which gas law?
\(B\)
→ The equation $\(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\)$represents Charles’ Law, which states that the volume of a gas is directly proportional to its temperature, provided the pressure and amount of gas remain constant.
\(\Rightarrow B\)
According to Gay-Lussac's Law, the pressure of a gas is directly proportional to which quantity, assuming the volume and the number of moles are constant?
\(B\)
→ Gay-Lussac’s Law states that the pressure of a gas is directly proportional to its temperature, provided the volume and number of moles remain constant.
→ This can be written mathematically like \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\).
\(\Rightarrow B\)
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a. \(\dfrac{dy}{dx}=2x\,\tan\,x + x^2\,\sec^{2}x\)
b. \(x^{2}\tan^{2}x-\dfrac{x^{3}}{3}+x+C\)
a. \(y=x^{2}\tan\,x\)
\(\text{By product rule:}\)
\(\dfrac{dy}{dx}=2x\,\tan\,x + x^2 \sec^{2}x\)
b. \(\displaystyle \int (x\,\tan\,x+1)^{2}\,dx\)
\[=\int x^2\tan^{2}x + 2x\,\tan\,x +1\ dx\]
\[=\int x^{2}(\sec^{2}x-1)+2x\,\tan\,x +1\,dx\]
\[=\int 2x\,\tan\,x + x^{2}\sec^{2}x-x^{2}+1\,dx\]
\[=x^{2}\tan\,x-\dfrac{x^{3}}{3}+x+C\]
The graph of the function \(g(x)\) is shown.
Using the graph, complete the table with the words positive, zero or negative as appropriate. (3 marks)
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\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{\(x\)-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of \(g(x)\) at \(x\)} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of \(g(x)\) at \(x\)} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=-3\)} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\rule{0pt}{2.5ex} \text{\(x=1\)} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\rule{0pt}{2.5ex} \text{\(x=5\)} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\end{array}
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{\(x\)-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of \(g(x)\) at \(x\)} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of \(g(x)\) at \(x\)} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=-3\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{negative} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=1\)} \rule[-1ex]{0pt}{0pt} & \text{zero} \rule[-1ex]{0pt}{0pt} & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=5\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\end{array}
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{\(x\)-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of \(g(x)\) at \(x\)} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of \(g(x)\) at \(x\)} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=-3\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{negative} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=1\)} \rule[-1ex]{0pt}{0pt} & \text{zero} \rule[-1ex]{0pt}{0pt} & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=5\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\end{array}
What is \( {\displaystyle \int(6 x+1)^3 d x} \) ?
\( A \)
\[ \int(6 x+1)^3 dx\] | \(=\dfrac{1}{4} \cdot \dfrac{1}{6}(6 x+1)^4+C\) | |
\(=\dfrac{1}{24}(6 x+1)^4+C\) |
\( \Rightarrow A \)
A piece of zinc weighing 3.20 grams is placed into a beaker containing 300.0 mL of 0.7500 mol/L hydrochloric acid.
\(\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}\)
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a. Zinc is the limiting reagent.
b. \(1.21\ \text{L}\)
a. \(\ce{n(Zn)} = \dfrac{m}{MM} = \dfrac{3.20}{65.38}=0.0489\ \text{mol}\)
\(\ce{n(HCl)} = c \times V = 0.75 \times 0.3 = 0.225\)
→ Based on the mole ratio, \(0.0489\ \text{mol}\) of Zinc would require \(0.0978\ \text{mol}\) of hydrochloric acid.
→ As there is excess hydrochloric acid, Zinc is the limiting reagent.
b. The Mole ratio of \(\ce{Zn:H2}\) is \(1:1\)
→ \(0.0489\ \text{mol}\) of \(\ce{H2(g)}\) is produced.
→ At \(25^{\circ}\text{C}\) and \(100\ \text{kPa}\), \(1\) mole of gas \(=24.79\ \text{L}\)
→ \(\ce{V(H2(g))} = 0.0489 \times 24.79 = 1.21\ \text{L}\)
Pia's marks in Year 10 assessments are shown. The scores for each subject were normally distributed.
\begin{array}{|l|c|c|c|}
\hline & \textit {Pia's mark} & \textit {Year 10 mean} & \textit {Year 10 standard} \\
&&&\textit {deviation}\\
\hline \text {English} & 78 & 66 & 6 \\
\hline \text {Mathematics} & 80 & 71 & 10 \\
\hline \text {Science} & 77 & 70 & 15 \\
\hline \text {History} & 85 & 72 & 9 \\
\hline
\end{array}
In which subject did Pia perform best in comparison with the rest of Year 10?
\(A\)
\(\text {Consider the z-score of each option:}\)
\(z \text {-score (English)}=\dfrac{78-66}{6}=2\)
\(z \text {-score (Maths)}=\dfrac{80-71}{10}=0.9\)
\(z \text {-score (Science) }=\dfrac{77-70}{15}=0.46 \ldots\)
\(z \text {-score (History})=\dfrac{85-72}{9}=1.4 \ldots\)
\(\Rightarrow A\)
The curves `y=(x-1)^2` and `y=5-x^2` intersect at two points, as shown in the diagram.
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a. `x=2\ \text{and}\ -1`
b. `9\ \text{units}^2`
a. `y=(x-1)^2, \ y=5-x^2`
\(\text{Intersection occurs when:}\)
`(x-1)^2` | `=5-x^2` | |
`x^2-2x+1` | `=5-x^2` | |
`2x^2-2x-4` | `=0` | |
`2(x-2)(x+1)` | `=0` |
`x=2\ \text{and}\ -1`
b. | `\text{Area}` | `= \int_{-1}^{2} (5-x^2)-(x-1)^2\ dx` |
`=\int_{-1}^{2} 5-x^2-x^2+2x-1\ dx` | ||
`=\int_{-1}^{2}4-2x^2+2x\ dx` | ||
`=[4x-\frac{2}{3}x^3+x^2]_{-1}^{2}` | ||
`=[(8-\frac{16}{3}+4)-(-4+\frac{2}{3}+1)]` | ||
`=\frac{20}{3}-(-\frac{7}{3})` | ||
`=9\ \text{u}^2` |
Find the sum of the terms in the arithmetic series
\(50 + 57 + 64 +\ ...\ +2024\) (3 marks)
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\(293\,471\)
\(\text{AP where}\ \ a=50, d=57-50=7\)
\(\text{Last term}\) | \(=a + (n-1)d\) | |
\(2024\) | \(=50+(n-1)d\) | |
\(n-1\) | \(=\dfrac{2024-50}{7}\) | |
\(n\) | \(=283\) |
\(S_{283}\) | \(=\dfrac{n}{2}(a + l)\) | |
\(=\dfrac{283}{2}(50+2024)\) | ||
\(=293\,471\) |
In a group of 60 students, 38 play basketball, 35 play hockey and 5 do not play either basketball or hockey.
How many students play both basketball and hockey?
\(B\)
\(\text{Method 1:}\)
\(\text{Method 2:}\)
\(n(B \cup H)\) | \(=n(B) + n(H)-n(B \cap H) \) | |
\(55\) | \(=38+35-n(B \cap H) \) | |
\(n(B \cap H) \) | \(=18\) |
\( \Rightarrow B \)
Consider the function shown.
Which of the following could be the equation of this function?
\(C\)
\(\text {Gradient is negative (top left } \rightarrow \text { bottom right)}\)
\(y \text{-intercept = 3 (only positive option)}\)
\(\Rightarrow C\)
If \(x=-2.531\), what is the value of \(x^2\) rounded to 2 decimal places?
\(D\)
\( x^2\) | \(=(-2.531)^2 \) | |
\(=6.405\ldots \) | ||
\(=6.41\) |
\(\Rightarrow D\)
Metal | Reaction when heated with oxygen | Reaction when heated with water |
\(\ce{Mg}\) | \(\text{burns readily if powered to form oxides}\) | \(\ce{\text{forms}\ OH\ \text{ions and hydrogen gas}}\) |
\(\ce{Al}\) | \(\text{reacts with steam to form oxide ions and hydrogen gas}\) | |
\(\ce{Zn}\) | ||
\(\ce{Fe}\) |
Using the table above, which of the following equations correctly represents the reaction of aluminium with water?
\(D\)
→ From the table, aluminium reacts with gaseous water to form aluminium oxide and hydrogen gas.
\(\Rightarrow D\)
The image below shows a dinosaur fossil found in South Africa believed to be 200 million years old.
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a. Diet: herbivore
→ The presence of flat teeth in a dinosaur species strongly indicates a herbivorous diet, as these teeth are well-suited for grinding and processing plant material.
b. Digestive tract features:
→ Herbivorous dinosaurs likely possessed specialised digestive tracts adapted for processing plant material.
→ One key feature would be an enlarged caecum, a pouch-like structure connected to the large intestine, which housed symbiotic bacteria to break down cellulose through fermentation. This process would have allowed dinosaurs to extract more nutrients from tough plant matter.
→ Additionally, these dinosaurs may have had elongated intestines to increase the surface area for nutrient absorption and provide more time for the digestion of fibrous plant material.
a. Diet: herbivore
→ The presence of flat teeth in a dinosaur species strongly indicates a herbivorous diet, as these teeth are well-suited for grinding and processing plant material.
b. Digestive tract features:
→ Herbivorous dinosaurs likely possessed specialised digestive tracts adapted for processing plant material.
→ One key feature would be an enlarged caecum, a pouch-like structure connected to the large intestine, which housed symbiotic bacteria to break down cellulose through fermentation. This process would have allowed dinosaurs to extract more nutrients from tough plant matter.
→ Additionally, these dinosaurs may have had elongated intestines to increase the surface area for nutrient absorption and provide more time for the digestion of fibrous plant material.
Examine the diagram provided, which depicts the biological relationships within an ecosystem.
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a. Food webs vs food chains:
→ Food webs and food chains both represent energy flow in ecosystems, but differ in complexity and scope.
→ While a food chain shows a single, linear path of energy transfer from one organism to another, a food web illustrates multiple interconnected food chains within an ecosystem.
→ Food webs thus provide a more comprehensive and realistic depiction of the complex feeding relationships and energy transfers that occur among various species.
b. Second trophic level (\Rightarrow\) first order consumer
Zebra or Impala
c. Ecological consequences:
→ The decline or extinction of impalas would impact lions and vultures as they are an important food source of both.
→ Zebras would benefit from increased available vegetation due to reduced competition from impalas, leading to potential population growth.
→ The potential increase in zebra population would provide a more abundant food source for lions and vultures eventually, mitigating but not replacing the loss of impalas to these predator.
→ This scenario illustrates the complex interdependencies within the ecosystem and the cascading effects of species loss on different trophic levels.
a. Food webs vs food chains:
→ Food webs and food chains both represent energy flow in ecosystems, but differ in complexity and scope.
→ While a food chain shows a single, linear path of energy transfer from one organism to another, a food web illustrates multiple interconnected food chains within an ecosystem.
→ Food webs thus provide a more comprehensive and realistic depiction of the complex feeding relationships and energy transfers that occur among various species.
b. Second trophic level (\Rightarrow\) first order consumer
Zebra or Impala
c. Ecological consequences:
→ The decline or extinction of impalas would impact lions and vultures as they are an important food source of both.
→ Zebras would benefit from increased available vegetation due to reduced competition from impalas, leading to potential population growth.
→ The potential increase in zebra population would provide a more abundant food source for lions and vultures eventually, mitigating but not replacing the loss of impalas to these predator.
→ This scenario illustrates the complex interdependencies within the ecosystem and the cascading effects of species loss on different trophic levels.
Consider a grassland ecosystem with a population of rabbits, foxes, and various grass species.
a. → Predation occurs when foxes hunt and eat rabbits.
b. Competition among rabbits could result from (choose two):
→ limited resources such as food or water
→ suitable burrow sites
→ mating partners
c. Removing foxes:
→ could lead to an increase in the rabbit population.
→ this growing population might then overgraze the grass species, potentially causing a decline in grass biodiversity and altering the ecosystem’s structure.
a. → Predation occurs when foxes hunt and eat rabbits.
b. Competition among rabbits could result from (choose two):
→ limited resources such as food or water
→ suitable burrow sites
→ mating partners
c. Removing foxes:
→ could lead to an increase in the rabbit population.
→ this growing population might then overgraze the grass species, potentially causing a decline in grass biodiversity and altering the ecosystem’s structure.
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a. Abiotic factor
b. Soil pH:
→ Affects plant growth by influencing nutrient availability and solubility.
→ Most plants thrive in slightly acidic to neutral soils (pH 6.0-7.0) where essential nutrients are most accessible.
c. Plants alter soil pH (choose 1):
→ through root exudates, which release organic acids into the soil.
→ through the decomposition of their leaf litter, which can increase soil acidity over time.
a. Abiotic factor
b. Soil pH:
→ Affects plant growth by influencing nutrient availability and solubility.
→ Most plants thrive in slightly acidic to neutral soils (pH 6.0-7.0) where essential nutrients are most accessible.
c. Plants alter soil pH (choose 1):
→ through root exudates, which release organic acids into the soil.
→ through the decomposition of their leaf litter, which can increase soil acidity over time.
Our actions as a human species are inadvertently altering the evolutionary trajectories of countless organisms.
Explain two distinct mechanisms by which human activities exert selection pressures on other species. For each mechanism, provide a specific example of a species affected by this pressure. (4 marks)
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Answers could include two of the following.
Mechanism: urbanisation
→ Human activities exert selection pressure on other species through urbanisation. As cities expand, isolated pockets of natural habitat are created, forcing species to adapt to smaller, disconnected areas.
→ For example, the San Diego deer mouse has shown rapid evolution in its size and fur colour to better camouflage against urban environments, with urban mice becoming larger and darker than their rural counterparts.
Mechanism: pollution from fossil fuel burning
→ Humans exert selection pressure on other species through pollution from fossil fuel burning.
→ The emission of sulfur-dioxide from coal-burning power plants has led to acid rain, which changes soil and water pH levels. This has resulted in strong selection pressure on aquatic organisms.
→ For instance, in some Scandinavian lakes, the European perch has evolved increased tolerance to acidic conditions in order to survive.
Mechanism: use of pesticides
→ Another mechanism is the widespread use of pesticides, which creates strong selection pressures for resistance. In agriculture, the overuse of pesticides has led to the evolution of resistance in many insect pest species.
→ A specific example is the green peach aphid, which has developed resistance to multiple classes of insecticides. This adaptation makes it increasingly difficult to control these crop pests without resorting to even more potent chemicals.
Answers could include two of the following.
Mechanism: urbanisation
→ Human activities exert selection pressure on other species through urbanisation. As cities expand, isolated pockets of natural habitat are created, forcing species to adapt to smaller, disconnected areas.
→ For example, the San Diego deer mouse has shown rapid evolution in its size and fur colour to better camouflage against urban environments, with urban mice becoming larger and darker than their rural counterparts.
Mechanism: pollution from fossil fuel burning
→ Humans exert selection pressure on other species through pollution from fossil fuel burning.
→ The emission of sulfur-dioxide from coal-burning power plants has led to acid rain, which changes soil and water pH levels. This has resulted in strong selection pressure on aquatic organisms.
→ For instance, in some Scandinavian lakes, the European perch has evolved increased tolerance to acidic conditions in order to survive.
Mechanism: use of pesticides
→ Another mechanism is the widespread use of pesticides, which creates strong selection pressures for resistance. In agriculture, the overuse of pesticides has led to the evolution of resistance in many insect pest species.
→ A specific example is the green peach aphid, which has developed resistance to multiple classes of insecticides. This adaptation makes it increasingly difficult to control these crop pests without resorting to even more potent chemicals.
Which of the following observations indicates a chemical change has occurred?
\(B\)
→ A chemical change is one where a new substance is created and cannot be reversed.
→ The changing temperature of a mixed solution is the result of bonds breaking and forming new products thus indicating a chemical change.
→ All other options can be reversed by physical processes and therefore represent physical changes only.
\(\Rightarrow B\)
Ecosystems are dynamic, shaped not only by physical forces but also by the living organisms within them. Including a specific example, explain one biotic factor that has significantly impacted past ecosystems:
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a. Short timescale biotic factor: introduction of invasive species
→ The introduction of cane toads in Australia in 1935 has rapidly altered local ecosystems within decades.
→ Cane toads have caused declines in native predator populations that attempt to eat the toxic toads, leading to cascading effects throughout the food web.
b. Geological timescale biotic factor: evolution of land plants
→ The development of land plants around 470 Mya led to increased oxygen production, soil formation, and the creation of new habitats.
→ This gradual but profound change altered atmospheric composition and weather patterns. This reshaped terrestrial ecosystems and paved the way for the evolution of terrestrial animal life.
a. Short timescale biotic factor: introduction of invasive species
→ The introduction of cane toads in Australia in 1935 has rapidly altered local ecosystems within decades.
→ Cane toads have caused declines in native predator populations that attempt to eat the toxic toads, leading to cascading effects throughout the food web.
b. Geological timescale biotic factor: evolution of land plants
→ The development of land plants around 470 Mya led to increased oxygen production, soil formation, and the creation of new habitats.
→ This gradual but profound change altered atmospheric composition and weather patterns. This reshaped terrestrial ecosystems and paved the way for the evolution of terrestrial animal life.
A student tested how soluble silver salts are by reacting a 0.1 mol L\(^{-1}\) silver nitrate solution with 0.1 mol L\(^{-1}\) solutions of calcium hydroxide, calcium chloride, and calcium sulfate. The results are shown below:
\begin{array} {|l|l|}
\hline \ \ \ \ \ \text{Compound} & \ \ \ \ \ \text{Observation} \\
\hline \text{calcium hydroxide} & \text{No reaction} \\
\hline \text{calcium chloride} & \text{White precipitate} \\
\hline \text{calcium sulfate} & \text{No reaction} \\
\hline \end{array}
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a. \(\ce{CaCl2(aq) + 2AgNO3(aq) -> Ca(NO3)2(aq) + 2AgCl(s)}\)
b. The white precipitate is \(\ce{AgCl}\) → silver chloride.
a. \(\ce{CaCl2(aq) + 2AgNO3(aq) -> Ca(NO3)2(aq) + 2AgCl(s)}\)
b. The white precipitate is \(\ce{AgCl}\) → silver chloride.
The graph below shows the concentration of \(\ce{CO2}\) in the earth's atmosphere over the last 800 years.
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a. Data would be obtained by gas analysis within ice cores.
b. Graph shape over the past 800 years:
→ Atmospheric levels of \(\ce{CO2}\) were very steady between the period 1200-1800 at approximately 280 ppm.
→ In the period 1800-2000, \(\ce{CO2}\) levels increased exponentially from 280 ppm to 370 ppm.
→ This increase coincided with industrialisation which saw the widespread use of fossil fuels like coal, oil, and natural gas for energy.
→ The use of these energy sources released large amounts of (\ce{CO2}\) into the atmosphere and is regarded as a major contributing factor to rise in atmospheric \(\ce{CO2}\) levels over this period.
a. Data would be obtained by gas analysis within ice cores.
b. Graph shape over the past 800 years:
→ Atmospheric levels of \(\ce{CO2}\) were very steady between the period 1200-1800 at approximately 280 ppm.
→ In the period 1800-2000, \(\ce{CO2}\) levels increased exponentially from 280 ppm to 370 ppm.
→ This increase coincided with industrialisation which saw the widespread use of fossil fuels like coal, oil, and natural gas for energy.
→ The use of these energy sources released large amounts of (\ce{CO2}\) into the atmosphere and is regarded as a major contributing factor to rise in atmospheric \(\ce{CO2}\) levels over this period.
Scientists analyse the ratio of \(\ce{^{16}O}\) to \(\ce{^{18}O}\) isotopes in various geological samples to reconstruct past climatic conditions.
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a. Scientists analyse gas trapped within ice cores.
b. \(\ce{^{16}O : ^{18}O}\) ratios
→ The ratio of \(\ce{^{16}O : ^{18}O}\) in geological samples reflects past temperature and precipitation patterns, which are crucial factors in shaping ecosystems.
→ During warmer periods, there is a higher proportion of the heavier \(\ce{^{18}O}\) isotope in precipitation.
→ During cooler periods, there is a relative increase in lighter \(\ce{^{16}O}\).
→ These isotopic signatures allow scientists to reconstruct past climate conditions and infer associated ecosystem changes.
a. Scientists analyse gas trapped within ice cores.
b. \(\ce{^{16}O : ^{18}O}\) ratios
→ The ratio of \(\ce{^{16}O : ^{18}O}\) in geological samples reflects past temperature and precipitation patterns, which are crucial factors in shaping ecosystems.
→ During warmer periods, there is a higher proportion of the heavier \(\ce{^{18}O}\) isotope in precipitation.
→ During cooler periods, there is a relative increase in lighter \(\ce{^{16}O}\).
→ These isotopic signatures allow scientists to reconstruct past climate conditions and infer associated ecosystem changes.
Which of the following are the products of the complete combustion of propane, \(\ce{C3H8}\)?
\(D\)
→ A reaction which under goes complete combustion will produce carbon dioxide and water.
→ Carbon monoxide is a product of incomplete combustion which occurs when there is a lack of oxygen available.
\(\Rightarrow D\)