Draw Lewis electron dot structures for
- a nitrogen molecule \( (\ce{N2})\) (1 mark)
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- an oxygen molecule \( (\ce{O2}) \) (1 mark)
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Aussie Maths & Science Teachers: Save your time with SmarterEd
Draw Lewis electron dot structures for
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i.
ii.
i.
ii.
When proteins are broken down in chemical digestion, they become
\(D\)
→ Proteins are broken down so as their beneficial components can be absorbed in the small intestine.
→ Amino acid chains (called polypeptide chains) make up all proteins, and are hence the smallest unit of a protein possible in chemical digestion.
\(\Rightarrow D\)
A soccer ball leaves the ground with a velocity of 10 ms\(^{-1}\) at 40° above the horizontal. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- i. ii. \(v_v=6.43\ \text{ms}^{-1}\) \(v_h=7.66\ \text{ms}^{-1}\) i. ii. \(v_v=10\sin40^{\circ}=6.43\ \text{ms}^{-1}\) \(v_h=10\cos40^{\circ}=7.66\ \text{ms}^{-1}\)
A ball is thrown vertically upwards from ground level, it gains 50 metres vertically and then falls back to the ground. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- i. \(u=31.3\ \text{ms}^{-1}\) ii. \(6.39\ \text{s}\). i. \(\text{At 50 m}\ \Rightarrow\ v=0\ \text{ms}^{-1}\) ii. \(\text{Time to highest point}\ = \dfrac{1}{2}\ \text{time of flight} \) \(\text{Total time of flight}\ = 2\times 3.194 = 6.39\ \text{s (2 d.p.)}\)
\(v^2\)
\(=u^2+2as\)
\(0^2\)
\(=u^2+2 \times -9.8 \times 50\)
\(u^2\)
\(=980\)
\(u\)
\(=31.3\ \text{ms}^{-1}\)
\(v\)
\(=u+at\)
\(0\)
\(=31.3-9.8t\)
\(9.8t\)
\(=31.3\)
\(t\)
\(=3.194\ \text{s}\)
Complete the table below by outlining one physical indication of each chemical change described below. (3 marks)
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}
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\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication} \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} & \text{Solid forms and falls out of solution} \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} & \text{Generation of heat and light} \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} & \text{Formation of gas (bubbles)} \\
\hline
\end{array}
\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication} \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} & \text{Solid forms and falls out of solution} \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} & \text{Generation of heat and light} \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} & \text{Formation of gas (bubbles)} \\
\hline
\end{array}
Outline both a physical and chemical property of the elements contained in the highlighted periodic table groups below. (4 marks)
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Group \(\text{A:}\) Alkali metals
→ Physical properties: Group 1 metals are soft and shiny.
→ Chemical properties: They are highly reactive and have low melting/boiling points.
Group \(\text{B:}\) Noble gases
→ Physical properties: Group 8 elements are colourless gases with no odour or taste.
→ Chemical properties: They are virtually unreactive and all have cryogenic boiling points.
Group \(\text{A:}\) Alkali metals
→ Physical properties: Group 1 metals are soft and shiny.
→ Chemical properties: They are highly reactive and have low melting/boiling points.
Group \(\text{B:}\) Noble gases
→ Physical properties: Group 8 elements are colourless gases with no odour or taste.
→ Chemical properties: They are virtually unreactive and all have cryogenic boiling points.
Name the following inorganic compounds:
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i. Sulfuric Acid
ii. Sulfurous Acid
Name the following inorganic compounds:
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i. Magnesium Carbonate
ii. Mercury Bromide
Calculate the mass of \(\ce{H}\) in 15.2 grams of \(\ce{H2O}\). Give your answer in grams correct to two decimal places. (2 marks)
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\(1.70\ \text{grams}\)
\(\ce{MM(H2O) = 2 \times 1.008 + 16.00 = 18.016\ \text{g mol}^{-1}}\)
\(\ce{\% of H = \dfrac{2.016}{18.016} \times 100 = 11.19\%}\)
\(\ce{m(H)\ \text{within 15.2 grams}\ H2O = 11.19\% \times 15.2 = 1.70\ \text{grams (to 2 d.p.)}} \)
Greenhouses have been used to generate higher crop yields than open-field agriculture. To encourage plant growth in greenhouses, the conditions required for photosynthesis are controlled. Commercial greenhouses, like the ones shown below, often use a lot of energy for heating, ventilation, lighting and water.
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a. Temperature maintenance:
→ Photosynthesis is regulated by different enzymes which are intrinsic to numerous processes involved in photosynthesis.
→ If the temperature of a greenhouse is changed, then the enzymes may decrease in activity or denature, causing a reduction in the rate of photosynthesis or even ceasing it all together.
→ This can cause problems in the plants such as a reduction in growth or plant death.
b. Crop yield increases:
→ Red wavelengths of light are used during photosynthesis, while green light is reflected off the plants.
→ This technology will allow an increase in crop yields by increasing the amount of red light which increases the rate of photosynthesis.
a. Temperature maintenance:
→ Photosynthesis is regulated by different enzymes which are intrinsic to numerous processes involved in photosynthesis.
→ If the temperature of a greenhouse is changed, then the enzymes may decrease in activity or denature, causing a reduction in the rate of photosynthesis or even ceasing it all together.
→ This can cause problems in the plants such as a reduction in growth or plant death.
b. Crop yield increases:
→ Red wavelengths of light are used during photosynthesis, while green light is reflected off the plants.
→ This technology will allow an increase in crop yields by increasing the amount of red light which increases the rate of photosynthesis.
A light ray from a laser passes from a glucose solution \((n=1.44)\) into the air \((n=1.00)\), as shown in Figure 12. --- 2 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) --- a. \(\theta_c=44^{\circ}\) → When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur. → As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer. c. Observer cannot see the laser: → When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur. → As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer.
a.
\(\sin\theta_c\)
\(=\dfrac{n_2}{n_1}\)
\(\theta_c\)
\(=\sin^{-1}\Big{(}\dfrac{1.00}{1.44}\Big{)}\)
\(=44^{\circ}\)
Ancient cave drawings contribute evidence of cognitive changes that are unique to modern humans, Homo sapiens. Biologists argue that these unique cognitive changes in \(H\). sapiens allowed cultural evolution that would have been impossible in other hominin species.
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A team of scientists dated the mineral deposits in the solid, transparent layer using uranium-thorium dating. They found that some of the mineral deposits are 65 000 years old.
What conclusion can be made about the age of the ancient drawings on the stalagmites? Justify your response. (2 marks)
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\( \textbf{Hominin species} \) |
\( \textbf{First appearance in Europe} \) \( \textbf{(years before present)} \) |
\( \textbf{Time of extinction} \) \( \textbf{(years before present)} \) |
\( \text{Homo sapiens} \) | \(45\ 000\) | \( \text{not extinct}\) |
\( \text{Homo neanderthalensis} \) | \(130\ 000\) | \(30\ 000\) |
\( \text{Homo heidelbergensis} \) | \(800\ 000-600\ 000\) | \(400\ 000-200\ 000\) |
\( \text{Homo evectus} \) | \(1\ 200\ 000-600\ 000\) | \(143\ 000\) |
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a. → Cave drawings show information being passed through generations.
→ The paintings often depict hunting techniques or dreaming stories.
→ Increasing complexity of these cave drawings over time shows the development of complex thought and symbolic relationships in H. sapien tribes.
b. → The drawings must be a minimum of 65,000 years old.
→ This is because they must be at least as old as the mineral layer on top of them.
c. → H. sapiens are believed to have arrived in Europe 45,000 years ago.
→ The dating shows the drawings are a minimum of 65,000 years old.
→ Therefore the data suggests that either H. neanderthalensis drew the art, or H. sapiens arrived earlier than currently believed.
a. → Cave drawings show information being passed through generations.
→ The paintings often depict hunting techniques or dreaming stories.
→ Increasing complexity of these cave drawings over time shows the development of complex thought and symbolic relationships in H. sapien tribes.
b. → The drawings must be a minimum of 65,000 years old.
→ This is because they must be at least as old as the mineral layer on top of them.
c. → H. sapiens are believed to have arrived in Europe 45,000 years ago.
→ The dating shows the drawings are a minimum of 65,000 years old.
→ Therefore the data suggests that either H. neanderthalensis drew the art, or H. sapiens arrived earlier than currently believed.
Lee listens while a police car with a loud siren comes towards her, travels past her and then continues on away from her.
Compared with the sound she would hear from the siren if the police car were stationary, the sound has.
\(A\)
→ By the application of the doppler effect, the frequencies of sounds increase when moving towards an object and decrease when moving away from an object.
\(\Rightarrow A\)
A model car of mass 2.0 kg is propelled from rest by a rocket motor that applies a constant horizontal force of 4.0 N, as shown below. Assume that friction is negligible.
With the same rocket motor, the car accelerates from rest for 10 seconds.
Calculate the the final speed of the model car? (2 marks)
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\(20\ \text{ms}^{-1}\)
\(a=\dfrac{F}{m} = \dfrac{4.0}{2.0} = 2\ \text{ms}^{-2} \)
\(v\) | \(=u+at\) | |
\(=0+2 \times 10\) | ||
\(=20\ \text{ms}^{-1}\) |
A model car of mass 2.0 kg is propelled from rest by a rocket motor that applies a constant horizontal force of 4.0 N, as shown below. Assume that friction is negligible.
Which one of the following best gives the magnitude of the impulse given to the car by the rocket motor in the first 5.0 seconds?
\(C\)
\(I\) | \(=Ft\) | |
\(=4 \times 5\) | ||
\(=20 \text{ N s}\) |
\(\Rightarrow C\)
Two large charged plates with equal and opposite charges are placed close together, as shown in the diagram below.
A distance of 5.0 mm separates the plates. The electric field between the plates is equal to 1000 N C\(^{-1}\).
Calculate the voltage difference between the plates. (2 marks)
Voltage difference is \(5\ \text{V}\).
\(V\) | \(=Ed\) | |
\(=1000 \times 5.0 \times 10^{-3}\) | ||
\(=5\ \text{V}\) |
Figure 13 shows two speakers, \(\text{A}\) and \(\text{ B}\), facing each other. The speakers are connected to the same signal generator/amplifier and the speakers are simultaneously producing the same 340 Hz sound.
Take the speed of sound to be 340 ms\( ^{-1}\).
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a. \(\lambda=1\ \text{m}\)
b. → At the centre of the speakers, there is an area of relative loudness
→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\).
→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive (loud) and destructive interference (quiet).
→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.
→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.
a. \(\lambda=\dfrac{v}{f}=\dfrac{340}{340}=1\ \text{m}\)
b. → At the centre of the speakers, there is an area of relative loudness
→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\).
→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive and destructive interference.
→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.
→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.
Alex hears the siren from a stationary fire engine.
Compared with the sound Alex hears from the stationary fire engine, the sound Alex will hear as the fire engine approaches him will have increased
\(C\) and \(D\)
→ \(D\) is correct by the doppler effect.
→ \(C\) is correct because as the fire engine approaches Alex, the sound will become louder which corresponds to an increase in wave amplitude.
A railway truck \(\text{X}\) of mass 10 tonnes, moving at 6.0 m s\(^{-1}\), collides with a stationary railway truck \(\text{Y}\) of mass 5.0 tonnes. After the collision the trucks are joined together and move off as one. The situation is shown below.
\(\text{Question 8}\)
The final speed of the joined railway trucks after the collision is closest to
\(\text{Question 9}\)
The collision of the railway trucks is best described as one where
\(\text{Question 8:}\ C\)
\(\text{Question 9:}\ B\)
\(\text{Question 8}\)
→ By the law of conservation of momentum:
\(m_Xu_X+m_Yu_Y\) | \(=m_Xv_X+m_Yv_Y\) | |
\(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \) | ||
\(10\ 000 \times 6 +0\) | \(= v(10\ 000 + 5000)\) | |
\(60\ 000\) | \(=15\ 000v\) | |
\(v\) | \(=4\ \text{ms}^{-1}\) |
\( \Rightarrow C\)
\(\text{Question 9}\)
→ By the law of conservation of momentum, momentum in the collision is conserved.
→ Kinetic energy conservation:
\(KE_{\text{init}}\) | \(=\dfrac{1}{2}m_Xu_X^2+\dfrac{1}{2}m_Yu_Y^2\) | |
\(=\dfrac{1}{2} \times 10\ 000 \times 6^2 + \dfrac{1}{2} \times 5000 \times 0^2\) |
||
\(=180\ 000\ \text{J}\) |
\(KE_{\text{final}}\) | \(=\dfrac{1}{2}(m_X+m_Y)v^2\) | |
\(=\dfrac{1}{2} \times 15\ 000 \times 4^2\) | ||
\(=120\ 000\ \text{J}\) |
→ The kinetic energy of the system decreases after the collision and so is not conserved.
\(\Rightarrow B\)
A straight wire carries a current of 10 A.
Which one of the following diagrams best shows the magnetic field associated with this current?
\(A\)
→ Using the right hand grip rule, the thumb points up and the fingers curl in an anticlockwise direction as seen from above.
\(\Rightarrow A\)
In an experimental set-up used to investigate standing waves, a 6.0 m length of string is fixed at both ends, as shown in Figure 12. The string is under constant tension, ensuring that the speed of the wave pulses created is a constant 40 ms\(^{-1}\). In an initial experiment, a continuous transverse wave of frequency 7.5 Hz is generated along the string. --- 3 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. \(\lambda=5.3\ \text{m}\) b. → A standing wave will not form. → A standing wave will form only if the wavelength of the wave is equal to an integer multiple \(\dfrac{\lambda}{2}\), as 5.3 m is not multiple of 3 m, a standing wave will not form. a. \(\lambda = \dfrac{v}{f} = \dfrac{40}{7.5} = 5.3\ \text{m} \) → A standing wave will only form if the length of the string is equal to an integer multiple of \(\dfrac{\lambda}{2}\). → Since 6.0 m is not an integer multiple of \(\dfrac{5.3}{2}\) m, a standing wave will not form.
b. → A standing wave will not form.
The graph below shows when the five major mass extinction events and several other mass extinction events occurred. It also shows the percentage of species that were lost in each event.
According to the graph, to be classified as a major mass extinction event, the percentage of species that are lost has to be at least
\(C\)
→ As the lowest % of species lost in a mass extinction event is 74% on this data set, then according the graph for an extinction event to be classified as major at least 74% of species must have been lost.
\(\Rightarrow C\)
In a series of experiments, the rate of photosynthesis in plant cells was measured in environments with different concentrations of carbon dioxide. All other variables were kept constant.
Which one of the following graphs reflects the trend that would be shown by the results of these experiments?
\(C\)
→ As carbon dioxide is a reactant in photosynthesis, increasing the concentration will lead to a higher rate of photosynthesis initially.
→ This rate of increase will plateau and eventually stop as all the active sites of involved enzymes are being used.
→ If this relationship was graphed, the shape seen would most accurately resemble that in graph \(C\).
\(\Rightarrow C\)
A 3.1g sample of \(\ce{CaCO3_{(s)}}\) decomposes into \(\ce{CaO_{(s)}}\) and \(\ce{CO2_{(g)}}\). Entropy values for these chemicals are given below and the molar enthalpy for the reaction is 360 kJ/mol.
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Substance}\rule[-1ex]{0pt}{0pt} & \text{Standard Entropy}\ (\Delta S) \\
\hline
\rule{0pt}{2.5ex}\ce{CaCO3}\rule[-1ex]{0pt}{0pt} & \text{92.88 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CaO(s)}\rule[-1ex]{0pt}{0pt} & \text{39.75 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CO2(g)}\rule[-1ex]{0pt}{0pt} & \text{213.6 J/K} \\
\hline
\end{array}
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i. \(\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}\)
ii. | \(\Delta S\) | \(=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}\) |
\(= 213.6 + 39.75-92.88\) | ||
\(= 160.47\ \text{J mol}^{-1}\ \text{K}^{-1}\) |
\(\ce{n(CaCO3)}= \dfrac{\text{m}}{\text{MM}} = \dfrac{3.1}{100.09} = 0.03097\ \text{mol} \)
\(\text{Entropy change}\ = 160.47 \times 0.03097 = 4.97\ \text{J K}^{-1}\)
iii. \(\text{Room Temperature = 298.15 K}\)
\(\Delta G\) | \(=\Delta H-T \Delta S\) | |
\(=360-(298.15 \times 0.16047) \) | ||
\(= 312.179\ \text{kJ}\) |
\(\text{Since}\ \Delta G > 0,\ \text{the reaction is not spontaneous.}\)
In the following reactions, predict whether entropy will increase or decrease, giving reasons. (3 marks)
i. → Decrease
→ 4 moles of gas becomes 2 moles of gas on the product side, causing the reaction to become more ordered and decreasing in entropy.
ii. → Increase
→ A solid decomposes into 2 moles of gas. This reaction becomes more disordered, and the phase change involves an increase in entropy.
iii. → Increase
→ An aqueous and solid reactant becomes an aqueous and gaseous product. The phase changes overall involve an increase in entropy.
i. → Decrease
→ 4 moles of gas becomes 2 moles of gas on the product side, causing the reaction to become more ordered and decreasing in entropy.
ii. → Increase
→ A solid decomposes into 2 moles of gas. This reaction becomes more disordered, and the phase change involves an increase in entropy.
iii. → Increase
→ An aqueous and solid reactant becomes an aqueous and gaseous product. The phase changes overall involve an increase in entropy.
Find the oxidation state of each element in the compound \(\ce{CuSO4}\). (2 marks)
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\(\ce{Cu +2, O -2, S +6}\)
\(\ce{Sulphate ion: SO4^{2-}}\)
\(\ce{Cu: +2\ \ \text{(compound is neutral)}}\)
\(\ce{O: -2}\)
\(\text{Let}\ \ x =\ \text{oxidation number of S}\)
\(2+x+(4 \times -2)\) | \(=0\) | |
\(x\) | \(= +6\) |
The oxidation state of phosphorus in the pyrophosphate ion \(\ce{P2O7^{4-}}\) is
\(B\)
\(\text{Let}\ x =\ \text{oxidation state of phosphorus} \)
\(2x + (7 \times -2)\) | \(=-4\) | |
\(2x-14\) | \(=-4\) | |
\(x\) | \(=+5\) |
\(\Rightarrow B\)
\(\ce{Cu(s) + 4HNO3(aq)\rightarrow Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)}\)
\(\text{Question 14}\)
Which one of the following will not increase the rate of the above reaction?
\(\text{Question 15}\)
In the above reaction, the number of successful collisions per second is a small fraction of the total number of collisions.
The major reason for this is that
\(\text{Question 14:}\ D\)
\(\text{Question 15:}\ D\)
\(\text{Question 14}\)
→ Options \(A, B\) and \(C\) will all increase the rate of the given chemical reaction.
\(\Rightarrow D\)
\(\text{Question 15}\)
→ Successful collisions occur only if the particles involved have at least the minimum kinetic energy required and the correct orientation.
\(\Rightarrow D\)
A helium balloon is inflated to a volume of 5.65 L and a pressure of 10.2 atm at a temperature of 25°C.
Calculate the amount of helium, in moles, in the balloon. (2 marks)
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\(2.36\ \text{mol}\)
\(\ce{n(He)}\) | \(= \dfrac{pV}{RT} \) | |
\(= \dfrac{10.2 \times 101.3 \times 5.65}{8.31 \times 298} \) | ||
\(= 2.36\ \text{mol} \) |
A home owner on a large property creates a backyard entertainment area. The entertainment area has a low-voltage lighting system. To operate correctly, the lighting system requires a voltage of 12 \(\text{V} _{\text{RMS}}\). The lighting system has a resistance of 12 \(\Omega\).
Calculate the power drawn by the lighting system. (3 mark) --- 4 WORK AREA LINES (style=lined) ---
\(12\ \text{W}\)
→ Using \(P=IV\) and \(I= \dfrac{I}{R} :\)
\(P\) | \(=\dfrac{V^2}{R}\) | |
\(=\dfrac{12^2}{12}\) | ||
\(=12\ \text{W}\) |
→ The power drawn from the lighting system is 12 \(\text{W}\).
Figure 2 shows two equal positive stationary point charges placed near each other.
Sketch on Figure 2 the shape and direction of the electric field lines. Use at least eight field lines. (2 marks)
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→ Field lines must not touch.
→ Fields will repel one another.
An aerosol can with a volume of 300.0 mL contains 2.80 g of propane gas as a propellant. The warning label says the aerosol may explode at temperatures above 60.0 °C.
What is the pressure in the can at a temperature of 60.0 °C? (2 marks)
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\(5.87 \times 10^{2}\ \text{kPa}\)
\(\ce{MM(C3H8) = 3 \times 12.0 + 8 \times 1.0 = 44.0\ \text{g mol}^{-1}} \)
\(\ce{n(C3H8) = \dfrac{2.80}{44.0} = 0.0636\ \text{mol}} \)
\(\ce{n(C3H8)}\) | \( = \dfrac{pV}{RT} \) | |
\(p(\ce{C3H8}) \) | \(= \dfrac{nRT}{V} \) | |
\(=0.0636 \times 8.31 \times \dfrac{(60.0 + 273)}{300.0 \times 10^{-3}} \) | ||
\(=5.87 \times 10^2\ \text{kPa} \) |
When concentrated sulfuric acid is added to dry sucrose, \(\ce{C12H22O11}\), a black residue of pure carbon is produced.
An equation for the reaction is
\(\ce{2C12H22O11(s)+ 2H2SO4(aq) + O2(g)\rightarrow 22C(s) + 2CO2(g) + 24H2O(g) + 2SO2(g)}\)
\(\text{Molar mass}\ \ce{(C12H22O11) = 342.0 \text{ g mol}^{–1}}\)
Calculate the mass of carbon residue that could be produced by the reaction of 50.0 g of sucrose with excess concentrated sulfuric acid. (2 marks)
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\(\text{19.3 grams}\)
\(\ce{n(C12H22O11)_{\text{react}} = \dfrac{50.0}{342.0} = 0.146\ \text{mol}} \)
\(\ce{n(C)_{\text{produced}} = 11 \times n(C12H22O11) = 11 \times 0.146 = 1.61\ \text{mol}} \)
\(\ce{m(C)_{\text{produced}} = 1.61 \times 12.0 = 19.3\ \text{g}} \)
The electric field between two parallel plates that are 1.0 × 10\(^{-2}\) m apart is 2.0 × 10\(^{-4}\) N C\(^{-1}\).
Calculate the voltage between the plates? (2 marks)
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\(2.0 \times 10^{-6}\ \text{V} \)
\(E\) | \(=\dfrac{V}{d}\) | |
\(V\) | \(=E \times d\) | |
\(=2.0 \times 10^{-4} \times 1.0 \times 10^{-2}\) | ||
\(=2.0 \times 10^{-6}\ \text{V} \) |
A 0.8 m long guitar string is set vibrating at a frequency of 250 Hz. The standing wave envelope created in the guitar string is shown in Figure 12.
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a. \(v=400\ \text{ms}^{-1}\)
b.
a. \(v\) | \(=f\lambda\) | |
\(=250 \times 1.6 \) | ||
\(=400\ \text{ms}^{-1}\) |
b.
Two blocks of mass 5 kg and 10 kg are placed in contact on a frictionless horizontal surface, as shown in the diagram below. A constant horizontal force, \(F\), is applied to the 5 kg block.
\(\text{Question 9}\)
Which one of the following statements is correct?
\(\text{Question 10}\)
If the force \(F\) has a magnitude of 250 N, what is the work done by the force in moving the blocks in a straight line for a distance of 20 m?
\(\text{Question 9:}\ C\)
\(\text{Question 10:}\ A\)
\(\text{Question 9}\)
Using Newton’s second Law: \(F=ma\ \ \Rightarrow\ \ a=\dfrac{F}{m}\).
→ The blocks will experience the same acceleration.
→ Both blocks will have the same force to mass ratio. Since the 5 kg block is half the mass of the 10 kg block, it will experience half the magnitude of the net force as the 10 kg block.
\(\Rightarrow C\)
\(\text{Question 10}\)
\(W\) | \(=F_{\parallel}s\) | |
\(=250 \times 20\) | ||
\(=5000\ \text{J}\) | ||
\(=5\ \text{kJ}\) |
\(\Rightarrow A\)
A distant fire truck travelling at 20 ms\(^{-1}\) to a fire has its siren emitting sound at a constant frequency of 500 Hz.
Chris is standing on the edge of the road. Assume that the fire truck is travelling directly towards him as it approaches and directly away from him as it goes past. The arrangement is shown in Figure 14.
a.
a.
A small sodium lamp, emitting light of wavelength 589 nm, is viewed at night through two windows from across a street. The glass of one window has a fine steel mesh covering it and the other window is open, as shown in Figure 18. Assume that the sodium lamp is a point source at a distance.
A Physics student is surprised to see a pattern formed by the light passing through the steel mesh but no pattern for the light passing through the open window. She takes a photograph of the observed pattern to show her teacher, who assures her that it is a diffraction pattern.
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a. Condition to satisfy:
→ The size of the gaps from the fine steel mesh must be of the same order of magnitude of the wavelengths of the light from the sodium lamp.
b. Open window differences:
→ The width of the window is significantly greater than wavelength of the light from the sodium lamp and the width of the glass molecules that make up the glass window is significantly smaller than the light from the sodium lamp.
→ These widths contrast greatly to the size of the steel mesh and are not of the same order of magnitude of the light from the sodium lamp, hence no diffraction will occur.
a. Condition to satisfy:
→ The size of the gaps from the fine steel mesh must be of the same order of magnitude of the wavelengths of the light from the sodium lamp.
b. Open window differences:
→ The width of the window is significantly greater than wavelength of the light from the sodium lamp and the width of the glass molecules that make up the glass window is significantly smaller than the light from the sodium lamp.
→ These widths contrast greatly to the size of the steel mesh and are not of the same order of magnitude of the light from the sodium lamp, hence no diffraction will occur.
Kym and Kelly are experimenting with trolleys on a ramp inclined at 25°, as shown in Figure 7. They release a trolley with a mass of 2.0 kg from the top of the ramp. The trolley moves down the ramp, through two light gates and onto a horizontal, frictionless surface. Kym and Kelly calculate the acceleration of the trolley to be 3.2 m s\(^{-2}\) using the information from the light gates.
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a.i. See Worked Solutions
a.ii. \(F_f=1.9\ \text{N}\)
b.i. \(2.0\ \text{ms}^{-1}\)
b.ii. For the collision to be elastic, the kinetic energy must be conserved.
\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)
\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)
→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.
a.i. The gravitational force down the slope:
\(F\) | \(=mg\, \sin \theta\) | |
\(=2.0 \times 9.8 \times \sin 25\) | ||
\(=8.3\ \text{N}\) |
a.ii. | \(F_{net}\) | \(=ma\) |
\(=2.0 \times 3.2\) | ||
\(=6.4\) |
\(6.4\) | \(=F-F_f\) | |
\(F_f\) | \(=8.3-6.4\) | |
\(=1.9\ \text{N}\) |
b.i. By the conservation of momentum:
\(m_1u_1+m_2u_2\) | \(=v(m_1+m_2)\) | |
\(2 \times 4 + 2 \times 0\) | \(=v(2 +2)\) | |
\(8\) | \(=4v\) | |
\(v\) | \(=2\ \text{ms}^{-1}\) |
b.ii. For the collision to be elastic, the kinetic energy must be conserved.
\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)
\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)
→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.
Which one of the following statements best describes transverse and longitudinal waves?
\(C\)
→ The energy transfer through transverse waves is perpendicular to the direction of vibrations.
→ The energy transfer through longitudinal waves is parallel to the direction of vibrations.
\( \Rightarrow C\)
Which one of the following statements best describes an observation of the Doppler effect for sound?
\(C\)
→ As a result of the Doppler effect, when there is relative movement towards a source of sound waves the frequency of the observed sound waves is increased.
\( \Rightarrow C\)
A railway truck \(\text{(X)}\) of mass 10 tonnes, moving at 3.0 m s\(^{-1}\), collides with a stationary railway truck \(\text{(Y)}\), as shown in the diagram below.
After the collision, they are joined together and move off at speed \(v= 2.0\ \text{m s}^{-1}\).
\(\text{Question 6}\)
Which one of the following is closest to the mass of railway truck \(\text{Y}\)?
\(\text{Question 7}\)
Which one of the following best describes the force exerted by the railway truck \(\text{X}\) on the railway truck \(\text{Y} \left(F_{\text { X on Y}}\right)\) and the force exerted by the railway truck \(\text{Y}\) on the railway truck \(\text{X} \left(F_{\text {Y on X}}\right)\) at the instant of collision?
\(\text{Question 6:}\ B\)
\(\text{Question 7:}\ C\)
\(\text{Question 6}\)
→ By the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.
\(m_Xu_X+m_Yu_Y\) | \(=m_Xv_X+m_Yv_Y\) | |
\(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \) | ||
\(10000 \times 3 +0\) | \(= 2(10000 + m_Y)\) | |
\(15000\) | \(=10000 +m_Y\) | |
\(m_Y\) | \(=5000\ \text{kg}\) | |
\(=5\ \text{tonnes}\) |
\( \Rightarrow B\)
\(\text{Question 7}\)
→ By Newton’s 3rd law of motion, each action has an equal and opposite reaction.
→ Hence, the force of \(F_{ \text { X on Y } }\) is equal in magnitude to \(F_{ \text { Y on X} }\) but opposite in direction which is indicated by the negative sign (–).
\( \Rightarrow C\)
Victoria’s fossil emblem Koolasuchus cleelandi, shown on the stamp above, was chosen in January 2022.
K. cleelandi fossils have been found in Boonwurrung country in Gippsland. Studies of K. cleelandi fossils have revealed many interesting findings. Several of these findings are summarised below.
K. cleelandi findings
\(\text{Question 32}\)
Which combination of findings from the list above is most likely to be associated with an increased chance of formation of K. cleelandi fossils?
\(\text{Question 33}\)
It is most likely that scientists determined the absolute age of K. cleelandi fossils
\(\text{Question 32:}\ B\)
\(\text{Question 33:}\ D\)
\(\text{Question 32}\)
→ K. cleelandi has large, dense bones (finding 2) which are less likely to decay overtime and hence have a higher chnace of being fossilised for a long period of time.
→ Because these bones are remaining in a river carrying many sediments (finding 4) the bones are more likely to be fossilised as they are covered quicker, reducing disturbance and hiding them from scavnegers.
\(\Rightarrow B\)
\(\text{Question 33}\)
→ Radioisotope dating of rocks is extremely accurate and reliable source of dating. Therefore it ios likely that scientists dated K. cleelandi fossils by dating the surrounding rock.
\(\Rightarrow D\)
Enzymes are associated with the biochemical pathways of cellular respiration and photosynthesis.
Consider a cell that carries out both cellular respiration and photosynthesis.
Which one of the following is a correct statement about the enzymes associated with these pathways?
\(B\)
By Elimination
→ While the net chemical equations for photosynthesis and cellular respiration appear to be the reverse of each other, each reaction is made up of various processes which require different reaction conditions, including different enzymes (Eliminate A).
→ Enzyme concentrations can increase/decrease the amount of available active sites which can change the rate of reaction rate (Eliminate C).
→ A large increase in temperature would not increase reaction rate but would denature the enzymes (Eliminate D).
→ Changes in various factors, including pH, can denature an enzyme if the change is outside the optimal range for that factor.
\(\Rightarrow B\)
The decomposition of ammonia is represented by the following equation.
\(\ce{2NH3(g)\rightleftharpoons N2(g) + 3H2(g) \quad \quad \Delta H = 92.4 \text{kJ mol}^{–1}}\)
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a.
b. → Tungsten will cause ammonia to decompose faster.
→ Faster rate due to lower activation energy.
→ Lower activation energy results in a greater proportion of successful collisions.
a.
b. → Tungsten will cause ammonia to decompose faster.
→ Faster rate due to lower activation energy.
→ Lower activation energy results in a greater proportion of successful collisions.
A reaction has the energy profile diagram shown below.
Which of the following represents the energy profile of the reverse reaction.
\(\text{Final product energy}\) \(\text{(kJ mol}^{−1})\) |
\(\Delta H\) \(\text{(kJ mol}^{−1})\) |
|
A. | \(40\) | \(+10\) |
B. | \(50\) | \(+10\) |
C. | \(50\) | \(-10\) |
D. | \(40\) | \(-10\) |
\(B\)
Reverse reaction \(\Rightarrow\) Endothermic
\(\Delta H \) | \(=\ \text{E(products) – E(reactants)}\) | |
\(= 50-40\) | ||
\(=10\ \text{kJ mol}^{-1}\) |
\(\Rightarrow B\)
The energy profile diagram below represents a particular reaction. One graph represents the uncatalysed reaction and
the other graph represents the catalysed reaction.
Which of the following best matches the energy profile diagram?
\(E_\text{a}\) \(\text{uncatalysed reaction}\) \(\text{(kJ mol}^{-1})\) |
\(\Delta H\) \(\text{catalysed reaction}\) \(\text{(kJ mol}^{-1}) \) |
|
A. | \(40\) | \(-140\) |
B. | \(90\) | \(-140\) |
C. | \(40\) | \(-50\) |
D. | \(90\) | \(-50\) |
\(D \)
→ The \(\text{E}_{a}\) of the uncatalysed reaction is higher (90) than the catalysed reaction (40).
→ The \(\Delta H\) energy changes are based off the zero reading (pre-reaction) to the final energy.
\(\Rightarrow D \)
The following energy profile shows the results obtained during an enzyme-catalysed reaction. Each stage of the reaction is labelled: M represents the initial reactants, N represents a stable intermediate and P represents the final products.
Which one of the following statements is correct?
\(D\)
\(\text{M → P}\) is endothermic since energy of products > energy of reactants.
\(\text{M → N}\) is endothermic
\(\text{N → P}\) is exothermic
\(\Rightarrow D\)
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a. Answers could include one of the following:
→ Heat loss into the surroundings from combustion process.
→ Specific examples of heat loss such as inefficient heating of water caused by proximity of the heat or not using a beaker lid.
b. \(0.68\ \text{g} \)
a. Answers could include one of the following:
→ Heat loss into the surroundings from combustion process.
→ Specific examples of heat loss such as inefficient heating of water caused by proximity of the heat or not using a beaker lid.
b. \(\Delta H = mC \Delta T = 0.200 \times 4.18 \times 10^{3} \times (45-21) = 20\ 064\ \text{J}\)
\(\ce{n(C2H6O)_{\text{req}} = \dfrac{20.064}{1367} = 0.014677\ \text{mol}} \)
\(\ce{MM(C2H6O) = 2 \times 12.01 + 6 \times 1.008 + 16 = 46.068\ \text{g mol}^{-1}} \)
\(\ce{m(C2H6O)_{\text{req}} = n \times MM = 0.014677 \times 46.068 = 0.68\ \text{g}} \)
Which statement explains why catalysts are often used in chemical reactions?
\(B\)
→ Catalysts increase the rate of reactions.
\(\Rightarrow B\)
Hydrogen peroxide, \(\ce{H2O2}\), in aqueous solution at room temperature decomposes slowly and irreversibly to form water, \(\ce{H2O}\), and oxygen, \(\ce{O2}\), according to the following equation. \(\ce{2H2O2(aq)\rightarrow 2H2O(l) + O2(g)}\) \(\Delta H < 0\) --- 6 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- a. → Increasing the temperature will increase the rate of \(\ce{O2}\) production. → An increase in temperature increases the kinetic energy of all molecules which causes both the frequency of collisions and the energy of each collision to increase. → The temperature increase results in more collisions, of which a higher proportion are successful (i.e. occur with greater energy than the activation energy threshold required to react). b. \(\ce{MnO2}\) acts as a catalyst in this reaction. a. → Increasing the temperature will increase the rate of \(\ce{O2}\) production. → An increase in temperature increases the kinetic energy of all molecules which causes both the frequency of collisions and the energy of each collision to increase. → The temperature increase results in more collisions, of which a higher proportion are successful (i.e. occur with greater energy than the activation energy threshold required to react). b. \(\ce{MnO2}\) acts as a catalyst in this reaction.
The correct equation for the incomplete combustion of ethanol is
\(C\)
→ The products of incomplete combustion are \(\ce{CO}\) and \(\ce{H2O}\).
\(\Rightarrow C\)
Methane gas, \(\ce{CH4}\), can be captured from the breakdown of waste in landfills. \(\ce{CH4}\) is also a primary component of natural gas. \(\ce{CH4}\) can be used to produce energy through combustion.
Write the equation for the incomplete combustion of \(\ce{CH4}\) to produce carbon monoxide, \(\ce{CO}\). (1 mark)
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\(\ce{2CH4(g) + 3O2 \rightarrow 2CO2(g) + 4H2O(l)\ \ \text{or}}\)
\(\ce{CH4(g) + \dfrac{3}{2}O2 \rightarrow CO2(g) + 2H2O(l)}\)
\(\ce{2CH4(g) + 3O2 \rightarrow 2CO2(g) + 4H2O(l)\ \ \text{or}}\)
\(\ce{CH4(g) + \dfrac{3}{2}O2 \rightarrow CO2(g) + 2H2O(l)}\)
Bioethanol, \(\ce{C2H5OH}\), is produced by the fermentation of glucose, \(\ce{C6H12O6}\), according to the following equation.
\(\ce{C6H12O6(aq)->2C2H5OH(aq) + 2CO2(g)}\)
The mass of \(\ce{C2H5OH}\) obtained when 5.68 g of carbon dioxide, \(\ce{CO2}\), is produced is
\(D\)
\(\ce{MM(CO2) = 12.01 + 2 \times 16 = 44.0\ \text{g}}\)
\(\ce{n(CO2) = \dfrac{5.68}{44.0} = 0.129\ \text{mol}} \)
\(\ce{n(C2H5OH) = n(CO2) = 0.129\ \text{mol}}\)
\(\ce{MM(C2H5OH) = 2 \times 12.01 + 6 \times 1.008 + 16 = 46.07\ \text{g}}\)
\(\ce{m(C2H5OH) = 0.129 \times 46.07 = 5.94\ \text{g}} \)
\(\Rightarrow D\)
--- 8 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- a. Timing of a falling mass. → Set up an electronic and automatic timing system with sensors to detect the presence of a small falling metal ball. → Heights for the ball should be set up between 0.2 m to 1 m with intervals every 0.2 m. To increase the reliability of the results, multiple trials should be conducted at each height and the average falling time for each height should be calculated which can then be used to graph the data. → The results should be plotted on a graph of height vs time\(^2\). This uses the equation \(s=ut +\dfrac{1}{2}at^2\) where \(u=0\) which becomes \(s=\dfrac{1}{2}at^2\). → After plotting the data, the acceleration due to gravity, \(a\), can be calculated using \(a=\dfrac{2s}{t^2}\), which will make it equal to 2 × the gradient of the line of best fit. b. → Look up known value on a reliable website (e.g. National Measurement Institute). → Ensure the value is for the location of the experiment (it can differ slightly). → Compare the known value to the value determined experimentally and the closer they are, the greater the accuracy of the experiment. c. → Conduct multiple trials at each height. → Use the average of the calculations as stated in the method above. d. → Compare the values obtained at a single height. → If there is a large variation in the calculations conducted at the same height, the data collected is less reliable. a. Timing of a falling mass. → Set up an electronic and automatic timing system with sensors to detect the presence of a small falling metal ball. → Heights for the ball should be set up between 0.2 m to 1 m with intervals every 0.2 m. To increase the reliability of the results, multiple trials should be conducted at each height and the average falling time for each height should be calculated which can then be used to graph the data. → The results should be plotted on a graph of height vs time\(^2\). This uses the equation \(s=ut +\dfrac{1}{2}at^2\) where \(u=0\) which becomes \(s=\dfrac{1}{2}at^2\). → After plotting the data, the acceleration due to gravity, \(a\), can be calculated using \(a=\dfrac{2s}{t^2}\), which will make it equal to 2 × the gradient of the line of best fit. b. → Look up known value on a reliable website (e.g. National Measurement Institute). → Ensure the value is for the location of the experiment (it can differ slightly). → Compare the known value to the value determined experimentally and the closer they are, the greater the accuracy of the experiment. c. → Conduct multiple trials at each height. → Use the average of the calculations as stated in the method above. → If there is a large variation in the calculations conducted at the same height, the data collected is less reliable.
d. → Compare the values obtained at a single height.
The diagram represents the electric field around a negative charge.
If the magnitude of the charge were doubled, which diagram would best represent the new electric field?
\(C\)
→ The density of the field lines corresponds to the strength of the electric field surrounding the point charge.
→ Hence a stronger charge produces a stronger field and so the field lines become denser.
\(\Rightarrow C\)
The industrial production of ammonia is represented by the Haber process reaction shown.
\( \ce{N2(g)} + 3 \ce{H2(g)} \rightleftharpoons \ce{2NH3(g)} \quad \Delta H=-91.8 \ \text{kJ} \ \text{mol}^{-1}\)
Factors such as temperature and pressure need to be considered in order to maximise yield.
Which of the following is correct?
\(B\)
→ Increasing the pressure would shift the position of equilibrium to the side of the equation with less moles (as per Le Chatelier’s principle).
→ This would increase the yield of \(\ce{NH3}\)
\(\Rightarrow B\)
Sodium chloride dissolves in water according to the following equation.
\( \ce{NaCl}(s) \rightleftharpoons \ce{Na^{+}} (aq) + \ce{Cl^{-}}(aq) \)
A saturated solution of \(\ce{NaCl}\) in water contains sodium and chloride ions at the following concentrations.
\( Ion \) | \( Concentration \) \( \left(\text{mol L}^{-1}\right) \) |
\( \ce{Na}^{+}\) | \(6.13\) |
\( \ce{Cl}^{-}\) | \(6.13\) |
\(D\)
\( \ce{NaCl}(s) \rightleftharpoons \ce{Na^{+}} (aq) + \ce{Cl^{-}}(aq) \)
\(K_{sp} = \dfrac{\ce{[Na+][Cl-]}}{\ce{[NaCl]}} \)
\(\ce{ \text{Since}\ NaCl\ \text{is a solid, its concentration is assumed to be 1}}\)
\(K_{sp}\) | \(= 6.13 \times 6.13\) | |
\(= 37.6\) |
\(\Rightarrow D\)
The structural formula of a compound is given.
What is the preferred IUPAC name of this compound?
\(B\)
→ The organic molecule has a 5-Carbon backbone, therefore the prefix will be “Pent”
→ The molecule contains no branches, so the rest of its name will be defined by the triple carbon bond “yne”
→ The rules of nomenclature state that the name must assign the triple carbon bond the lowest carbon number, this is achieved by numbering the carbons from right to left, placing the triple carbon bond at carbon 2
→ Hence the name of the compound is “Pent-2-yne”
\(\Rightarrow B\)