013 Matching No. 1 Below you will find 5 locations in Derbyshire. Each location sits on a river. Please match the correct river to the location it flows through. The correct answers are as follows: Over Haddon is above the River Lathkill, Bakewell is on the River Wye, Matlock is on the River Derwent, South Wingfield is on the River Amber and the River Dove flows through Beresford Dale. 1 0.33 0 1 Bakewell Wye Matlock Derwent Beresford Dale Dove Over Haddon Lathkil South Wingfield Amber Erewash Trent 0 0.33 001f Examples description formative This short assessment showcases the question types that are available in Moodle 1.9. They are classified by the type of response required and fall into three categories <ul><li>numeric </li> <li>text </li> <li>selection </li> </ul>The rubric that precedes each of these categories explains a little more about the possibilities and constraints offered by each question type within the category. In all cases you are allowed one try at each question, though you can return to your response and amend it as many times as you wish prior to submission. As this is a formative assessment designed for you to learn about Moodle's facilities, you are allowed to make multiple attempts at the whole assessment. Click on 'Next' to move to the next screen. Alternatively you may move directly to individual questions by clicking on the corresponding numbers. You may end the assessment at any point by clicking on 'End test...' which will lead you to a Summary page from where you may 'Submit' your answers and finish. Information screens such as this appear in the Questions panel as an 'i'. 0 0 0 0 003A Multiple choice with two hints Estimate the gradient of the curve shown in the plot at x = 2. <img vspace="0" hspace="0" height="200" src="http://learn.open.ac.uk/file.php/2448/website/mcgraph.png" alt="curved graph of x against y" title="curved graph of x against y" /> Click on the answer closest to your estimate. <img vspace="0" hspace="0" height="197" src="http://learn.open.ac.uk/file.php/2448/website/mcgraph2.png" alt="A graph of x against y showing tangent passing through (0.5, 0) and (4, 14)." title="A graph of x against y showing tangent passing through (0.5, 0) and (4, 14)." /> Gradient of tangent = $$\frac{\Delta y}{\Delta x} = \frac{14 - 0}{4.0 - 0.5} = \frac{14}{3.5} = 4.0$$ The gradient of a curve at any point is the same as the gradient of a tangent drawn at that point, so the gradient of the curve is 4.0 at x = 2. 1 0.33 0 0 true false Your answer is correct. Your answer is incorrect. none 1 This value is too low. 2 This value is too low. 3 This value is too low. 4 The gradient is indeed 4. 5 This value is too high. 6 This value is too high. 0 0.33 Start by imagining a tangent drawn against the curve at x = 2. Your tangent should be exactly parallel to the curve at x = 2. Where does your tangent, drawn at x = 2, cross the x and y axes? Estimate these points then calculate the gradient as the difference in y values divided by the difference in x values. 001A Numeric Solve a quadratic with two hints Question: Solve $$x^2 + 3x - 28 = 0$$ Enter either of the possible answers. The quadratic equation is $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a = 1, b = 3 and c = -28.$$x = \frac{-3 \pm \sqrt{3^2 - 4\times 1\times(-28)}}{2\times a}$$ $$x = \frac{-3 \pm \sqrt{9 + 112}}{2}$$ $$x = \frac{-3 \pm \sqrt{121}}{2}$$ $$x = \frac{-3 \pm 11}{2}$$ $$x = \frac{-14}{2}$$ or $$\frac{8}{2}$$ and the solutions are x = -7 and x = 4. We can prove these are correct by substituting the values into the original equation. $$(-7)^2 + 3\times(-7) - 28 = 49 - 21 - 28 = 0$$ $$4^2 + 3\times4 - 28 = 16 + 12 -28 = 0$$ 1 0.33 0 0 -7 0 Your answer is correct. 4 0 Your answer is correct. * 0 Your answer is incorrect. 0 0.33 You may need to be reminded of the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ You will need to substitute for a, b and c in the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a is the coefficient for $$x^2$$, b is coefficient for $$x$$ and c is the constant. 002A Short-answer Increase the rate of a reaction by adding a catalyst with two hints There are three different ways of increasing the rate of a chemical reaction. Two of them are <ul><li>to increase the temperature</li> <li>to increase the concentration of one or more of the reactants.</li> </ul>What is the third way? The rate of a chemical reaction can be increased by any of <ul><li>adding a catalyst</li> <li>increasing the temperature</li> <li>increasing the concentration of one or more of the reactants.</li> </ul> 1 0.33 0 0 0 add*catalyst* Your answer is correct. use*catalyst* Your answer is correct. *catalyst* Your answer appears to be on the right lines but perhaps you could phrase it differently. * Your answer is incorrect. 0 0.33 Every chemical reaction has an energy barrier that must be overcome before reactant molecules can form products. One way to increase the rate is to lower this barrier. How might you achieve this? You will find more on lowering the energy barrier in Block 8 Section 10.5. Perhaps you would care to read that now before trying again. 005 True false HCl vibration frequencies The vibration frequency of the O2 molecule is higher than that of the HF molecule. O2 contains a double bond, but the lighter molecule HF contains a single bond. The effects of mass and bond strength are acting in opposite senses and it is not easy to predict the answer. However higher vibrations involving hydrogen are typically at higher frequency than other vibrations. The vibrational numbers are O2 = 1556 cm-1 HF = 3958 cm-1 1 0.1 0 0 true Your answer is incorrect. false Your answer is correct. 014 Essay No. 1 Please compare and contrast the quality of the water in the River Derwent as it leaves Lady Bower with the quality found downstream of the Cromford sewage works. It is not anticipated that this essay question type will be widely used (if it is used at all). Authors wishing to assess students' written work should use the eTMA system. 1 0 0 0