Preliminary Biology Assessment Task #1 Assessment Overview: There are THREE (3) parts to this assessment. Part 1: Research and planning; To be done in own time. Part 1 is to be completed and handed in before the start of period 1 on Friday 13/05/2016. The task is to individually design a first - hand investigation (experiment) to identify the relationship between surface area and rate of reaction. Please write up your procedure on the scaffold sheets provided. At the end of 20 minutes, you will hand in your procedure and will be provided with a method which can be used to carry out the first-hand investigation. Note: To carry out this investigation you will be provided with soluble aspirin tablets. Background information: Soluble aspirin is meant to easily dissolve in water The rate at which soluble aspirin dissolves is related to the rate of diffusion of the aspirin particles into the water. Equipment available for use: Soluble aspirin x 4 Mortar and pestle Stopwatch Safety goggles 100 ml measuring cylinder 250 ml beakers x 2 Safety gloves Distilled water Note: Additional equipment can be provided on request. The practical task is to be designed and carried out individually you may need to wait for equipment availability. Draft paper available on request.
a) Risk Assessment: (3 marks) Potential Danger Safety Precaution Risk Management beaker Mortar and pestle Aspirin tablets If dropped, glass can cause cuts to skin. Bruises and pain can be experienced if not aware of what is being grinded. If consumed by a person who already has a low ph, it can lower it to levels of discomfort. Do not keep beakers on the far edges of the bench. Keep fingers away when grinding and do not keep on edges of the bench. Do not consume or ingest. b) Title: (2 marks) Effect of surface area on rate of reaction. c) Aim: (2 marks) To identify the relationship between surface area and the rate of reaction. d) Hypothesis: (2 marks) The crushed tablet will dissolve faster when placed in water due to its increased surface area.
e) Scientific Diagram: (4 marks)
f) Step by step procedure (METHOD) (6 marks) 1. Using an electronic balance, take equal weights of 2 aspirin tablets, one whole and the other grinded using a mortar and pestle. 2. Place the whole aspirin tablet into the beaker labelled A and the crushed tablet into the beaker labelled B. 3. Using a measuring cylinder, draw up 100mL of distilled water. 4. Pour the water simultaneously into the two beakers. 5. Use the stopwatch to measure how long each tablet fizzes, until they are completely dissolved. 6. Repeat again under the same conditions. g) In a scientific experiment we try to achieve validity, reliability and accuracy. Outline how your experimental design achieves: (6 marks) Validity (Valid) The method actually tests the aim by only changing the surface area of the tablet (independent variable) and measuring the time it takes to dissolve (dependent variable). All other variables are controlled including the volume of distilled water, same distilled water used and the type of tablet. Reliability (Reliable) The method involves repeating the experiment more than once under the same conditions which ultimately tests whether consistent results are obtained. Then, by averaging the results, outliers are removed and the effect of random errors are minimised such as not knowing when the tablet is dissolved completely the first time. Accuracy This is achieved by using the same equipment that are calibrated to give precise measurements, especially for measuring such as the same measuring cylinder, electronic balance and stopwatch which allows for accurate readings. To maintain accuracy, the same person obtains all measurements especially when reading a meniscus which can be viewed at differently at different angles and can affect the outcome of the investigation.
PART 2: The actual first-hand experiment, the results and discussion questions all to be completed in class on the 13/05/2016. (22 marks total.) The paper and all equipment will be distributed on the day. PART 3: Application questions. There are to be completed in class on the 13/05/2016 after the experiment. They need to be completed on the paper provided. The exact questions are printed below so that the answers can be research and learnt before the day of the assessment. No notes can be used on the day. The questions have to be answered during the assessment. (20 marks total.) Application Questions 1. Using a 3cm x 3cm x 3cm cube as your example, show how the same volume of 1cm x 1cm x 1cm cubes can have a different surface area from the larger cube. Use a diagram to assist with your explanation. (3 marks) Answer: By calculating the SA:V ratios of both cubes, the amount of SA per one cubic centimetre is found. As a result, the SA:V ratio of the large cube is 2:1 and 6:1 for the small cube. It can be seen that the small cube has a larger amount of surface area to the same volume, which is actually a preferred higher SA:V ratio. 2. Explain, using your practical results as evidence, the effect surface area has on the rate of diffusion. (3 marks) Answer: It was seen in the practical that the crushed tablet had a faster rate of diffusion with a time of 57sec than the whole tablet which had a time of 1min 18sec. This difference in times is due to differences in surface area, where the greater the surface area, the faster the rate of diffusion.
3. Great care is taken to avoid a build up of coal dust in coal mines because if a fire occurs there is a strong chance of an explosion. Explain why coal dust is potentially explosive while coal is not. (3 marks) Answer: Coal in the form of dust is more explosive and smaller than coal. This means that coal dust has much more surface area and hence, a higher SA:V ratio than that of coal. Increasing the SA:V ratio of coal dust leads to more collisions as there is more contact area for the other reactant to be exposed to. Therefore, the increased SA:V of coal dust increases the rate of reaction and ultimately, increasing the chances of an explosion. 4. Explain the importance of teeth in the digestive process with regard to the exposure of complex molecules to digestive chemicals. (3 marks) Answer: Complex molecules are first broken down mechanically upon entering the mouth. The action of teeth greatly increases the surface area of these complex foods to expose them to enzymes that will help to break them down. As a result, there is more contact area for digestive chemicals to be exposed to, leading to more collisions, hence, making the process of digestion fast and efficient. Without teeth, the digestive process would not be efficient and functional enough to digest such complex molecules, proving detrimental to the functioning of an organism. 5. Human lungs are broken into millions of tiny air-sacs called alveoli. Why are the alveoli such an efficient means of increasing gaseous exchange in the lungs? (3 marks) Answer: Human lungs are broken into a number of smaller structures called alveoli in order to create an efficient means of increasing gaseous exchange in the lungs. Due to the small size of alveoli, a high SA:V ratio is achieved, making rate of exchange rapid as there is more surface area for contact and less space to get through for the materials. With all these gases being exchanged simultaneously, it allows for a much faster and efficient process of gaseous exchange in the lungs. 6. Explain in terms of surface area to volume, why cells are so small? (3 marks) Answer: Cells need to be small because they rely on diffusion for getting substances into and out of their cells. They are limited on how big they can be because cells need to fulfil a high SA:V ratio. This is imperative so that the rate of exchange is fast. It means that the cells can obtain essential nutrients and get rid of waste quickly to ultimately maintain the functioning of the organism. If cells
were any larger, their SA:V ratio would become smaller, decreasing efficiency, proving detrimental to the functioning of an organism. 7. Outline the process of diffusion (2 marks) Answer: Diffusion is the passive movement of molecules from a region where the concentration of those molecules are high to a region where the concentration is low, until the two concentrations are equal. This process does not require a membrane and can occur in liquids and gases.