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Sarah Jane H.
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Question:

How do I best structure a scientific report?

Sarah Jane H.
Answer:

The first step is to look at the question or the objectives of the assessment and any marking criteria that are given. We need to break down what you want to say into manageable components that relate to each other. It is always best to look at scientific work like a funnel with broad ideas in the beginning that get narrower and more specific down the bottom. It will all depend on the assesment and subject material how we tackle this strcture of the report.

Environmental Science
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Question:

Justify the need to ban a specific pesticide in terms of its impacts on non-target species and human health.

Sarah Jane H.
Answer:

A pesticide such as DDT is non-selective and impacts on a broad range of species, such as the many insect species. It is used widely to control pest species in agriculture. Insects affected may be both beneficial, like bees that are responsible for the pollination of many flowering plants, and harmful like grasshoppers that feed on crops. Bees are non-target species whereas grasshoppers are target species. Residual pesticides and the effect on non-target species can impact on human health as the residual pesticides can be accumulated in the food chains and, in some cases, toxic levels in humans can be reached.

Chemistry
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Question:

An unattended car is stationary with its engine running in a closed workshop. The workshop is 5.0 m × 5.0 m × 4.0 m and its volume is 1.0 × 10^5 L. The engine of the car is producing carbon monoxide in an incomplete combustion according to the following chemical equation: C8H18(l) + 17/2 O (g) 8CO(g) + 9H2O(l) Exposure to carbon monoxide at levels greater than 0.100 g L^–1 of air can be dangerous to human health. 6.0 kg of octane was combusted by the car in this workshop. Using the equation provided, determine if the level of carbon monoxide produce in the workshop would be dangerous to human health. Support your answer with relevant calculations.

Sarah Jane H.
Answer:

Volume of garage is 1.0 × 10^5L = 100000 L 6.0 kg octane = 6000 g, n=m/M = 600/114.224 = 52.53 moles Molar ratio from equation is 1 : 8 ∴ 8 × 52.53 moles of carbon monoxide are produced = 420.23 moles. m = n × M = 420.23 × 28.01 = 11771 g [CO] =11771/100000 = 0.118 g L−1 Therefore the level of carbon monoxide is dangerous as it is more than 0.100 g L–1.

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