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Provide four differences between NaCl and C6H12O6, list four fundamental units of measurement and their symbols, define a derived unit, and list two examples of worms that infest cattle and two precau

Chemistry

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1 (ii) Write down four differences between NaCl and C6_6H12_{12}O6_6.

  • NaCl is an ionic compound, while C6_6H12_{12}O6_6 (glucose) is a covalent compound.
  • NaCl has a high melting point (801 °C), while C6_6H12_{12}O6_6 has a lower melting point (146 °C).
  • NaCl conducts electricity when molten or dissolved in water, while C6_6H12_{12}O6_6 does not conduct electricity.
  • NaCl is an inorganic compound, while C6_6H12_{12}O6_6 is an organic compound.

1 (c) (i) State four fundamental units of measurement and their symbols.

  • Length: meter (m)
  • Mass: kilogram (kg)
  • Time: second (s)
  • Electric current: ampere (A)

1 (c) (ii) What is a derived unit?
A derived unit is a unit of measurement that is obtained by multiplying or dividing fundamental (base) units of the International System of Units (SI). Examples include square meter (m2^2) for area or meter per second (m/s) for speed.

1 (d) (i) List two examples of worms that infest cattle.

  • Tapeworms
  • Roundworms (e.g., Ascaris, hookworms)

1 (d) (ii) State two precautions that could be taken to reduce worm infestation in cattle.

  • Regularly deworming cattle with appropriate anthelmintics.
  • Practicing rotational grazing to break the life cycle of worms and reduce pasture contamination.

1 (d) (iii) State four ways in which these worms can harm their host.

  • They compete for nutrients with the host, leading to malnutrition.
  • They can cause damage to host tissues and organs, such as the intestines or liver.
  • Some worms cause blood loss, leading to anemia.
  • They can reduce growth rates and overall productivity (e.g., milk yield) in cattle.

2 (a) (i) State four ways in which excessive alcohol consumption is harmful to humans.

  • Causes liver damage, such as fatty liver, alcoholic hepatitis, and cirrhosis.
  • Leads to brain damage and cognitive impairment.
  • Increases the risk of developing certain cancers, including those of the mouth, throat, esophagus, and liver.
  • Can result in addiction and dependence, leading to social and psychological problems.

2 (a) (ii) List two effects of cigarette smoke on humans.

  • Causes lung cancer and other respiratory diseases like emphysema and chronic bronchitis.
  • Increases the risk of heart disease, stroke, and other cardiovascular problems.

2 (b) (i) Give the functional groups present in

  • (α) Alcohol: Hydroxyl group (-OH)
  • (β) Ethylene: Carbon-carbon double bond (C=C)
  • (γ) Carboxylic acid: Carboxyl group (-COOH)
  • (δ) Esters: Ester group (-COO-)

2 (b) (ii) Write down four differences between neutralization and esterification.

  • Reactants: Neutralization involves an acid and a base, while esterification involves a carboxylic acid and an alcohol.
  • Products: Neutralization produces a salt and water, while esterification produces an ester and water.
  • Reaction type: Neutralization is typically an ionic reaction, while esterification is a covalent reaction.
  • Catalyst: Neutralization usually does not require a catalyst, while esterification typically requires an acid catalyst (e.g., concentrated H2_2SO4_4).

2 (c) Define the following terms as applied to a simple machine, stating the effect of friction on each of them

  • (i) Mechanical advantage: The ratio of the load (resistance force) to the effort (applied force). Friction reduces the mechanical advantage of a machine because more effort is needed to overcome friction.
  • (ii) Velocity ratio: The ratio of the distance moved by the effort to the distance moved by the load. Friction does not affect the velocity ratio, as it is a theoretical value determined by the machine's design.
  • (iii) Efficiency: The ratio of the useful work output to the total work input, often expressed as a percentage. Friction reduces the efficiency of a machine because some of the input energy is converted into heat due to friction, thus reducing the useful work output.

2 (d) (i) Name two tools used in the preparation of a nursery bed.

  • Spade
  • Rake

2 (d) (ii) State four factors you would consider in selecting a site for a nursery.

  • Soil type and fertility: Needs well-drained, fertile soil.
  • Water availability: Easy access to a reliable water source for irrigation.
  • Sunlight exposure: Adequate sunlight for seedling growth.
  • Protection: Shelter from strong winds, excessive rain, and pests.

2 (d) (iii) Mention two reasons for nursing seeds at the nursery.

  • It allows for better control over environmental conditions (e.g., watering, pest control) during the critical early stages of plant growth.
  • It helps to protect young seedlings from harsh weather conditions, pests, and diseases, leading to higher survival rates.

3 (a) (i) State four beneficial effects of microorganisms to humans.

  • Food production: Used in making bread, cheese, yogurt, and fermented foods.
  • Digestion: Gut bacteria aid in the digestion of food and synthesis of vitamins.
  • Decomposition: Break down organic waste and dead organisms, recycling nutrients in the environment.
  • Medicine production: Used to produce antibiotics and vaccines.

3 (a) (ii) State two reasons why organisms communicate with one another.

  • To attract mates for reproduction.
  • To warn others of danger or predators.

3 (b) (i) (α) What is a hydrocarbon?
A hydrocarbon is an organic compound that consists entirely of hydrogen and carbon atoms.

3 (b) (i) (β) Name three sources of hydrocarbons?

  • Crude oil
  • Natural gas
  • Coal

3 (b) (ii) Determine how much water should be added to 30cm3^3 of 0.2 mol/dm3^3 NaCl solution to obtain 0.02 mol/dm3^3.

  1. Identify the given values for initial concentration (C1C_1), initial volume (V1V_1), and desired final concentration (C2C_2).

    C1=0.2 mol/dm3C_1 = 0.2 \text{ mol/dm}^3 V1=30 cm3V_1 = 30 \text{ cm}^3 C2=0.02 mol/dm3C_2 = 0.02 \text{ mol/dm}^3
  2. Use the dilution formula C1V1=C2V2C_1V_1 = C_2V_2 to find the final volume (V2V_2) of the solution.

    (0.2 mol/dm3)×(30 cm3)=(0.02 mol/dm3)×V2(0.2 \text{ mol/dm}^3) \times (30 \text{ cm}^3) = (0.02 \text{ mol/dm}^3) \times V_2
  3. Solve for V2V_2.

    V2=0.2 mol/dm3×30 cm30.02 mol/dm3V_2 = \frac{0.2 \text{ mol/dm}^3 \times 30 \text{ cm}^3}{0.02 \text{ mol/dm}^3} V2=60.02 cm3V_2 = \frac{6}{0.02} \text{ cm}^3 V2=300 cm3V_2 = 300 \text{ cm}^3
  4. Calculate the amount of water that needs to be added by subtracting the initial volume from the final volume.

    Volume of water added=V2−V1\text{Volume of water added} = V_2 - V_1 Volume of water added=300 cm3−30 cm3\text{Volume of water added} = 300 \text{ cm}^3 - 30 \text{ cm}^3 Volume of water added=270 cm3\text{Volume of water added} = \boxed{\text{270 cm}^3}
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