Question

course activity: observing a chemical reaction
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• the chemical reaction in this experiment can overflow. to avoid overflow, do not use more than ½ teaspoon of baking soda or ½ cup of vinegar at any time during the experiment.

question 1: planning and data collection
part a
this investigation is a controlled experiment, so you can change only one variable at a time. decide whether you’ll change the amount of baking soda or the amount of vinegar. give a reason for your choice.

Explanation:

To answer this, we can choose to change the amount of baking soda (or vinegar) and explain the reason. Here's an example:

If we choose to change the amount of baking soda: The reason could be that baking soda is a solid reactant, and its quantity can be measured precisely in small increments (like fractions of a teaspoon), which allows for a more detailed exploration of how the amount of a solid reactant affects the chemical reaction (the reaction between baking soda and vinegar). Alternatively, if we choose vinegar (a liquid), we can note that liquids are easy to measure in cups and we can observe how different volumes of the liquid reactant impact the reaction. For simplicity, let's pick changing the amount of baking soda. The reason is that baking soda can be added in small, controlled amounts (e.g., 1/4 tsp, 1/2 tsp, 3/4 tsp) to observe how the rate or extent of the reaction (like how much it fizzes, if it overflows) changes, and since it's a solid, measuring small variations is manageable.

Answer:

I would choose to change the amount of baking soda. Reason: Baking soda is a solid, so it can be measured in small, precise increments (e.g., 1/4 tsp, 1/2 tsp) to study how the quantity of this reactant affects the chemical reaction with vinegar. This allows for a detailed investigation of the relationship between reactant amount and reaction outcomes (like fizzing intensity, overflow potential).

(Note: If you choose to change vinegar, a similar explanation can be given about liquid measurement and reaction observation.)