Unit 1 Higher Chemistry Questions

[Maren Ruminski]

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The AP Chemistry exam assesses your ability to understand science practices in relation to four big ideas: scale, proportion and quantity; structure and properties; transformations; and energy. The exam is 3 hours and 15 minutes long and is comprised of two separate sections:

CollegeBoard is also known to provide answer options that are meant to confuse you into choosing the wrong answer instead of the correct one. But fear not! We will show you how to carefully read the question and make sure that you are confident in the answer you choose!

In this first question, we see that a base Mg(OH)2 is being titrated by an acid HNO3. In order to determine the mass of the initial base, we need to perform a series of titration calculations. This is a great example of when your labs come in handy, since the lab can help you visualize what is going on.

First, we need to determine the reaction that is occurring when Mg(OH)2 is reacted with HNO3. We first write out the equation we know, which is that the base and acid react to create water and an aqueous solution of the remaining ions:

We now need to balance the above reaction. Hydrogen is already balanced, so the next element we move to is Oxygen. We see that Oxygen is unbalanced with five oxygen on the left and four oxygen on the right. We can balance Oxygen by:

Now we return to the titration problem. Since we know that the acid is being used as the titration solution, we know that HNO3 is in the buret. We also know that 20 mL is needed to get to the end point since:

Now that we know the molar mass of the base (58.3 grams), the molarity of the acid solution, and the amount of acid needed to titrate the bass and the balanced equation, we can perform a simple stoichiometric equation. Make sure to balance all your units and note that it takes 2 moles of HNO3 for every one mole of Mg(OH)2!

Geometry of a molecule or compound is an important topic covered on the AP Chemistry exam so it is important that you know your tetrahedrals from your trigonal bipyramidials! An easy hint is that you can almost always determine what type of shape a compound is based on the number of lone electron pairs and molecules that are attached to the central atom.

Now this is a tricky compound. At first glance, you would assume that ClF3 is not planar due to the impact of the two electron pairs present. However, the chlorine trifluoride compound is actually a T-shaped geometrical structure. In this shape, the 3 Fluorine atoms are in the same plane and form a T shape with the Chlorine atom. Surprisingly, this is not the correct answer.

This leaves us with phosphorus trichloride. The electron pair causes a disruption in what would be expected of a planar compound. Each individual chlorine atom experiences great repulsion from the lone electron pair, thereby causing the Chlorine atoms to bend closer together. This gives this compound a trigonal pyramidal shape instead of a planar geometry.

This is a tricky question since we are given the initial mass of both reactants and are asked to find the maximum mass of a product, in this case water. This means this is a limiting reactant problem in which a limiting reactant limits how much a product can be produced.

Here we see that while N2O4 may produce the most amount of water, this is an inaccurate representation of what happens. N2H4 acts as a limiting reactant and limits how much water is produced. N2O4 is in excess. Therefore, the maximum amount of water produced is 9 grams.

The optimum buffer action occurs when the amounts of acid and its conjugate base are approximately equal. In the case of our situation, this happens when half of the weak acid has been neutralized to form its conjugate base.

We are looking for the point when the buffer action is optimal, which as we defined before, is when there are equal amounts of acid and its conjugate base. This point occurs approximately midway between the equivalence point (at which the slope of the graph is nearly infinite) and the starting point. In relation to the graph this occurs at point V.

This type of question is best solved as reduction and oxidation half reactions To do this, we need to determine which elements are being reduced and which elements are being oxidized. We see that the chromium ion gains electrons, therefore becoming reduced, while sulfur loses electrons when it becomes a solid, therefore becoming oxidized.

From here, we must determine how many electrons are needed to balance out the charges of either side, which we can do by adding 6 electrons to the left side of the reduction equation and 2 electrons the the right side of the oxidation equation:

This type of question seems difficult to look at since you are immediately presented with numerous compounds you may have not seen before. With this type of question, it is important to remember characteristics about specific compounds such as halides and sulfates.

One fact to remember about carbonates (CO32-) is that they are most often insoluble. They are especially insoluble when paired with Group II atoms such as Barium, Calcium and Strontium. Therefore, BaCO3 is the least soluble in water of the groupings.

This is another one of those questions where it is extremely helpful if you are able to remember labs shown or conducted in your class. This question also requires you to remember qualitative characteristics of compounds to deduce what ions could be present in an unknown solution.

The initial white precipitate formed corresponds to a chloride precipitate since hydrochloric acid was used to treat the unknown solution. However, we know that BaCl2 is soluble, so we can deduce that Barium is not in the initial solution since a solid forms after the treatment.

This leaves us with AgCl, HgCl2 and PbCl2 as potential precipaties. Since a bright yellow precipitate formed after the introduction of the potassium iodide solution, we know that this is a characteristic of lead iodide. Therefore, lead is present in the solution. This leaves us with answer choice (B) as the right answer.

To verify that B is correct, we must determine that silver is also present in the unknown solution. We know this to be true because silver easily dissolves in ammonia, as described in the problem statement.

You may know the concepts very well, but you also need to be able to apply the concepts in a timed setting. Make sure you take multiple practice tests administered through CollegeBoard or through your school. You might be able to answer every question on the test perfectly, but you need to be sure that you can finish the multiple choice section in 90 minutes! Remember, that leaves just about one and a half minutes to read the question and determine the answer. Over time, you will get better at quickly answering questions and determining what the question is asking for on your first try!

People will always tell you to get a good eight hours of sleep and eat a healthy breakfast. While this may seem like basic info that is not relevant to the test material you are memorizing, it is important that you are awake and ready to sit through a three hour exam!

Chemistry is about the molecules all around us! It allows us to study and understand how matter changes and behaves in the way it does. Everything from frying an egg to how soap makes you feel clean can be explained by chemistry.

Chemistry is an important qualification if you intend to enter a career in pharmaceuticals, engineering, mathematics, science and some medical sciences. Furthermore, it is a recommended course for study for other careers, including law, military service, accountancy and management.

There is no external exam at National 3 and 4 level, although there will be written tests throughout the course, a scientific investigation and a short research task. The course is graded PASS or FAIL.

Each of the units has a written assessment based on knowledge and understanding that must be passed. There is a final exam set by the SQA lasting 2 hours 30 minutes (worth 100 marks) and an externally assessed research project worth 30 marks.

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