### All GRE Subject Test: Chemistry Resources

## Example Questions

### Example Question #1 : Limiting Reagent

Given the unbalanced equation below, how many grams of carbon dioxide will be produced from one mole of glucose and three moles of oxygen?

**Possible Answers:**

**Correct answer:**

The first step to solve will be to balance the chemical reaction:

We see that we see that for every one mole of glucose used, six moles of carbon dioxide will be made. Similarly, for every six moles of oxygen used, six moles of carbon dioxide will be formed. For the reaction to carry out to completion, however, there must exist six moles of oxygen for every one mole of glucose. In the problem's circumstances, one of these compounds becomes the limiting reactant, in this case it is oxygen.

We only have three moles of oxygen, but we would need six to react all the given glucose, making oxygen the limiting reagent. We need to find the carbon dioxide produced from the limited amount of oxygen present. Use the molar ratio between oxygen and carbon dioxide and the molar mass of carbon dioxide to solve.

### Example Question #1 : Gravimetric Analysis

Household vinegar contains the organic compound acetic acid with chemical formula, . If a vinegar sample contains of acetic acid, calculate the percent (mass/mass) of the in the vinegar sample.

Density of vinegar is the following:

**Possible Answers:**

**Correct answer:**

Convert the moles of to grams:

To calculate the percentage of in the vinegar we need to use the following formula:

Therefore,

### Example Question #6 : Solubility And Equilibrium

Consider the following balanced equation for the solubility of barium hydroxide in an aqueous solution.

What is the solubility of barium hydroxide?

**Possible Answers:**

**Correct answer:**

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction.

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated and , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

### Example Question #41 : Analytical Chemistry

Write the solubility product, , for .

**Possible Answers:**

**Correct answer:**

The solubility product constant () is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution. The equilibrium constant for can be written as:

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

The product, , can be considered a constant expression called the solubility product constant () and the equation can be written in the form:

### Example Question #42 : Analytical Chemistry

Write the solubility product, , for .

**Possible Answers:**

**Correct answer:**

The solubility product constant () is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution. The equilibrium constant for can be written as:

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

The product, , can be considered a constant expression called the solubility product constant () and the equation can be written in the form:

### Example Question #43 : Analytical Chemistry

Write the expression for the following equilibria:

**Possible Answers:**

None of these

**Correct answer:**

The solubility product constant () is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution.The equilibrium between undissolved and its ions dissolved in solution is:

The equilibrium constant for can be written as:

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

The product, , can be considered a constant expression called the solubility product constant () and the equation can be written in the form:

### Example Question #44 : Analytical Chemistry

Considering that for the equilibrium reaction:

What would be the concentration for a solution buffered at a pH of 8.5?

**Possible Answers:**

**Correct answer:**

The pH of the solution is , therefore the pOH would be considering the relationship:

Therefore ^{ }based on the equation

The solubility product expression for is :

By inserting the knowns in to the expression, gives:

Rearrange this equation

### Example Question #6 : Analyzing Solids

The K_{sp} of is . What is the molar solubility of _{ }in water?

**Possible Answers:**

**Correct answer:**

The equation for the dissolution of in water is below:

The K_{sp }for the above equation is:

Due to the solubility of _{ }of in water, the concentration of and ^{-} should be equal:

Plug into the K_{sp} equation:

Therefore:

The solubility of in water is

### Example Question #45 : Analytical Chemistry

The K_{sp} of is . What is the molar solubility of in_{ }water?

**Possible Answers:**

**Correct answer:**

The equation for the dissolution of in water is below:

The K_{sp} for the above equation is:

Due to the solubility of of in water, the concentration of and should be equal:

Plugging X into the K_{sp} equation gives:

Therefore:

The solubility of CuBr in water is

### Example Question #46 : Analytical Chemistry

Which salt is insoluble in water?

**Possible Answers:**

**Correct answer:**

Solubility rules must be followed for substances in aqueous media. Below are some of the solubility rules and they must be followed in the order given. For example, rule 1 should have precedence over rule 2.

1. All alkali metal and compounds are soluble.

2. Nitrate , perchlorate , chlorate, , and acetate salts are soluble.

3. Silver, lead, and mercury salts are insoluble.

Thus, we see that silver chloride is insoluble due to rule 3.

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