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GRE Subject Test: Chemistry : GRE Subject Test: Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

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Example Questions

Example Question #33 : Analytical Chemistry

Which of the following would be considered a Lewis base?

Possible Answers:

$CH_{2}Cl_{2}$

$Mn^{2+}$

$Br^{-}$

$H^{+}$

Zn(s)

Correct answer:

$Br^{-}$

Explanation:

A lewis base is an electron pair donor. $Br^{-}$ would be considered a lewis base based on the definition. Because $Br^{-}$ is electron rich based on its oxidation number, it is able to donate an electron pair to an electron deficient ion.

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Example Question #34 : Analytical Chemistry

Which of the following types of reactions best describes the following reaction:

$Ce^{4+}+Fe^{2+}\rightarrow\ Ce^{3+}+Fe^{3+}$

Possible Answers:

Redox reaction

Combustion

Double replacement

Catalytic

Combination

Correct answer:

Redox reaction

Explanation:

A redox reaction is also known as an oxidation/reduction reaction. This type of reaction involves the transfer of electrons from on reactant to another. In this reaction, an electron is transferred from $Fe^{2+}$ to $Ce^{4+}$ forming the products $Ce^{3+}$ and $Fe^{3+}$ . The substance that gains an electron is referred to as being reduced. The substance that loses an electron is referred to as being oxidized.

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Example Question #303 : Gre Subject Test: Chemistry

What is the oxidation state of manganese in $KMnO_{4}$ ?

Possible Answers:

Correct answer:

Explanation:

In this compound oxygen has an oxidation number of . There are 4 oxygens, therefore the total oxidation number for the oxygens in $KMnO_{4}$ is: $Total\ oxygen\ oxidation\ number=(-2) \times 4=-8$ . Potassium,, is a group 1 metal and has an oxidation number of . The sum of all oxidation number in a neutral compound such as $KMnO_{4}$ is zero. We will give the oxidation number of equal to :

$((-2) \times 4)+(+1)+x=0$

Simplifying the above equation:

(-8)+(+1)+x=0

(-7)+x=0

Rearranging to solve for :

x=+7

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Example Question #32 : Analytical Chemistry

What is the oxidation state of manganese in $K_{3}MnO_{4}$ ?

Possible Answers:

Correct answer:

Explanation:

In this compound oxygen has an oxidation number of . There are 4 oxygens, therefore the total oxidation number for the oxygens in $K_{3}MnO_{4}$ is: $Total\ O\ oxidation\ number=(-2) \times 4=-8$ . Potassium, , is a group 1 metal and has an oxidation number of . There are 3 potassium ions present in $K_{3}MnO_{4}$ , therefore the total oxidation number for the potassium ions is:

$Total\ K\ oxidation\ number=(+1) \times 3=+3$ .The sum of all oxidation number in a neutral compound such as $K_{3}MnO_{4}$ is zero. We will give the oxidation number of equal to :

(-8)+(+3)+x=0

Simplifying the above equation:

(-5)+x=0

Rearranging to solve for :

x=+5

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Example Question #303 : Gre Subject Test: Chemistry

250mL of 2N HClO_4 is added to 100mL of 5N NH_3 . An indicator in the solution is known to be yellow at any pH greater than 8.3 and green at any pH less than 8.3. Which of the following best describes the solution once it reaches equilibrium?

Possible Answers:

Neutral and green

Acidic and green

Basic and yellow

Basic and green

Acidic and yellow

Correct answer:

Acidic and green

Explanation:

Each of these compounds requires one equivalent of H⁺ is added to 100mL of 5N NH₃mol, so for HClO₄, 2N = 2M, and for NH₃, 5N = 5M.

Using the concentrations and volumes, we can find the moles, finding $\dpi{100} \small 0.25\times 2=0.5mol$ HClO₄ and $\dpi{100} \small 0.10\times 5=0.5mol$ NH₃.

In this case, equivalents of acid are equal to the equivalents of base, meaning that we are at the equivalence point in a titration. HClO₄ is a strong acid, and NH₃ is a weak base.

Thus, the acid will fully dissociate, while the base will not, resulting in a greater concentration of H⁺ than OH^– in the solution. This means the resulting solution will be acidic. We know that the indicator changes from yellow to green at 8.3, which is a basic pH. Our initial solution is basic, and we must pass through the pH of 8.3 to reach our final acidic solution, with pH < 8.3, meaning that the indicator must change from yellow to green during the reaction. This gives out final answer that the solution will be acidic and green.

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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?

$C_6H_{12}O_6+O_2\rightarrow CO_2+H_2O$

Possible Answers:

224g

132g

148g

72g

4.98g

Correct answer:

132g

Explanation:

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

$C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O$

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.

$1mol\ C_6H_{12}O_6*\frac{6mol\ O_2}{1mol\ C_6H_{12}O_6}=6mol\ O_2$

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.

$3mol\ O_2*\frac{6mol\ CO_2}{6mol\ O_2}*\frac{44g\ CO_2}{1mol\ CO_2}=132g\ CO_2$

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Example Question #1 : Gravimetric Analysis

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . If a 10.00 g vinegar sample contains $0.1024\ moles$ of acetic acid, calculate the percent (mass/mass) of the $CH_{3}COOH$ in the vinegar sample.

Density of vinegar is the following:

$CH_{3}COOH=1.05\frac{g}{mL}$

Possible Answers:

$3.3\%$

$12.8\%$

$75\%$

$61.5\%$

Correct answer:

$61.5\%$

Explanation:

Convert the moles of $CH_{3}COOH$ to grams:

$0.1024\ moles \times \frac{60.05g}{mole}=6.1491g\ CH_{3}COOH$

To calculate the percentage of $CH_{3}COOH$ in the vinegar we need to use the following formula:

$\%CH_{3}COOH=\frac{grams\ of\ CH_{3}COOH}{grams\ of\ vinegar}$

Therefore,

$\frac{6.15g}{10.00g}\times 100=61.5\%\ CH_{3}COOH$

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Example Question #1 : Solubility And Precipitation

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

$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

$\small K_{sp} = 5*10^{-3}$

What is the solubility of barium hydroxide?

Possible Answers:

$\small 0.035M$

$\small 0.05M$

$\small 0.071M$

$\small 0.108M$

Correct answer:

$\small 0.108M$

Explanation:

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 $\small +x$ and $\small +2x$ , respectively.

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

$\begin{matrix} Ba(OH)_2 & Ba^{2+} &OH^- \\ /& 0 & 0\\ / & x & 2x \end{matrix}$

$K_{sp}=[Ba^{2+}][OH^-]^2$

$\small 5*10^{-3} = [x][2x]^{2}$

$\small x = 0.108M$

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Example Question #304 : Gre Subject Test: Chemistry

Write the solubility product, $K_{sp}$ , for $BaSO_4_{(s)}$ .

Possible Answers:

$K_{sp} = [Ba][SO_{4}]^{2-}$

$K_{sp} = [Ba][SO_{4}]$

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

$K_{sp} = [Ba]^{4+}[SO_{4}]^{2-}$

$K_{sp} = [Ba]^{2+}[SO_{4}]^{2-}$

Correct answer:

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

Explanation:

The solubility product constant ( $K_{sp}$ ) 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 $BaSO_4_{(s)}$ can be written as:

$K_{eq} = \frac{[Ba^{2+}][SO_{4^}^{}^{2-}]}{[BaSO_{4}(s)]}$

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

$K_{eq}*[BaSO_{4}]=[Ba^{2+}][SO_{4^}^{}^{2-}]$

The product, $K_{eq}*[BaSO_{4}]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

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Example Question #2 : Solubility And Precipitation

Write the solubility product, $K_{sp}$ , for $AgCN_{(s)}$ .

Possible Answers:

$K_{sp} = [Ag^{2+}][CN^{2-}]$

$K_{sp} = [Ag][CN^{-}]$

$K_{sp} = [Ag][CN]$

$K_{sp} = [Ag^{+}][CN^{-}]$

$K_{sp} = [Ag]^{+}[CN]^{-}$

Correct answer:

$K_{sp} = [Ag^{+}][CN^{-}]$

Explanation:

$K_{eq} = \frac{[Ag^{+}][CN^{-}]}{[AgCN(s)]}$

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

$K_{eq}*[AgCN(s)]=[Ag^{+}][CN^{-}]$

The product, $K_{eq}*[AgCN]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Ag^{+}][CN^{-}]$

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GRE Subject Test: Chemistry : GRE Subject Test: Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #33 : Analytical Chemistry

Example Question #34 : Analytical Chemistry

Example Question #303 : Gre Subject Test: Chemistry

Example Question #32 : Analytical Chemistry

Example Question #303 : Gre Subject Test: Chemistry

Example Question #1 : Limiting Reagent

Example Question #1 : Gravimetric Analysis

Example Question #1 : Solubility And Precipitation

Example Question #304 : Gre Subject Test: Chemistry

Example Question #2 : Solubility And Precipitation

All GRE Subject Test: Chemistry Resources

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