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GRE Subject Test: Chemistry : General Chemistry

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

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1 Diagnostic Test 152 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #151 : General Chemistry

What happens when a solution of NaCl is added to a solution of AgNO_3 ?

Possible Answers:

A decomposition reaction

A synthesis reaction

A combination reaction

A double displacement reaction

A combustion reaction

Correct answer:

A double displacement reaction

Explanation:

The chemical reaction that would occur is provided below:

$AgNO_{3(aq)}+NaCl_{(aq)}\rightarrow NaNO_{3(aq)} + AgCl_{(s)}$

This type of reaction is called a double displacement reaction in which there is an exchange of ions between two compounds. This type of reaction usually results in the formation of a precipitate or gas. In terms of the positive ions, the $Ag^{+}$ switch places $Na^{+}$ to form two new products.

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Example Question #2 : Reactions With Inorganic Compounds

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

$C_{3}H_{8}\ +\ 5O_{2}\rightarrow\ 3CO_{2}\ +\ 4H_{2}O$

Possible Answers:

Synthesis

Alkylation

Combustion

Catalytic

Correct answer:

Combustion

Explanation:

The type of reaction given is termed a combustion reaction. A combustion reaction is one that involves a substance reacting with molecular oxygen. For combustion reactions involving a hydrocarbon as given in the example, the major products are carbon dioxide and water. The word combustion is a synonym for burn.

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Example Question #152 : General Chemistry

For the reaction given, identify the type of reaction taking place.

Possible Answers:

Combination reaction

Synthesis reaction

Displacement reaction

Decomposition reaction

Correct answer:

Displacement reaction

Explanation:

A displacement reaction is one in which an atom, ion or molecule moves from one compound to another to replace an atom, ion or molecule in another compound. In the reaction given, $SO_{4}^{2-}$ ion moves from $CuSO_{4}$ to Zn(s) to form $ZnSO_{4}$ .

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Example Question #3 : Reactions With Inorganic Compounds

Determine the percentage yield of the reaction given if 12.6 grams of $Ba(NO)_{3}$ is reacted excess $Na_{2}SO_{4}$ to give 10.4 grams of $BaSO_{4}$ .

Possible Answers:

$92.0 \%$

$83.5 \%$

$77.0 \%$

$2.0 \%$

Correct answer:

$92.0 \%$

Explanation:

In chemistry, percentage yield is a term used to quantify the efficiency of a chemical reaction. In order to calculate this quantity, the value for the actual yield and theoretical yield is needed. The actual yield is the amount of product obtained during a chemical reaction. The theoretical yield is the maximum amount of product possible to be obtain by a chemical reaction which is based on the amount of reactants used. Below is the formula needed to calculate the percentage yield.

Convert the number of grams of $Ba(NO_{3})_{2}$ to moles:

$12.6 g\ Ba(NO_{3})_{2}\times \frac{mole}{261.37g}=0.0482\ moles\ Ba(NO_{3})_{2}$

Based on the chemical equation 1 mole of $BaSO_{4}$ is produced for every 1 mole of $Ba(NO_{3})_{2}$ reacted.

$moles\ of Ba(NO_{3})_{2}=moles\ of\ BaSO_{4}$

Convert the moles of $BaSO_{4}$ to grams:

$0.0482\ moles\ BaSO_{4} \times\ \frac{233.43g}{moles}=11.3g\ BaSO_{4}$

$Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$Percentage\ Yield =\frac{10.4g}{11.3g}\times100=92.0 \%$

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Example Question #4 : Reactions With Inorganic Compounds

Determine the percentage yield of the reaction given if 21.0 grams of NaOH is reacted with excess HCl to give 10.2 grams of NaCl .

Possible Answers:

$64.3 \%$

$33.2 \%$

$16.6 \%$

$53.0 \%$

Correct answer:

$33.2 \%$

Explanation:

Convert the number of grams of NaOH to moles:

$21.0 g\ NaOH\times \frac{mole}{40.0g}=0.525\ moles\ NaOH$

Based on the chemical equation 1 mole of NaCl is produced for every 1 mole of NaOH reacted.

$moles\ of\ NaOH=moles\ of\ NaCl$

Convert the moles of NaCl to grams:

$0.525\ moles\ \times\ \frac{58.44g}{moles}=30.7g\ NaCl$

$Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$Percent\ Yield =\frac{10.2g}{30.7g}\times100=33.2 \%$

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Example Question #5 : Reactions With Inorganic Compounds

Determine the theoretical yield in grams of $Al_{2}(SO_{4})_{3}$ produced if 5.6 grams of $Al(OH)_{3}$ is reacted with excess $H_{2}SO_{4}$ .

Possible Answers:

12.3

4.6

4.7

1.1

Correct answer:

12.3

Explanation:

Convert the number of grams of $Al(OH)_{3}$ to moles:

$5.6 g\ Al(OH)_{3}\times \frac{mole}{78.00g}=0.0718\ moles\ Al(OH)_{3}$

Based on the chemical equation 1 mole of $Al_{2}(SO_{4})_{3}$ is produced for every 2 mole of $Al(OH)_{3}$ reacted.

$2\ moles\ of Al(OH)_{3}=1\ mole\ of\ Al_{2}(SO_{4})_{3}$

Therefore: $0.0718\ moles\ Al(OH)_{3}\times \frac{1\ moles\ Al_{2}(SO_{4})_{3}}{2\ moles\ Al(OH)_{3}}=0.0359\ moles\ Al_{2}(SO_{4})_{3}$

Convert the moles of $Al_{2}(SO_{4})_{3}$ to grams:

$0.0359\ moles\ Al_{2}(SO_{4})_{3} \times\ \frac{342.15g}{moles}=12.3g\ Al_{2}(SO_{4})_{3}$

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Example Question #6 : Reactions With Inorganic Compounds

For the reaction given, identify the type of reaction taking place.

Possible Answers:

Synthesis reaction

Double displacement reaction

Combination reaction

Decomposition reaction

Correct answer:

Double displacement reaction

Explanation:

A displacement reaction is one in which an atom, ion or molecule moves from one compound to another to replace an atom, ion or molecule in another compound. In the reaction given, $NO_{3}^{-}$ ion moves from $AgNO_{3}$ to $Na_{2}CO_{3}$ . Also, $CO_{3}^{2-}$ ion moves from $Na_{2}CO_{3}$ to $AgNO_{3}$ .

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Example Question #3 : Reactions With Inorganic Compounds

Determine the percentage yield of the reaction given if 40.0 grams of $CaCO_{3}$ is heated to give 15.0 grams of CaO .

Possible Answers:

$68.2 \%$

$50.1 \%$

$25.1 \%$

$76.4 \%$

Correct answer:

$68.2 \%$

Explanation:

Convert the number of grams of $CaCO_{3}$ to moles:

$40.0 g\ CaCO_{3}\times \frac{mole}{100.09g}=0.3996\ moles\ CaCO_{3}$

Based on the chemical equation 1 mole of CaO is produced for every 1 mole of $CaCO_{3}$ reacted.

$moles\ of\ CaCO_{3}=moles\ of\ CaO$

Convert the moles of CaO to grams:

$0.3996\ moles\ \times\ \frac{56.08g}{moles}=22.0g\ CaO$

$Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$Percentage\ Yield =\frac{15.0g}{22.0g}\times100=68.2 \%$

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Example Question #153 : General Chemistry

For the reaction given, identify the type of reaction taking place.

Possible Answers:

Combination reaction

Synthesis reaction

Decomposition reaction

Displacement reaction

Correct answer:

Displacement reaction

Explanation:

A displacement reaction is one in which an atom, ion or molecule moves from one compound to another to replace an atom, ion or molecule in another compound. In the reaction given, 2 $Cl^{-}$ ions move from 2 moles of HCl to Cu(s) to form $CuCl_{2}$ .

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Example Question #1 : Equilibrium Constant And Reaction Quotient

$CO(g) + 2H_{2}(g) \rightarrow CH_{3}OH(g)$

$K_{eq} = 14.5$

The initial concentrations of this reaction are listed below.

$[CH_{3}OH] = 2.5M, [CO] = 1.3M, [H_{2}] =0.6M.$

Based on these initial concentrations, which statement is true?

Possible Answers:

The reaction will shift left

$Q=\frac{[CO][H_2]^2}{[CH_3OH]}$

The reaction is in equilibrium

The forward rate will be greater than the reverse rate

The ratio of products to reactants is greater than the equilibrium constant

Correct answer:

The forward rate will be greater than the reverse rate

Explanation:

When given initial concentrations, we can determine the reaction quotient (Q) of the reaction. The reaction quotient is given by the same equation as the equilibrium constant (concentration of products divided by concentration of reactants), but its value will fluctuate as the system reacts, whereas the equilibrium constant is based on equilibrium concentrations.

By comparing the reaction quotient to the equilibrium constant, we can determine in which direction the reaction will proceed initially.

$Q\ \text{or}\ K_{eq}=\frac{[CH_3OH]}{[H_2]^2[CO]}$

The reaction quotient with the beginning concentrations is written below.

$\frac{[2.5]}{[0.6]^{2}[1.3]} = 5.3$

This shows that the ratio of products to reactants is less than the equilibrium constant. Since Q is less than K_eq in the beginning, we conclude that the reaction will proceed forward until Q is equal to K_eq. In this manner, the denominator (reactants) will decrease and the numerator (products) will increase, causing Q to become closer to K_eq.

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All GRE Subject Test: Chemistry Resources

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GRE Subject Test: Chemistry : General Chemistry

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #151 : General Chemistry

Example Question #2 : Reactions With Inorganic Compounds

Example Question #152 : General Chemistry

Example Question #3 : Reactions With Inorganic Compounds

Example Question #4 : Reactions With Inorganic Compounds

Example Question #5 : Reactions With Inorganic Compounds

Example Question #6 : Reactions With Inorganic Compounds

Example Question #3 : Reactions With Inorganic Compounds

Example Question #153 : General Chemistry

Example Question #1 : Equilibrium Constant And Reaction Quotient

All GRE Subject Test: Chemistry Resources

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