Chapter 3 Chemistry Quiz

1. ... $straight C subscript 3 straight H subscript 8$ (g) + ... $straight O subscript 2$ (g) --> ... $straight H subscript 2 straight O$ (g) + ... $CO subscript 2$ (g)When the equation for the reaction represented above is balanced and all coefficients are reduced to the lowest whole-number terms, the coefficient for $straight O subscript 2$ (g) is

1

2

3

5

6

In a balanced chemical equation, the coefficients represent the relative number of molecules or moles of each substance involved in the reaction. In this equation, there are two reactant molecules of (g) and two product molecules of (g). To balance the equation, we need to ensure that the number of (g) molecules is equal on both sides. By adding a coefficient of 5 in front of the (g) on the left side, we can achieve this balance. Therefore, the coefficient for (g) is 5.

Explanation

In a balanced chemical equation, the coefficients represent the relative number of molecules or moles of each substance involved in the reaction. In this equation, there are two reactant molecules of (g) and two product molecules of (g). To balance the equation, we need to ensure that the number of (g) molecules is equal on both sides. By adding a coefficient of 5 in front of the (g) on the left side, we can achieve this balance. Therefore, the coefficient for (g) is 5.

2. In 1.00 mol of potassium zirconium sulfate trihydrate, $straight K subscript 4 Zr left parenthesis SO subscript 4 right parenthesis subscript 3 times straight H subscript 2 straight O$ , there are

3 x 6.02 x 10 to the power of 23 hydrogen atoms

6.02 x 10 to the power of 23 sulfur atoms

4 x 6.02 x 10 to the power of 23 potassium atoms

4 moles of oxygen atoms

4 moles of zirconium atoms

In 1.00 mol of potassium zirconium sulfate trihydrate, there are 4 x 6.02 x 10 to the power of 23 potassium atoms. This is because the mole ratio of potassium atoms to the compound is 4:1. Therefore, for every 1 mole of the compound, there are 4 moles of potassium atoms. Since we have 1.00 mol of the compound, we can multiply this by the mole ratio of potassium to the compound to find the number of potassium atoms, which is 4 x 6.02 x 10 to the power of 23.

Explanation

In 1.00 mol of potassium zirconium sulfate trihydrate, there are 4 x 6.02 x 10 to the power of 23 potassium atoms. This is because the mole ratio of potassium atoms to the compound is 4:1. Therefore, for every 1 mole of the compound, there are 4 moles of potassium atoms. Since we have 1.00 mol of the compound, we can multiply this by the mole ratio of potassium to the compound to find the number of potassium atoms, which is 4 x 6.02 x 10 to the power of 23.

3. How many carbon atoms are contained in 2.8 g of $straight C subscript 2 straight H subscript 4$ ?

1.2 x 10 to the power of 23

3.0 x 10 to the power of 23

4.6 x 10 to the power of 23

1.2 x 10 to the power of 24

6.0 x 10 to the power of 23

The given answer, 1.2 x 10 to the power of 23, suggests that there are 1.2 x 10 to the power of 23 carbon atoms contained in 2.8 g of a substance. This answer is based on the concept of Avogadro's number, which states that one mole of any substance contains 6.022 x 10 to the power of 23 particles. Since the question asks for the number of carbon atoms, we can assume that the substance mentioned is a pure carbon compound. Therefore, 2.8 g of this substance would contain 1.2 x 10 to the power of 23 carbon atoms.

Explanation

The given answer, 1.2 x 10 to the power of 23, suggests that there are 1.2 x 10 to the power of 23 carbon atoms contained in 2.8 g of a substance. This answer is based on the concept of Avogadro's number, which states that one mole of any substance contains 6.022 x 10 to the power of 23 particles. Since the question asks for the number of carbon atoms, we can assume that the substance mentioned is a pure carbon compound. Therefore, 2.8 g of this substance would contain 1.2 x 10 to the power of 23 carbon atoms.

4. The atomic mass of copper is 63.55. Given that there are only two naturally occurring isotopes of copper. $blank to the power of 63 Cu$ and $blank to the power of 65 Cu$ , the naturally abundance of the $blank to the power of 65 Cu$ isotope must be approximately

90%

70%

50%

25%

10%

Based on the given information that there are only two naturally occurring isotopes of copper, we can deduce that the sum of their abundances must be 100%. Since the atomic mass of copper is 63.55, and the naturally abundant isotope is not specified, we can assume that the other isotope has a mass of 63.55 - atomic mass of the abundant isotope. Therefore, the naturally abundant isotope must be approximately 63.55 - (63.55 - atomic mass of the abundant isotope), which simplifies to approximately 63.55 - 63.55 + atomic mass of the abundant isotope. So, the naturally abundant isotope must be approximately 100% - atomic mass of the abundant isotope%. Since the answer is 25%, it means that the atomic mass of the abundant isotope is 75%.

Explanation

Based on the given information that there are only two naturally occurring isotopes of copper, we can deduce that the sum of their abundances must be 100%. Since the atomic mass of copper is 63.55, and the naturally abundant isotope is not specified, we can assume that the other isotope has a mass of 63.55 - atomic mass of the abundant isotope. Therefore, the naturally abundant isotope must be approximately 63.55 - (63.55 - atomic mass of the abundant isotope), which simplifies to approximately 63.55 - 63.55 + atomic mass of the abundant isotope. So, the naturally abundant isotope must be approximately 100% - atomic mass of the abundant isotope%. Since the answer is 25%, it means that the atomic mass of the abundant isotope is 75%.

5. ... $Ca subscript 3 left parenthesis PO subscript 4 right parenthesis subscript 2$ (s) + ... $straight H subscript 3 PO subscript 4$ (l) $rightwards arrow$ ... $Ca left parenthesis straight H subscript 2 PO subscript 4 right parenthesis subscript 2$ (s)When the equation above is balanced and all coefficients are reduced to lowest whole-number terms, what is the coefficient of $straight H subscript 3 PO subscript 4$ (l)?

1

2

3

4

5

not-available-via-ai

Explanation

not-available-via-ai

6. What is the maximum number of moles of $Al subscript 2 straight O subscript 3$ that can be produced by the reaction of 0.40 mol of Al with 0.40 mol of $straight O subscript 2$

0.10 mol

0.20 mol

0.27 mol

0.33 mol

0.40 mol

The maximum number of moles of a product that can be produced in a chemical reaction is determined by the limiting reactant. In this case, the limiting reactant is Al, since both Al and are present in equal amounts. The balanced equation for the reaction is not provided, so it is not possible to calculate the exact amount of moles of produced. Therefore, without additional information, it is not possible to determine the maximum number of moles of that can be produced.

Explanation

The maximum number of moles of a product that can be produced in a chemical reaction is determined by the limiting reactant. In this case, the limiting reactant is Al, since both Al and are present in equal amounts. The balanced equation for the reaction is not provided, so it is not possible to calculate the exact amount of moles of produced. Therefore, without additional information, it is not possible to determine the maximum number of moles of that can be produced.

7. If 0.40 mol of $straight H subscript 2$ and 0.15 mol of $straight O subscript 2$ were to react as completely as possible to produce $straight H subscript 2 straight O$ , what mass of reactant would remain?

0.20 g of H subscript 2

0.40 g of H subscript 2

3.2 g of H subscript 2

4.0 g of H subscript 2

4.4 g of H subscript 2

The balanced chemical equation for the reaction between H2 and O2 to produce H2O is 2H2 + O2 -> 2H2O. According to the stoichiometry of the equation, 2 moles of H2 are required to react with 1 mole of O2 to produce 2 moles of H2O. Since there is an excess of H2 in this case (0.40 mol H2 vs 0.15 mol O2), all the O2 will be used up and 0.30 mol of H2 will react. The remaining 0.10 mol of H2 will not react and will remain. To calculate the mass of the remaining H2, we can use the molar mass of H2 (2 g/mol). Therefore, the mass of the remaining H2 is 0.10 mol * 2 g/mol = 0.20 g.

Explanation

The balanced chemical equation for the reaction between H2 and O2 to produce H2O is 2H2 + O2 -> 2H2O. According to the stoichiometry of the equation, 2 moles of H2 are required to react with 1 mole of O2 to produce 2 moles of H2O. Since there is an excess of H2 in this case (0.40 mol H2 vs 0.15 mol O2), all the O2 will be used up and 0.30 mol of H2 will react. The remaining 0.10 mol of H2 will not react and will remain. To calculate the mass of the remaining H2, we can use the molar mass of H2 (2 g/mol). Therefore, the mass of the remaining H2 is 0.10 mol * 2 g/mol = 0.20 g.

8. In which of the following compounds is the mass ratio of chromium to oxygen closest to 1.62 to 1.00?

CrO3

CrO2

CrO

Cr2O

Cr2O3

The compound CrO2 has the closest mass ratio of chromium to oxygen to 1.62 to 1.00. This is because in CrO2, there is one chromium atom and two oxygen atoms, resulting in a ratio of 1:2. The other compounds listed have different ratios, such as CrO3 with a ratio of 1:3 and Cr2O3 with a ratio of 2:3. Therefore, CrO2 is the closest to the desired ratio.

Explanation

The compound CrO2 has the closest mass ratio of chromium to oxygen to 1.62 to 1.00. This is because in CrO2, there is one chromium atom and two oxygen atoms, resulting in a ratio of 1:2. The other compounds listed have different ratios, such as CrO3 with a ratio of 1:3 and Cr2O3 with a ratio of 2:3. Therefore, CrO2 is the closest to the desired ratio.

9. A compound that contains only C and H was burned in excess oxygen. The following masses were obtained after the complete combustion of the compound in oxygen: 22.0 g of $CO subscript 2$ and 18.0 g $straight H subscript 2 straight O$ was produced by the combustion. Identify the correct empirical formula for the hydrocarbon.

CH

CH subscript 2

CH subscript 3

CH subscript 4

The empirical formula of a compound represents the simplest whole number ratio of the atoms present in the compound. In this case, the compound contains only carbon (C) and hydrogen (H). The given information states that 22.0 g of carbon dioxide (CO2) and 18.0 g of water (H2O) were produced by the combustion of the compound. From this information, we can determine the ratio of carbon to hydrogen in the compound. The molar mass of CO2 is 44.01 g/mol, and the molar mass of H2O is 18.02 g/mol. By dividing the mass of carbon dioxide by its molar mass and the mass of water by its molar mass, we find that the ratio of carbon to hydrogen is approximately 1:4. Therefore, the correct empirical formula for the hydrocarbon is CH4.

Explanation

The empirical formula of a compound represents the simplest whole number ratio of the atoms present in the compound. In this case, the compound contains only carbon (C) and hydrogen (H). The given information states that 22.0 g of carbon dioxide (CO2) and 18.0 g of water (H2O) were produced by the combustion of the compound. From this information, we can determine the ratio of carbon to hydrogen in the compound. The molar mass of CO2 is 44.01 g/mol, and the molar mass of H2O is 18.02 g/mol. By dividing the mass of carbon dioxide by its molar mass and the mass of water by its molar mass, we find that the ratio of carbon to hydrogen is approximately 1:4. Therefore, the correct empirical formula for the hydrocarbon is CH4.

10. A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula of the compound?

SF subscript 2

SF subscript 3

SF subscript 4

SF subscript 5

SF subscript 6

The empirical formula of a compound represents the simplest whole number ratio of the atoms present in the compound. To determine the empirical formula, we need to find the ratio of sulfur to fluorine in the given sample. The given sample contains 3.21 g of sulfur and 11.4 g of fluorine. To find the ratio, we divide the mass of each element by its molar mass. The molar mass of sulfur is 32.06 g/mol and the molar mass of fluorine is 18.99 g/mol. Dividing the mass of sulfur by its molar mass gives us 0.1 mol, and dividing the mass of fluorine by its molar mass gives us 0.6 mol. The ratio of sulfur to fluorine is 1:6, which corresponds to the empirical formula SF subscript 6.

Explanation

The empirical formula of a compound represents the simplest whole number ratio of the atoms present in the compound. To determine the empirical formula, we need to find the ratio of sulfur to fluorine in the given sample. The given sample contains 3.21 g of sulfur and 11.4 g of fluorine. To find the ratio, we divide the mass of each element by its molar mass. The molar mass of sulfur is 32.06 g/mol and the molar mass of fluorine is 18.99 g/mol. Dividing the mass of sulfur by its molar mass gives us 0.1 mol, and dividing the mass of fluorine by its molar mass gives us 0.6 mol. The ratio of sulfur to fluorine is 1:6, which corresponds to the empirical formula SF subscript 6.