Chemistry 103 - Chapter 10

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Chemistry 103 - Chapter 10 - Quiz

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Questions and Answers
  • 1. 

    If 20L of H gas are heated from 298.15 K to 552.50 C calculate the new volume.

    • A.

      52.7 L

    • B.

      69.98 L

    • C.

      90.22 L

    • D.

      55.37 L

    Correct Answer
    D. 55.37 L
    Explanation
    The given question provides the initial volume of H gas as 20L and the initial temperature as 298.15K. The final temperature is given as 552.50C. To calculate the new volume, we can use the ideal gas law equation: V1/T1 = V2/T2. Rearranging the equation, we get V2 = (V1 * T2) / T1. Plugging in the values, we get V2 = (20 * 825.65) / 298.15 = 55.37 L. Therefore, the correct answer is 55.37 L.

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  • 2. 

    Which of the following has the highest molar density ? 

    • A.

      N2

    • B.

      O2

    • C.

      CO2

    • D.

      H2

    Correct Answer
    C. CO2
    Explanation
    CO2 has the highest molar density compared to the other gases listed. Molar density refers to the number of moles of a substance per unit volume. CO2 has a molar mass of 44 g/mol, which is higher than the molar masses of N2 (28 g/mol), O2 (32 g/mol), and H2 (2 g/mol). Since molar density is directly proportional to molar mass, CO2 will have the highest molar density among the given options.

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  • 3. 

    What is the V in L occupied by 25g of HCL at STP ? ( R = 0.0821 L.atm/mol.K ) 

    • A.

      14.9 L

    • B.

      15.3 L

    • C.

      7.9 L

    • D.

      26.0 L

    Correct Answer
    B. 15.3 L
    Explanation
    The V in L represents the volume of the gas, which is occupied by 25g of HCL at STP. To find the volume, we can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. At STP, the pressure is 1 atm and the temperature is 273 K. We can calculate the number of moles of HCL using its molar mass, which is 36.461 g/mol. By rearranging the ideal gas law equation and plugging in the given values, we can solve for V, which is approximately 15.3 L.

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  • 4. 

    How many moles of NO2 gas we have if V=5L at 50c and 375 torr ? ( R = 0.0821 L.atm/mol.K )

    • A.

      0.183 mole

    • B.

      0.122 mole 

    • C.

      2.56 mole

    • D.

      4.00 mole 

    Correct Answer
    A. 0.183 mole
    Explanation
    The question is asking for the number of moles of NO2 gas given the volume, temperature, and pressure. To solve this, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation, we have n = PV / RT. Plugging in the given values, we get n = (375 torr * 5L) / (0.0821 L.atm/mol.K * 323K) = 0.183 mole. Therefore, the correct answer is 0.183 mole.

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  • 5. 

    CH4 gas and SO2 gas, the CH4 gas Diffuse ............... times at that of SO2 . If the M.W(CH4) =16 g/mole, M.W(SO2)=64g/mole.

    • A.

      4

    • B.

      8

    • C.

      2

    • D.

      0.4 

    Correct Answer
    C. 2
    Explanation
    The diffusion rate of a gas is inversely proportional to the square root of its molar mass. Since the molar mass of CH4 is 16 g/mole and the molar mass of SO2 is 64 g/mole, the ratio of their diffusion rates would be the square root of the ratio of their molar masses. The square root of 16/64 is 1/2, so the diffusion rate of CH4 gas is 2 times that of SO2 gas.

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  • 6. 

    Which of the gases diffuse faster ?

    • A.

      H2

    • B.

      Ne 

    • C.

      Ar

    • D.

      O2

    Correct Answer
    A. H2
    Explanation
    H2 gas diffuses faster compared to Ne, Ar, and O2 gases. This is because H2 gas has a smaller molecular size and lower molecular weight compared to the other gases. The smaller size allows H2 molecules to move more easily through small openings and gaps, resulting in faster diffusion. Additionally, the lower molecular weight of H2 means that it has higher average kinetic energy, leading to faster molecular motion and diffusion.

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  • 7. 

    A mixture of Ne , Xe and Ar has a total pressure 8.3 atm , what is the mole fraction of Ar if the partial pressure of Ne and Xe are 2.3 atm and 1.4 atm , respectively ?

    • A.

      0.55 

    • B.

      0.28

    • C.

      0.45

    • D.

      0.17

    Correct Answer
    A. 0.55 
    Explanation
    The mole fraction of a gas in a mixture is calculated by dividing the partial pressure of that gas by the total pressure of the mixture. In this case, the partial pressure of Ar is not given directly, but it can be calculated by subtracting the partial pressures of Ne and Xe from the total pressure. Therefore, the mole fraction of Ar can be calculated by dividing the partial pressure of Ar by the total pressure. The partial pressure of Ar is 8.3 atm - 2.3 atm - 1.4 atm = 4.6 atm. Therefore, the mole fraction of Ar is 4.6 atm / 8.3 atm = 0.55.

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  • 8. 

    Arrange the following gases in order of increasing average molecular speed at 25 C . He , O2 , CO2 , N2

    • A.

      He < O2 < N2 < CO2 

    • B.

      He < N2 < O2 < CO2 

    • C.

      CO2 < O2 < N2 < He

    • D.

      N2 < O2 < He < CO2 

    Correct Answer
    C. CO2 < O2 < N2 < He
    Explanation
    The average molecular speed of a gas is determined by its molar mass. He (helium) has the lowest molar mass, followed by N2 (nitrogen), O2 (oxygen), and CO2 (carbon dioxide) which has the highest molar mass. Therefore, the correct order of increasing average molecular speed at 25 C is CO2 < O2 < N2 < He.

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  • 9. 

    Density of nobel gas 4.42g/L at  303K and 1.31 atm ,what is the chemical formula of nobel gas ?

    • A.

      Ne

    • B.

      Ar

    • C.

      Xe

    • D.

      Kr

    Correct Answer
    D. Kr
    Explanation
    The given information states that the density of the noble gas is 4.42g/L at 303K and 1.31 atm. Among the options provided, Kr (Krypton) is the only noble gas that has a density close to 4.42g/L. Therefore, the chemical formula of the noble gas is Kr.

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  • 10. 

    If the pressure of confined gas sample doubled , while the volume was held constant , what happened to the temperature? 

    • A.

      It halved 

    • B.

      It went up by a factor of four

    • C.

      It doubled

    • D.

      Cannot be determined with information given 

    Correct Answer
    C. It doubled
    Explanation
    When the pressure of a confined gas sample doubles while the volume is held constant, according to Boyle's Law, the temperature of the gas also doubles. Boyle's Law states that the pressure and volume of a gas are inversely proportional when the temperature is kept constant. Therefore, if the pressure doubles, the temperature must also double to maintain the constant volume.

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  • 11. 

    The pressure of a 4.0 L N2 is decreased to one-third of its original pressure , and its absolute T is decreased by one half . The volume is now :

    • A.

      8.0 L

    • B.

      6.0 L

    • C.

      4.0 L

    • D.

      1.0 L 

    Correct Answer
    B. 6.0 L
    Explanation
    When the pressure of a gas is decreased to one-third of its original pressure, and the absolute temperature is decreased by one-half, according to the combined gas law, the volume of the gas will increase. The combined gas law states that the product of pressure and volume is directly proportional to the absolute temperature. Therefore, when the pressure is decreased and the temperature is decreased, the volume will increase. In this case, since the pressure is decreased to one-third and the temperature is decreased by one-half, the volume will increase by a factor of 3 and 2 respectively. Therefore, the volume is now 6.0 L.

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  • 12. 

    The rate of effusion of a gas A is 1.14 times that of SO2 . The density of gas A ( in g/L ) at 572 mmHg at 100 C is : 

    • A.

      1.171

    • B.

      1.256

    • C.

      1.212

    • D.

      1.131 

    Correct Answer
    A. 1.171
    Explanation
    The rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since the rate of effusion of gas A is 1.14 times that of SO2, we can conclude that the molar mass of gas A is 1.14 times that of SO2. Given that the density of gas A is directly proportional to its molar mass, we can also conclude that the density of gas A is 1.14 times that of SO2. Therefore, the density of gas A at 572 mmHg at 100 C is 1.171 g/L.

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  • Current Version
  • Mar 14, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Dec 03, 2019
    Quiz Created by
    Pharmacgy
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