Know About Chemical Quiz

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| By Ashley Vickers
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Ashley Vickers
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Quizzes Created: 7 | Total Attempts: 5,318
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Know About Chemical Quiz - Quiz

Chemists deserve respect, as the content that they study is not a walk in the park. Among the many things that they are expected to master are chemical elements and compounds. The chemical quiz below tests on this and more.


Questions and Answers
  • 1. 

    The volume of water that must be added in order to dilute 40 mL of 9.0 M HCl to a concentration of 6.0 M is closest to

    • A.

      10 mL

    • B.

      20 mL

    • C.

      30 mL

    • D.

      40 mL

    • E.

      60 mL

    Correct Answer
    B. 20 mL
    Explanation
    To dilute a solution, the formula C1V1 = C2V2 can be used, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. In this case, the initial concentration is 9.0 M, the initial volume is 40 mL, the final concentration is 6.0 M, and the final volume is unknown. Plugging these values into the formula, we get (9.0 M)(40 mL) = (6.0 M)(V2). Solving for V2, we find that V2 is equal to 60 mL. Since we already have 40 mL of the 9.0 M HCl solution, we need to add an additional volume of 60 mL - 40 mL = 20 mL of water to dilute it to a concentration of 6.0 M. Therefore, the closest answer is 20 mL.

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  • 2. 

    A chemical supply company sells a concentrated solution of aqueous  (molar mass 98 g/mol) that is 50. percent by mass.  At 25C, the density of the solution is 1.4 g/mL.  What is the molarity of the  solution at 25C?

    • A.

      1.8 M

    • B.

      3.6 M

    • C.

      5.1 M

    • D.

      7.1 M

    Correct Answer
    D. 7.1 M
    Explanation
    The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we need to find the moles of solute in the solution. The solution is 50% by mass, so we can assume that 100 g of the solution contains 50 g of the solute. Using the molar mass of the solute (98 g/mol), we can calculate the moles of solute: 50 g / 98 g/mol = 0.51 mol. The density of the solution is given as 1.4 g/mL, which means that 1000 mL of the solution has a mass of 1400 g. Using the density and mass, we can calculate the volume of the solution: 1400 g / 1.4 g/mL = 1000 mL = 1 L. Now we can calculate the molarity by dividing the moles of solute by the volume of the solution: 0.51 mol / 1 L = 0.51 M. Therefore, the molarity of the solution is 0.51 M, which is closest to the answer 7.1 M.

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  • 3. 

    A 360. mg sample of aspirin, , (molar mass 180 g), is dissolved in enough water to produce 200. mL of solution.  What is the molarity of aspirin in a 50. mL sample of this solution?

    • A.

      0.0800 M

    • B.

      0.0400 M

    • C.

      0.0200 M

    • D.

      0.0100 M

    • E.

      0.00250 M

    Correct Answer
    D. 0.0100 M
    Explanation
    The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we have a 360 mg sample of aspirin, which is equivalent to 0.360 g. To find the moles of aspirin, we divide the mass by the molar mass: 0.360 g / 180 g/mol = 0.002 mol. The volume of the solution is given as 200 mL, which is equivalent to 0.200 L. To find the molarity, we divide the moles of aspirin by the volume of the solution: 0.002 mol / 0.200 L = 0.0100 M. Therefore, the molarity of aspirin in a 50 mL sample of this solution is 0.0100 M.

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  • 4. 

    Appromixately what mass of (250 g/mol) is required to prepare 250 mL of 0.10 M copper (II) sulfate solution?

    • A.

      4.0 g

    • B.

      6.2 g

    • C.

      34 g

    • D.

      85 g

    • E.

      140 g

    Correct Answer
    B. 6.2 g
    Explanation
    To calculate the mass of copper (II) sulfate required, we can use the formula:

    Mass = Molarity x Volume x Molar mass

    Given that the molar mass of copper (II) sulfate is 250 g/mol, the volume is 250 mL (which is equivalent to 0.250 L), and the molarity is 0.10 M, we can substitute these values into the formula:

    Mass = 0.10 mol/L x 0.250 L x 250 g/mol = 6.25 g

    Rounding to the nearest tenth, the mass of copper (II) sulfate required is approximately 6.2 g.

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  • 5. 

    A yellow precipitate forms when 0.5M NaI (aq) is added to a 0.5 M solution of which of the following ions? 

    • A.

      Pb to the power of plus 2 end exponent (aq)

    • B.

      Zn to the power of plus 2 end exponent (aq)

    • C.

      CrO subscript 4 to the power of minus 2 end exponent (aq)

    • D.

      SO subscript 4 to the power of minus 2 end exponent (aq)

    • E.

      OH to the power of minus 1 end exponent (aq)

    Correct Answer
    A. Pb to the power of plus 2 end exponent (aq)
    Explanation
    When 0.5M NaI (aq) is added to a solution containing Pb2+ ions, a yellow precipitate forms. This indicates the formation of lead iodide (PbI2), which is insoluble in water. The yellow color of the precipitate is characteristic of lead iodide. The other ions listed do not form a yellow precipitate when reacted with NaI. Therefore, the correct answer is Pb2+ ions.

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  • 6. 

    Oxygen is acting as an oxidizing agent in all of the following reactions EXCEPT

    • A.

      2 C (s) + O2 (g) --> 2 CO (g)

    • B.

      S (s) + O2 (g) --> SO2 (g)

    • C.

      2 F2 (g) + O2 (g) --> 2 OF2 (g)

    • D.

      2 Na (s) + O2 (g) --> Na2O2 (s)

    • E.

      2 Mg (s) + O2 (g) --> 2 MgO (s)

    Correct Answer
    C. 2 F2 (g) + O2 (g) --> 2 OF2 (g)
    Explanation
    In all the given reactions, oxygen is acting as an oxidizing agent except for the reaction 2 F2 (g) + O2 (g) --> 2 OF2 (g). In this reaction, oxygen is not accepting any electrons or causing oxidation. Instead, it is being reduced by accepting electrons from fluorine to form OF2. Therefore, oxygen is not acting as an oxidizing agent in this reaction.

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  • 7. 

    If 200. mL of 0.60 M (aq) is added to 400. mL of distilled water, what is the concentration of (aq) in the resulting solution?  (Assume volumes are additive.)

    • A.

      0.20 M

    • B.

      0.30 M

    • C.

      0.40 M

    • D.

      0.60 M

    • E.

      1.20 M

    Correct Answer
    A. 0.20 M
    Explanation
    When 200 mL of 0.60 M (aq) is added to 400 mL of distilled water, the volumes of the two solutions are added together to give a total volume of 600 mL. The moles of the solute, (aq), remains the same before and after the dilution. Therefore, the concentration of (aq) in the resulting solution can be calculated by dividing the moles of (aq) by the total volume of the solution. Since the moles of (aq) remain the same and the total volume is 600 mL, the concentration of (aq) in the resulting solution is 0.20 M.

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  • 8. 

    What mass of Au is produced with 0.0500 mol of is reduced completely with excess ?

    • A.

      9.85 g

    • B.

      19.7 g

    • C.

      24.5 g

    • D.

      39.4 g

    • E.

      48.9 g

    Correct Answer
    B. 19.7 g
    Explanation
    The molar mass of Au (gold) is 196.97 g/mol. Given that 0.0500 mol of Au is reduced completely, we can calculate the mass of Au produced by multiplying the number of moles by the molar mass: 0.0500 mol * 196.97 g/mol = 9.85 g. Therefore, the correct answer is 9.85 g.

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  • 9. 

    Which of the following is a weak acid in aqueous solution?

    • A.

      HCl

    • B.

      HClO subscript 4

    • C.

      HNO subscript 3

    • D.

      H subscript 2 S

    • E.

      H subscript 2 SO subscript 4

    Correct Answer
    D. H subscript 2 S
    Explanation
    H2S is a weak acid because it only partially dissociates in water, releasing a small amount of H+ ions. Weak acids have a low tendency to donate protons, and H2S falls into this category. In contrast, strong acids like HCl, HClO4, HNO3, and H2SO4 fully dissociate in water, releasing a large amount of H+ ions.

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  • 10. 

    A student mixes equal volumes of 1.0 M solutions of tin(II) chloride and copper(II) sulfate and observes that no precipitate forms.  Then the student mixes equal volumes of 1.0 M solutions of zinc(II) sulfate and tin(II) fluoride and observes the formation of a precipitate.  The formula of the precipitate must be

    • A.

      SnF subscript 2

    • B.

      SnSO subscript 4

    • C.

      Sn(SO subscript 4 ) subscript 2

    • D.

      ZnF

    • E.

      ZnF subscript 2

    Correct Answer
    E. ZnF subscript 2
    Explanation
    When equal volumes of 1.0 M solutions of zinc(II) sulfate and tin(II) fluoride are mixed, a precipitate is formed. This indicates a chemical reaction between the two solutions. The formula of the precipitate is ZnF2 because zinc(II) sulfate contains the Zn2+ ion and tin(II) fluoride contains the Sn2+ ion. The ions combine to form a solid compound with a 1:2 ratio of zinc to fluoride ions, resulting in the formula ZnF2.

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  • Current Version
  • Feb 23, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Sep 08, 2014
    Quiz Created by
    Ashley Vickers
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