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Hydrogen Test - 25

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Hydrogen Test - 25
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  • Question 1
    1 / -0
    Hydrogen has the tendency to lose one electron and form $$H^{+}$$. In this respect, it resembles:
    Solution
    Alkali metals also have a tendency to lose one electron and form cations, so this similar property of hydrogen matches with alkali metals.
    Carbon has a tendency to form covalent bonds.
    Alkaline earth metals have a tendency to lose two electrons and Halogens have a tendency to gain electrons.

    Option A is correct.
  • Question 2
    1 / -0
    Hydrogen is:
    Solution
    Hint: Hydrogen has an electronegativity of $$2.1$$

    Correct Answer: Option C

    Explanation: 
    Hydrogen is one of the most unique elements of the periodic table. It can act as both an electropositive element as well as an electronegative element depending on which element it is reacting with.

    When it is reacting with alkali metals and alkaline earth metals, it will be acting as an electronegative element and forms ionic hydrides:

    $$2Na + H_2 \rightarrow 2Na^+H^-$$

    When it is reacting with halogens, it will be acting as an electropositive element. Hence a partial positive charge gets developed on it.

    $$H_2 + F_2 \rightarrow 2 \ H^{\delta+}-F^{\delta -}$$
  • Question 3
    1 / -0
    The conversion of atomic hydrogen into molecular hydrogen is______.
    Solution
    Reaction:

    $$2H\rightarrow H_2+heat$$ 

    So, its a exothermic change.

  • Question 4
    1 / -0
    Which property of water is used while washing clothes?
    Solution
    Water is preferred substance for cleaning due to its odourless and colourless properties. But what makes it actually good at cleaning is the fact that it's an excellent solvent meaning that hydrophilic dirt will dissolve in water.
  • Question 5
    1 / -0
    Amongst the following the total number of elements, which produce $$H_2$$ gas with sodium hydroxide is:
    Zn, Al, Sn, Pb, P, Si
    Solution
    $$Zn+2NAOH\rightarrow Na_2ZnO_2+H_2\uparrow$$
    $$2Al+2NaOH+H_2O\rightarrow 2NaAlO_2+3H_2\uparrow$$
    $$Sn+2NaOH+H_2O\rightarrow Na_2SnO_3+2H_2\uparrow$$
    $$Pb+2NaOH+H_2O\rightarrow Na_2PbO_3+2H_2\uparrow$$
    $$Si+2NaOH+H_2O\rightarrow Na_2SiO_3+2H_2\uparrow$$
    $$P_4+3NaOH+3H_2O\rightarrow PH_2\uparrow +3NaH_2PO_2$$
  • Question 6
    1 / -0
    Hydrogen does not combine with:
    Solution
    Hint: $$He$$ is a noble element. It does not combine with any other element.

    Correct Answer: Option D

    Explanation:
    Hydrogen is an element that combines with almost every other element in the periodic table, except noble gases. 
    It reacts with $$s-block$$ elements to form ionic hydrides, with $$p-block$$ elements to form covalent hydrides, and with $$d-block$$ elements to form interstitial hydrides. 
    However it doesn't combine with noble gases,

    A) $$Sb + H_2 \xrightarrow{\Delta} SbH_3$$
    B) $$Na + H_2 \xrightarrow{\Delta} NaH$$
    C) $$Bi + H_2 \xrightarrow{\Delta} BiH_3$$
    D) $$He + H_2 \rightarrow no\  reaction$$
  • Question 7
    1 / -0
    In the Merck's process, the reagents involved for the preparation of hydrogen peroxide are:
    Solution
    In laboratory, $$H_2O_2$$ is prepared by Merck's process. It is prepared by adding calculated amounts of sodium peroxide to ice cold dilute $$(20\%)$$ solution of $$H_2SO_4$$.

    $$Na_2O_2+ H_2SO_4 \rightarrow Na_2SO_4+ H_2O_2$$.
  • Question 8
    1 / -0
    Under what conditions of temperature and pressure, the formation of molecular hydrogen from atomic hydrogen will be favoured most?
    Solution
    Reaction:
    $$2H\rightarrow H_2+\text{heat}$$
    So, according to Le Chatelier's principle, low temperature and high pressure is favored.
  • Question 9
    1 / -0
    $$H_{2}O_{2}$$ can be obtained when following reacts with $$H_{2}SO_{4}$$ except with_______.
    Solution
    The laboratory preparation of hydrogen peroxide was based on the technique that Thenard used. In this technique, barium nitrate, purified by recrystallization, was decomposed by heating in air in a porcelain retort. The resulting oxide was further oxidized by heating in a stream of oxygen to a dull red heat. The barium peroxide which formed was then dampened, ground, and dissolved in hydrochloric acid. A slight excess of sulfuric acid was then added to precipitate barium sulfate and regenerate hydrochloric acid. The procedure of barium peroxide solution and sulfate precipitation was repeated several times in the same solution to increase the peroxide concentration. But $$PbO_{2}$$ can not be used in this process instead of $$BaO_{2}$$.
    Reaction:
    $$BaO_2.8H_2O + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 + 8H_2O$$.
  • Question 10
    1 / -0
    In laboratory, $$H_{2}O_{2}$$ is prepared by the action of:
    Solution
    In laboratory, $$H_{2}O_{2}$$ is preparedby the following reaction:

    $$BaO_2.8H_2O + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 + 8H_2O$$
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