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Hydrogen Test - 68

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Hydrogen Test - 68
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  • Question 1
    1 / -0
    A good quality of water will have 
    A) high DO
    B) high BOD
    C) high COD
  • Question 2
    1 / -0
    In water-gas shift reaction, hydrogen gas is produced from the reaction of steam with: 
    Solution

    Water-gas shift reaction explains the reaction of carbon monoxide and water vapour to form carbon dioxide and hydrogen:

    $$CO+H_2O\rightarrow  CO_2 + H_2$$

    Hence option C is correct.

  • Question 3
    1 / -0
     High purity $$(>99.95\%)$$ dihydrogen is obtained by: 
    Solution

  • Question 4
    1 / -0
    A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of $$10^{-6}$$ M hydrogen ions. The EMF of the cell is 0.118 V at $$25^{\circ}C$$. The concentration of hydrogen ions at the positive electrode is
    Solution
    For a Concentration cell Nernst Equation is 

    $$Ecell= \frac{0.059}{n}log\frac{C_2}{C_1}$$

    Given $$Ecell= 0.118V$$

    Concentration of hydrogen ion at the positive electrode$$\left [ C_2 \right ]=Unknown$$
    Concentration of hydrogen ion at the negetive elctrode $$\left [ C_1 \right ]=10^{-6}M$$
     
    $$Ecell= \frac{0.059}{n}log\frac{C_2}{C_1}$$
    25∘C
    $$0.118= \frac{0.059}{1}log\frac{C_2}{C_1}$$

    $$log\frac{C_2}{C_1}=\frac{0.118}{0.059}$$

    $$\frac{C_2}{C_1}= 10^{2}= 100$$

    $$C_2=100\times 10^{-6}M=10^{-4}M$$

    Hence the correct answer is [C]

  • Question 5
    1 / -0
    A solution of $$H_{2} O_{2}$$ is titrated with a solution of $$KMnO_{4}.$$ The reaction is
    $$ 2MnO_{4}-+5H_{2}O_{2}+6H^{+}$$
    $$ \rightarrow 2Mn^{2+}+5O_{2}+8H_{2}O$$
    It requires $$50$$ ml of $$0.1$$ $$ M - KMnO_{4} $$ to oxidize $$10$$ ml of $$ H_{2}O_{2}.$$ The strength of $$ H_{2}O_{2} $$ solution is

    Solution
    we have balanced equation
    $$ 2MnO_{4}^{\circleddash} + 5H_{2}O_{2}+ 6H^{\oplus } \rightarrow 2Mn^{2+}+5O_{2}+8H_{2}O$$

    moles of oxidizing agent = moles of reducing  $$ \times $$ stoichiometric Ratio
    $$ 0.05 \times 0.1 = 0.01 \times M \times \left ( \dfrac{2}{5} \right )$$

    $$ M = 1.25 \, molar $$

    (ml.wt. of $$ H_{2}O_{2} = 34 $$ )

    wt. of $$ H_{2}O_{2} = 1.25 \times 34 = 42.5 gm$$

    strength (gm/m) $$= \dfrac{42.5 gm}{1000 ml} \times 100 = 4.25 \%$$

    $$ = 4.25\% $$

    Ans- option (A)
  • Question 6
    1 / -0
    Hydrogen can be prepared by mixing steam and water gas at $$673 K$$ in the presence of $$Fe_2O_3$$ and $$Cr_2O_3$$. This process is called: 
    Solution
    $$CO_{(g)}+H_2O_{(g)}\xrightarrow[Fe_2O_3/Cr_2O_3]{673K}CO_{2_{(g)}}+H_{2_{(g)}}$$ 
                                (Catalyst)

    This is called water-gas shift reaction or Bosch's process.
  • Question 7
    1 / -0
     $$H_2$$ gas is usually prepared by the reaction of 
    Solution

  • Question 8
    1 / -0
    $$H_2O_2$$ is manufactured these days...
    Solution
    Electrolysis of $$50\%$$ sulphuric acid gives per disulphuric acid, represented as $$H_2S_2O_8$$, which on distillation with water yields $$30\%$$ solution of hydrogen peroxide as follows :

    $$H_2S_2O_8 + 2H_2O\rightarrow 2H_2SO_4 + H_2O_2$$

    Hence, option (C) is correct.
  • Question 9
    1 / -0
    Which of the following statement is correct for ionic hydrides?
    Solution

  • Question 10
    1 / -0
    Which of the following properties of water leads in comparison to $$H_2S$$ and $$H_2Se$$ due to H-bonding? 
    Solution

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