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Chemistry Test-40

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Chemistry Test-40
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  • Question 1
    4 / -1

    Maximum number of electrons in an orbit of an atom is determined by:

    (n = Orbit Number)

    Solution

    The maximum number of electrons in an orbit of an atom is determined by 2n2 (n = Orbit Number).

    According to Bohr's model of an atom, the maximum number of electrons in an orbit/shell is given by 2n2.

    Here 'n' is orbit number/ shell number/ energy level; which is 1 for K shell,  2 for L shell and so on.

  • Question 2
    4 / -1

    The half-cell potentials for the metallic elements \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) are \(0.8 \mathrm{~V},-0.74 \mathrm{~V}, 1.1 \mathrm{~V}\) and \(+0.34 \mathrm{~V}\), respectively. Arrange these in the order of decreasing metallic character:

    Solution

    Given,

    The half-cell potentials for the metallic elements \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) are \(0.8 \mathrm{~V},-0.74 \mathrm{~V}, 1.1 \mathrm{~V}\) and \(+0.34 \mathrm{~V}\), respectively.

    Lower (less positive or more negative) the electrode potential, greater the metallic character. Therefore, the metallic character of these elements follows the order:

    \(\mathrm{B}>\mathrm{D}>\mathrm{A}>\mathrm{C}\)

  • Question 3
    4 / -1

    Faraday constant:

    Solution

    The charge of \(1\) mole of electrons is called the Faraday constant, \(F\).It is calculated by multiplying the charge of one electron by the Avogadro constant. i.e,

    \(F = eN A\)

    \(=\left(1.6 \times 10^{-19}\right)\left(6.023 \times 10^{23}\right)\)

    \(=9.648 \times\) \(10^4\) C.mol \(^{-1}\)

    It is a universal constant.

  • Question 4
    4 / -1

    A radioactive sample contains \(10^{-3} kg\) each of two nuclear species \(A\) and \(B\) with half life \(4\) days and \(8\) days, respectively. The ratio of the amounts of \(A\) and \(B\) after period of \(16\) days is:

    Solution

    Given,

    Half-life of species\(A =4\) days

    Half-life of species \(B =8\) days

    For species \(A , N = N _{0}\left(\frac{1}{2}\right)^{ n } ; n =\frac{16}{4}=4\)

    Now, \(N _{1}= N _{0}\left(\frac{1}{2}\right)^{4}\)

    Similarly, for species \(B\)

    \(N _{2}= N _{0}\left(\frac{1}{2}\right)^{2} \text {....(ii) }\)

    On dividing Eq. (i) from Eq. (ii), we get

    \(\frac{ N _{1}}{ N _{2}}=\frac{(2)^{2}}{(2)^{4}} \Rightarrow \frac{ N _{1}}{ N _{2}}=\frac{1}{4}\)

    \(\Rightarrow N _{1}: N _{2}=1: 4\)

  • Question 5
    4 / -1

    A secondary amine is:

    Solution

    An amine in which the amino group is directly bonded to two carbons of any hybridization and this is happen when a organic compound with \(2\) amine group.

  • Question 6
    4 / -1

    Rate of production of \(\text{NO}_2\) is ___________ whennitric oxide reacts with ozone in the stratosphere.

    Solution
    Rate of production of \(\text{NO}_2\) is faster when nitric oxide reacts with ozone in the stratosphere.
    NO (g) + \(\text{O}_3\)(g) → \(\text{NO}_2\)(g) + \(\text{O}_2\)(g)
    The irritant red haze in the traffic and congested places is due to oxides of nitrogen. Higher concentrations of NO2 damage the leaves of plants and retard the rate of photosynthesis. Nitrogen dioxide is a lung irritant that can lead to an acute respiratory disease in children. It is toxic to living tissues also. Nitrogen dioxide is also harmful to various textile fibres and metals.
  • Question 7
    4 / -1

    Which of the following antibiotics is used to cure typhoid?

    Solution

    (A). Novalgin: Also called Metamizole, is an analgesic and antipyretic.

    (B). Quinine: Used to treat Malaria.

    (C). Chloromycetin: Also called Chloramphenicol, is an antibiotic used to treat typhoid.

    (D). Paracetamol: Also called Acetaminophen, is an analgesic and antipyretic.

  • Question 8
    4 / -1

    A liquid that scatters a beam of light, but leaves no residue when passed through filter paper is a/an:

    Solution

    A liquid that scatters a beam of light, but leaves no residue when passed through filter paper is a/ancolloidal solution.

  • Question 9
    4 / -1

    The following reaction is an example of a:

    \(4 \mathrm{NH_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)}\)

    Solution

    \(4 \mathrm{NH_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)}\)

    The above reaction is a displacement reaction and redox reaction.

    Redox is a chemical reaction that involves changing the oxidation states of atoms. Redox reactions are defined as the actual or formal movement of electrons between chemical entities, with one species often experiencing oxidation while the other experiences reduction.

    Ammonia's hydrogen is displaced by oxygen. With the addition of oxygen, the nitrogen in ammonia is oxidised. The above reaction is known as a redox reaction because it involves both oxidation and reduction at the same time. It is also a displacement reaction because the \(\mathrm{H}\) of \(\mathrm{NH}_{3}\) is displaced by oxygen.

  • Question 10
    4 / -1

    A polyamide synthetic polymer prepared by prolonged heating of caprolactum, is:

    Solution

    Nylon 6 is obtained by heating of caprolactam with water at higher temperature.

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