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Chemistry Test - 20

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Chemistry Test - 20
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  • Question 1
    1 / -0

    The elements A, B and C belong to group 2, 14 and 16 respectively, of the periodic table. Which of the two elements will form covalent bonds?

    Solution

    The covalent bond is formed by the sharing of electrons between two atoms. As the element B (which belongs to group 14) has 4 valence electrons which it can share with two elements of C type (from group 16) electrons to complete the octet of each included atom.

  • Question 2
    1 / -0

    Arrange the alkyl in decreasing order of their mass.

    Solution

    For \(\mathrm{{CH}_{3} Br}\):

    Atomic mass:

    \(=(12\times 1)+(1\times3)+(80\times1)\)

    \(=95\)

    For \(\mathrm{CH}_{3} \mathrm{Cl}\):

    Atomic mass:

    \(=(12\times1)+(1\times3)+(35.5\times1)\)

    \(=50.5\)

    For \(\mathrm{H}_{2} \mathrm{O}\):

    Atomic mass:

    \(=(16\times1)+(2\times1)\)

    \(=18\)

    According to above data regarding atomic masses of compounds, the decreasing order will be:

    \(\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{Cl}>\) \(\mathrm{H}_{2} \mathrm{O}\)

  • Question 3
    1 / -0

    A commercially sold conc. \(\mathrm{HCl}\) is \(35 \% \mathrm{HCl}\) by mass. If the density of this commercial acid is \(1.46 \mathrm{~g} / \mathrm{mL}\), the molarity of this solution is:

    (Atomic mass: \(\mathrm{Cl}=35.5 \mathrm{amu} . \mathrm{H}=1 \mathrm{amu}\) )

    Solution

    \(35  \% \mathrm{HCl}\) by mass means in \(100 \mathrm{gm} \mathrm{HCl}\) solution \(35 \mathrm{gm} \mathrm{HCl}\) present.

    Now, volume of \(100 \mathrm{gm} \mathrm{HCl}\) solution

    \(=\frac{100}{1.46} \mathrm{ml} \)

    \(=\frac{\frac{100}{1.46}}{1000} \mathrm{I} \)

    \(\text { Moles of } \mathrm{HCl}=\frac{35}{36.5} \quad \text { moles of solute } \)

    \( \text { Now, molarity }=\frac{\frac{35}{36.5}}{\text { volume of solution ( in } \mathrm{L})} \)

    \(=\frac{\frac{35}{36.5}}{\frac{100}{1.46}}=14\)

  • Question 4
    1 / -0

    Which of the following has a regular repeated molecular pattern in three dimensional space?

    Solution

    Solids have a regular repeated molecular pattern in three dimensional space and this is due to their property of high inter-molecular forces. But in liquids and gases, the arrangement of molecular pattern is irregular.

  • Question 5
    1 / -0

    ___________ is a quantity that represents the total energy of the system.

    Solution

    Internal energyis a quantity that represents the total energy of the system.

  • Question 6
    1 / -0

    The cathode ray experiment was done for the first time by:

    Solution

    In 1897, great physician J.J. Thompson, conducted his first cathode ray tube experiment.

    Thompson, conducted this experiment to prove that rays emitted from an electron gun are inseparable from the latent charge. He built his cathode ray tube with a metal cylinder on the other end. The metal had two small diversions(slits), leading to electrometer that could measure small electric charge.

  • Question 7
    1 / -0

    Which of the following cannot be a ligand?

    Solution

    \(\mathrm{Ni}^{2+}\)cannot be a ligand.

    The ions/molecules bound to the central atom/ion is called a ligand. \(\mathrm{Ni}^{2+}\) is a metal ion, and according to Werner, the secondary valences can be satisfied only by neutral molecules or negative ions. \(\mathrm{Cl}^{-}, \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{NH}_{3}\) are all possible ligands.

  • Question 8
    1 / -0

    Glycerol is:

    Solution

    Glycerol is Trihydric alcohol.

    Compounds having one −OH group is called monohydric alcohol.

    Compounds having two −OH group is called dihydric alcohol.

    Compounds having three −OH group is called trihydric alcohol.

    Image of glycerol is given below, it has three −OH group

  • Question 9
    1 / -0

    Electronic configuration of a transition element \(X\) in \(+3\) oxidation state is \([A r] 3 d^{5} .\) What is its atomic number?

    Solution

    In oxidation, an element looses its electrons.

    There is \(+3\) oxidation state means no. of loss of electrons is 3, and that 2 electrons are released from \(4\mathrm{s}^2\) orbital and 1 electron from \(3 \mathrm{d}\) orbital to attain half filled stability.

    Therefore, \(\left[\mathrm{Ar}\right]\) has 18 electrons and 5 electrons in \(3 \mathrm{d}\) orbital as per given.

    Therefore, actual configuration will be \([A r]^{18} 3 d^{6} 4 s^{2}\). Therefore, atomic number will be \(18+6+2=26\).

    This is Fe having atomic number 26.

  • Question 10
    1 / -0
    If the solubility of \({Ag}_{2} \mathrm{CrO}_{4}\) is \({s}\) moles/litre, its solubility product will be ___________.
    Solution
    Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as \(K_{s p}\).
    The equation for the ionization of the silver chromate is given as:
    \(A g_{2} C r O_{4} \leftrightharpoons 2 A g^{+}+C r O_{4}^{2-}\)
    We are given:
    Solubility of \(A g_{2} C r O_{4}=s \mathrm{~mol} / \mathrm{L}\)
    By stoichiometry of the reaction:
    \(1\) mole of \(A g_{2} C r O_{4}\) gives 2 moles of \(A g^{+}\) and 1 mole of \(C r O_{4}^{2-}\)
    When the solubility of \(A g_{2} C r O_{4}\) is \(s\) moles/liter, then the solubility of \(A g^{+}\) will be \(2 {~s}\) moles/liter and solubility of \(\mathrm{Cr} \mathrm{O}_{4}^{2-}\) will be \({s}\) moles/liter.
    Expression for the equilibrium constant of \(A g_{2} C r O_{4}\) will be:
    \(\begin{array}{l}
    K_{s p}=\left[A g^{+}\right]^{2}\left[C r O_{4}^{2-}\right] \\
    K_{s p}=[2 s]^{2}[s]=4 s^{3}
    \end{array}\)
    Thus, the solubility product of \(A g_{2} C r O_{4}\) is \(4 s^{3}\).
     
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