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Chemistry Test 213
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Chemistry Test 213
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  • Question 1/10
    4 / -1

    XeF6 on partial hydrolysis with water, produces a compound X. The same compound X is formed when XeF6 reacts with silica. The compound X is:

    Solutions

    During the hydrolysis the oxidation of xenon not changes in xenon hexafluoride.

     

  • Question 2/10
    4 / -1

    How many EDTA (ethylenediaminetetraacetate) unit(s) are required to make an octahedral complex with a Ca2+ ion?

    Solutions

    As EDTA is hexadentate ligand. Hence, only one EDT (ethylenediamine tetraacetate) molecule is required to make an octahedral complex with Ca2+.

     

  • Question 3/10
    4 / -1

    Electrometallurgical process is used to extract

    Solutions

    Because Na is very reactive and can not be extracted by means of the reduction by C, CO etc. So, it is extracted by electrolysis of molten NaCl Solution.

     

  • Question 4/10
    4 / -1

    For one mole of a Van der Waals gas, when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the Van der Waals constant a (atm litre2 mol-2) is:

    Solutions

     

  • Question 5/10
    4 / -1

    Cr2+ and Mn3+ both have d4 configuration. Thus

    Solutions

     

  • Question 6/10
    4 / -1

    Three elements X, Y, and Z have atomic numbers 19, 37, and 55 respectively. Then the correct statements(s) is/are

    Solutions

    The elements X (19), Y (37) and Z (55) are the elements of same group (IA). As atomic number, increases, the ionization potential decrease down the group. Since, the position of Y is in between X and Z. Thus, the ionization potential of Y will also be in between X and Z.

     

  • Question 7/10
    4 / -1

    An electron in an atom jumps in such a way that its kinetic energy changes from x to x/9. The change in its potential energy (magnitude) will be-

    Solutions

    Change in kinetic energy is

    For Bohr model of atom

    The kinetic energy in a state is equal to half of the potential energy in magnitude.

    ∴ Potential energy

    = 2 Kinetic energy

     

  • Question 8/10
    4 / -1

    The amount (in grams) of sucrose (mol wt. = 342 g) that should be dissolved in 100 g water in order to produce a solution with a 105.0 oC difference between the freezing point and boiling point is

    (Given that Kf = 1.86 K kg mol−1 and Kb = 0.51 K kg mol−1 for water)

    Solutions

    Boiling point (Tb) = 100 + ΔTb = 100 + kbm

    Freezing point (Tf) = 0 − ΔTf = −kfm

    Tb − Tf = (100 + kbm) − (−kfm)

    105 = 100 + 0.51 m + 1.86 m

     

  • Question 9/10
    4 / -1

    The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k1 and k2 respectively. The energy of activation for the two reactions are 152.30 kJ mol−1 and 157.7 kJ mol−1 as well as frequency factors are 1013 and 1014 respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be the same for the two reactions.

    Solutions

     

  • Question 10/10
    4 / -1

    Choose from the indicated protons, the one that is most acidic.

    Solutions

    Deprotonation of hydrogen labelled 44 produces a conjugate base which has more stable resonating structure and a resonance structure in which negative charge is present on the electronegative oxygen atom. Therefore, hydrogen labelled 44 is most acidic.

     

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