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Chemistry Test - 40

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Chemistry Test - 40
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  • Question 1
    1 / -0

    What are Oxo-Acids?

    Solution

    Acid containing Oxygen is Oxo-Acids.

    An oxyacid, oxoacid, or ternary acid is an acid that contains oxygen. Specifically, it is a compound that contains hydrogen, oxygen, and at least one other element, with at least one hydrogen atom bond to oxygen that can dissociate to produce the H+ cation and the anion of the acid. e.g., carbonic acid (H2CO3), sulphuric acid (H2SO4).

  • Question 2
    1 / -0

    Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state? 

    Solution

    In Ag2SO4, silver is present as Ag+, in CuF2, copper is present as Cu2+, in ZnF2, zinc is present as Zn2+, in Cu2Cl2, the copper is present as Cu+1

    As they are metal cations, the electrons are removed from their actual configuration to form ions.

    The respective transition elements exists in:

    Ag+[Ar]4d105s0, (have completely filled d orbital).

    Cu+2[Ar]3d94s0, (have incompletely filled d orbital).

    Zn+2[Ar]3d104s0, (have completely filled d orbital).

    Cu+1[Ar]3d104s9, (have completely filled d orbital). 

    Since, Cu+2 has unpaired electron, it will show electron transitions. Therefore, will be coloured.

  • Question 3
    1 / -0

    The enthalpy change on freezing of 1 mol of water at 5C to ice at 5Cis :

    ( Given Δfis H=6 kJ mol1 at 0C

    Cp(H2O,1)=75.3 J mol1 K1

    Cp(H2O,s)=36.8 J mol1 K1)

    Solution

    In order to calculate the enthalpy change for H2O at 5C to ice at 5C, we need to calculate the enthalpy change of all the transformation involved in the process.

    (a) Energy change of 1 mol,H2O(1), at 5C

    1 mol,H2O(I),0C

    (b) Energy change of 1 mol,H2O(1), at 0C

    1 mol,H2O (s)(ice), 0C

    (c) Energy change of 1 mol, ice (s), at 0C

    1 mol, ice (s)25C

     Total ΔH

    =Cp[H2O(1)]ΔT+ΔH freezing +Cp[H2O(s)]ΔT

    =(75.3 J mol1 K1)(05)K+(6×103 J mol1)

    +(36.8 J mol1 K1)(50)K

    ΔH=6.56 kJ mol1 (cxothermic process) 

  • Question 4
    1 / -0

    How many structural isomers are possible for C3H9 N ?

    Solution

    4isomers are possible for C3H9 N.

    1.CH3CH2CH2NH2
    Propanamine
    2.CH3CHCH3|NH2
    Isopropylamine
    3.CH3CH2NHCH3
    N-Methylethanamine
      CH3  |4.CH3NCH3
    N,N-Dimethylmethanamine
     
  • Question 5
    1 / -0

    Calculate energy of one mole of photons of radiation whose frequency is 5×1014 Hz.

    Solution

    Given,

    Frequency, ν=5×1014 Hz

    As we know, Planck constant is given as,

    h=6.626×1034Js

    Energy of one photon is given by theexpression,

    E=hν

    E=(6.626×1034 J s)×(5×1014 s1)

    =3.313×1019 J

    As we know,

    Avogadro's number =6.022×1023 mol1

    Energy of one mole of photons,

    =(3.313×1019 J)×(6.022×1023 mol1)

    =199.51 kJ mol1

  • Question 6
    1 / -0

    A system is said to be in thermodynamic equilibrium if the system is in:

    Solution

    A system is said to be in thermodynamic equilibrium if the conditions for the following all three equilibrium is satisfied:

    Mechanical equilibrium:

    When there are no unbalanced forces within the system and between the system and the surrounding, the system is said to be under mechanical equilibrium.

    Chemical equilibrium:

    The system is said to be in chemical equilibrium when there are no chemical reactions going on within the system or there is no transfer of matter from one part of the system to another due to diffusion.

    Thermal equilibrium:

    When the temperature of the system is uniform and not changing throughout the system and also in the surroundings, the system is said to be in thermal equilibrium.

  • Question 7
    1 / -0

    Hydrogen gas is not obtained when zinc reacts with:

    Solution

    Hydrogen gas is not obtained when zinc reacts with cold water.

    Zinc is a low reactive metal which does not react with cold water and hence no gas is evolved. However, Zn reacts with dil. HCl, dil. H2SO4 and hot NaOH solution.

  • Question 8
    1 / -0

    In which of the preparation process of dihydrogen, the syn gas is produced?

    Solution

    Reaction (a) gives

    2H2O(l)Traces of acid /baseElectrolysis2H2O(g)+O2(g)

    Reaction (b) gives

    By electrolysis of brine solution

    At anode: 2Cl-aq  Cl2g+2e-

    At Cathode: 2H2Ol+2e-H2Og+2OH-aq

    Overall reaction:

    Reaction (c) gives

    CnH2n+2+ nH2O Ni1270 KnCO +3n+1H2

    The mixture of CO and water is termed as syn gas.

  • Question 9
    1 / -0

    A solid has a ‘BCC’ structure. If the distance of nearest approach between two atoms is 1.73Å, the edge length of the cell is:

    Solution

    If a is the edge length of the cube and r is the radius of each atom.

    Distance of nearest approach between two atoms =2r=1.73Å.

    Therefore,

    r=1.732=0.865Å

    For BCC lattice, the relation of edge length and radius of the atom is,

    3a=4r

    a=4r3

    a=4×0.8653

    a=1.99=2Å =200pm

    The edge length of the BCC lattice is 200pm.

  • Question 10
    1 / -0

    The oxidation number of Cr in CrO5 which has the following structure, is:

     

    Solution

    In the structure of CrO5, it has four O atoms as peroxide which have oxidation number =1 and one O as oxide (double bonded O) atom with oxidation number =2.

    Let the oxidation number of Cr is x.

    Then, x+4×(1)+1×(2)=0 or, x=+6

    The oxidation number of Cr in CrO5 is +6.

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