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Chemistry Test 84

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Chemistry Test 84
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Weekly Quiz Competition
  • Question 1
    4 / -1

    When NaNO3 is heated in a closed vessel, O2 is liberated and NaNO2 is left behind. At equilibrium,

    (i) Addition of NaNO3 favours the forward reaction.

    (ii) Addition of NaNO2 favours the backward reaction.

    (iii) Increasing pressure favours reverse reaction.

    (iv) Increasing temperature favours the forward reaction.

    Which one of the following is the correct option?

    Solution

    The reaction proceeds as:

    NaNO3(s)⇌NaNO2(s)+ 1/2 O2(g),△H>0

    NaNO3 and NaNO2 are in so lid state, changing their a mount has no effect on equilibrium. Increasing temperature will favour forward reaction due to endotherm ic nature of reaction. Also, increasing pressure will favour backward reaction in which some O2(g) will com bine with NaNO2(x) forming NaCO3.

    Hence, the correct option is (D).

     

  • Question 2
    4 / -1

    In a collection of H -atoms, all the electrons jump from n=5 to the ground level finally (directly or indirectly), without emitting any line in the Balmer series. The number of maximum possible different radiations is:

    Solution

     

  • Question 3
    4 / -1

    XeF6 dissolves in anhydrous HF to give a conducting solution which contains:

    Solution

     

  • Question 4
    4 / -1

    Which of the following process does not occur at the interface of phases?

    Solution

    When the reactants and the catalyst are in the same phase (i.e., liquid or gas), the process is said to be homogeneous catalysis. Therefore, the homogenous catalysis process does not occur at the interface of phases as they do not have separation of phases and their distribution is uniform throughout.

    Hence, the correct option is (C).

     

  • Question 5
    4 / -1

    The Poling process is used for:

    Solution

    Poling is a method used to purify metals that have oxidized impurities. It is typically used to purify metals like copper (Cu) or tin (Sn) that are in the impure form of a copper oxide (Cu2O) or tin oxide. This is the old method of extraction of copper metal from its oxide.

    Hence, the correct option is (A).

     

  • Question 6
    4 / -1

    Which one of the following statements regarding photochemical smog is not correct?

    Solution

    Photochemical smog is formed in warm and sunny climates during the daytime by the action of sunlight on primary pollutants. It contains nitrogen oxides, ozone, PAN, etc., which are oxidizing in nature. So, photochemical smog is an oxidizing agent in character. It causes irritation in the eyes and throat.

    Hence, the correct option is (D).

     

  • Question 7
    4 / -1

    Which of the following is/are a phospholipid(s)?

    Solution

    Lecithin and cephalin both are phospholipids. Lecithin is probably the most common phospholipid which contains the amino alcohol, choline. It is found in egg yolks, wheat germ, and soybeans. Lecithin is extracted from soybeans for use as an emulsifying agent in foods. Lecithin is an emulsifier because it has both polar and non-polar properties, which enable it to cause the mixing of other fats and oils with water components. Cephalins are phosphoglycerides that contain ehtanolamine or the amino acid serine attached to the phosphate group through phosphate ester bonds. These are found in most cell membranes, particularly in brain tissues. They also important in the blood clotting process as they are found in blood platelets.

    Hence, the correct option is (C).

     

  • Question 8
    4 / -1

    Which of the following would not form upon electrolysis of aqueous solution of potassium propanoate?

    Solution

    Ethyl ethanoate CH3COOCH2CH3 would not form upon electrolysis of aqueous solution of potassium propanoate. Instead, butane, ethyl propanoate, ethene and ethane will be obtained.

     

  • Question 9
    4 / -1

    A system absorbs 300 cal of heat as a result that, the volume of the system becomes double the initial volume and temperature changes from 273K to 546K. the work done by the system on the surroundings is 200.0 cal. Calculate ΔE:

    Solution

    According to the thermodynamics first law, the energy can neither be created nor destroyed but can be transformed from one form to the other form, change in internal energy of a system is related to its heat, and work is done as follows:

    ΔE = Q + W

    Where,

    ΔE is a change in the internal energy of the system.

    Q is a heat absorbed by the system = + 300 cal

    W is the work done by the system on the surroundings = −200 cal

    Calculating the change in internal energy of the system from done using the equation,

    ΔE = Q + W

    ⇒ ΔE = 300 cal + (−200 cal)

    ⇒ ΔE = 100 cal

    Therefore, the change in internal energy of the system is 100 cal.

    Hence, the correct option is (D).

     

  • Question 10
    4 / -1

    Inert pair effect is maximum in:

    Solution

    Inert pair effect is absent in Si due to the absence of d− orbitals. Ge and Sn exhibit inert pair effect due to the ineffective shielding of d-orbitals. Due to weak shielding of valence shell electrons by d− and f− orbitals, Pb shows maximum inert pair effect.

    Hence, the correct option is (A).

     

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