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Chemistry Test 110
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Chemistry Test 110
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  • Question 1/10
    4 / -1

    Which of the following causes laxative effect when present above  500 ppm in water :

  • Question 2/10
    4 / -1

    Which is not a pair of reducing sugars-

  • Question 3/10
    4 / -1

    Given the value of their van der Waals’ constant ‘a’ arrange the following gases in the order of their expected liquification pressure  at a temperature T. T is below the critical point  of all the

    gases.

    Gas                     CH4       Kr       N2         Cl2

    ‘a’(atmL2.mol–2) 2.283   2.439   1.408   6.579

  • Question 4/10
    4 / -1

    Which of the following d-block element has highest stable oxidation state.

  • Question 5/10
    4 / -1

    Find D

    Solutions

     

     

  • Question 6/10
    4 / -1

    Calculate ΔSuniverse for following chemical reaction 

    C (graphite) +  2H2(g) → CH4(g);

    ΔfH° = –74.81 kJ at 298 K.

    The standard entropies of C (graphite), H2(g) and CH4(g) are 5.740, 130.684 and 186.264 J/K-mol, respectively. 

    Solutions

    ΔSReaction = [186.264] – [5.740 + 2 ×130.684)] or ΔSSystem = – 80.844 J/K

    ΔSSurrounding =  + ΔH/T = 74.81 × 103 / 298 = 251 j/k

    ΔSUniverse = – 80.844 + 251 = 170.1 J/K

     

  • Question 7/10
    4 / -1

    Some white colourless crystals are heated. A cracking sound is heard and brown fumes are given off and the residue left is yellow-brown in colour. When a glowing splinter is held in the fumes, it is relighted. The fumes consist of : 

    Solutions

    Pb(NO3)2   PbO + 2 NO2 + 1/2 O2

  • Question 8/10
    4 / -1

    Solutions

    Vicinal diol compounds breakdown 

    by HIO4.

     

  • Question 9/10
    4 / -1

    Which of the following structure is correctly drawn according to fundamental idea of VSEPR theory-

    Solutions

    (A) θ incorrect less than 109º28'

    (B) incorrect position of lone pair

    (C) correct square planar structure with θ = 90º        

    (D) θ incorrect because it is square planar with two lone pair at Xe and θ = 90º

     

  • Question 10/10
    4 / -1

    The end products of following reaction would be,  



     

    Solutions

    3 moles of HCHO give cross aldol condensation with aromatic ketone one mole of HCHO under goes cross Cannizaro reaction with the aldol product.

     

     

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