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Manipal Medical Manipal Medical Solved Paper-2000

  • question_answer
    The weight of a residue obtained by heating 2.76 g of silver carbonate is:

    A)  2.76 g            

    B)  2.98 g

    C)  2.16 g            

    D)  2.44 g

    Correct Answer: C

    Solution :

     \[2A{{g}_{2}}C{{O}_{3}}\xrightarrow[{}]{Heat}4Ag+2C{{O}_{2}}+{{O}_{2}}\]
    \[2[(2\times 108)+12+48]\] \[4\times 108\]
    \[2(216+12+48)\] \[4\times 108\]
    \[2\times 276=552\] \[4\times 108=32\]
    \[\because \]552gm silver carbonate gives silver \[=432\text{ }g\] \[\therefore \] 2.76 g silver carbonate gives \[\frac{432\times 2.76}{552}=2.16g\]


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