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    Chemistry-
    General
    Easy

    Question

    In the given reaction

    1. CH subscript 3 minus CH subscript 2 minus OH and 
    2. CH subscript 3 minus CH subscript 2 OH and     
    3. CH subscript 2 equals CH subscript 2 and     
    4.    

    The correct answer is: CH subscript 3 minus CH subscript 2 OH and 

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    ; A&B are given by the set -

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    In the reaction,  the product obtained will be -

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    blank subscript 83 end subscript superscript 214 end superscript B i stack ⟶ with alpha text -emission  end text on top A stack ⟶ with beta text -emission  end text on top B stack ⟶ with beta minus text  cmission  end text on top C stack ⟶ with beta text -emission  end text on top D stack ⟶ with alpha text -emission  end text on top E E' is.an element of stable nucleus What is the element 'E'?

    blank subscript 83 end subscript superscript 214 end superscript B i stack ⟶ with alpha text -emission  end text on top A stack ⟶ with beta text -emission  end text on top B stack ⟶ with beta minus text  cmission  end text on top C stack ⟶ with beta text -emission  end text on top D stack ⟶ with alpha text -emission  end text on top E E' is.an element of stable nucleus What is the element 'E'?

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    Consider the following nuclear reactions: table row cell blank subscript 92 end subscript superscript 238 end superscript M ⟶ blank subscript y end subscript superscript x end superscript N plus 2 blank subscript 2 end subscript superscript 4 end superscript H e end cell row cell blank subscript y end subscript superscript x end superscript N ⟶ blank subscript B end subscript superscript A end superscript L plus 2 beta to the power of plus end exponent end cell end table the number of neutrons present in the element 'L' is:

    Consider the following nuclear reactions: table row cell blank subscript 92 end subscript superscript 238 end superscript M ⟶ blank subscript y end subscript superscript x end superscript N plus 2 blank subscript 2 end subscript superscript 4 end superscript H e end cell row cell blank subscript y end subscript superscript x end superscript N ⟶ blank subscript B end subscript superscript A end superscript L plus 2 beta to the power of plus end exponent end cell end table the number of neutrons present in the element 'L' is:

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    A radioactive element decays by parallel path as given below :
    table attributes columnalign right end attributes row cell A stack ⟶ with lambda subscript 1 end subscript on top B lambda subscript 1 end subscript end cell cell blank equals 1.8 cross times 10 to the power of negative 2 end exponent s e c to the power of negative 1 end exponent end cell row cell 2 A stack ⟶ with lambda subscript 2 end subscript on top B lambda subscript 2 end subscript end cell cell blank equals 10 to the power of negative 3 end exponent s e c to the power of negative 1 end exponent end cell end table
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    (A) blank subscript 92 end subscript superscript 238 end superscript U left parenthesis I I I B right parenthesis stack ⟶ with negative alpha on top A stack ⟶ with negative alpha on top B stack ⟶ with negative beta on top C
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