Identify the missing nuclide in the following nuclear equation:

Chemistry

1 answer:

ra1l [238]3 years ago

7 0

Answer:

B. Bi-214.

Explanation:

The equation shows beta particle emission of 214/82 Pb which result into 214/83 Bi, in which the mass remain same but the the atomic number increases by one.

During this emission neutron get split into an electron and a proton which are represented as 0/e/-1.

So, the final nuclear equation becomes : 214/82 Pb => 0 e -1 + 214/83 Bi

Hence, the correct answer is "B. Bi-214."

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Draw the alkane formed when 4,5,5-trimethyl-1-hexyne is treated with two equivalents of hbr.

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<h3>Answer:</h3>

The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).

<h3>Explanation:</h3>

Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.

The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.