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noname [10]
3 years ago
6

Which isotope will spontaneously decay and emit particles with a charge of +2?

Chemistry
2 answers:
VikaD [51]3 years ago
7 0

Answer: The correct option is 4.

Explanation: All the options will undergo some type of radioactive decay processes. There are 3 decay processes:

1) Alpha decay: It is a decay process in which alpha particle is released which has has a mass number of 4 and a charge of +2.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

2) Beta-minus decay: It is a decay in which a beta particle is released. The  beta particle released has a mass number of 0 and a charge of (-1).

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

3) Beta-plus decay: It is a decay process in which a positron is released. The positron released has a mass number of 0 and has a charge of +1.

_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0\beta

For the given options:

Option 1: This nuclei will undergo beta-plus decay process to form _{25}^{53}\textrm{Mn}

_{26}^{53}\textrm{Fe}\rightarrow _{25}^{53}\textrm{Mn}+_{+1}^0\beta

Option 2: This nuclei will undergo beta-minus decay process to form _{80}^{198}\textrm{Hg}

_{79}^{198}\textrm{Au}\rightarrow _{80}^{198}\textrm{Hg}+_{-1}^0\beta

Option 3: This nuclei will undergo a beta minus decay process to form _{56}^{137}\textrm{Ba}

_{55}^{137}\textrm{Cs}\rightarrow _{56}^{137}\textrm{Ba}+_{-1}^0\beta

Option 4: This nuclei will undergo an alpha decay process to form _{85}^{216}\textrm{At}

_{87}^{220}\textrm{Fr}\rightarrow _{85}^{216}\textrm{At}+_2^4\alpha

Hence, the correct option is 4.

Citrus2011 [14]3 years ago
7 0
It is 220Fr that would spontaneously decay and emit particles with a charge of +2.
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With a percent yield of 87.5%.

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Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:

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3.86 g of KCl would react, so the amount remaining would be:

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