The element Uuq (Ununquadium,atomic number Z | vedclass

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(D) The atomic number $(Z)$ of $Uuq$ is $114$,so the number of protons $(p)$ is $114$.Let the number of neutrons be $n$.According to the problem,the n

Vedclass · Accepted Answer

$243$

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(D) The atomic number $(Z)$ of $Uuq$ is $114$,so the number of protons $(p)$ is $114$.Let the number of neutrons be $n$.According to the problem,the number of neutrons is $12.8 \%$ more than the number of protons:$n = p + 0.128 	imes p = 1.128 	imes p$$n = 1.128 	imes 114 = 128.592 \approx 129$.The mass number $(A)$ is the sum of protons and neutrons:$A = p + n = 114 + 129 = 243$.Therefore,the approximate mass number is $243$.

$(A)$. $^{24}_{12}Mg$	$(I)$. $5$
$(B)$. $^{27}_{13}Al$	$(II)$. $12$
$(C)$. $^{9}_{4}Be$	$(III)$. $6$
$(D)$. $^{12}_{6}C$	$(IV)$. $14$

The element $Uuq$ (Ununquadium,atomic number $Z = 114$) contains $12.8 \%$ more neutrons compared to protons. What is the approximate mass number of it?

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