Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left\{{}\begin{matrix}P=E\\\dfrac{N}{N+P}=\dfrac{11}{20}\\P+E+N=58\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2P+N=58\\20N-11N-11P=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2P+N=58\\9N-11P=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}P=E=Z=18\\N=22\end{matrix}\right.\\ \Rightarrow A=Z+N=18+22=40\left(đ.v.C\right)\\ \Rightarrow KH:^{40}_{18}Ar\)
Ta có: \(\left\{{}\begin{matrix}p+e+n=34\\p=e\\\dfrac{n}{n+p}=\dfrac{12}{23}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=11\\n=12\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}p+e+n=155\\p=e\\p+e-n=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=47\\n=61\end{matrix}\right.\)
\(\Rightarrow A=p+n=47+61=108\left(u\right)\)
\(KHNT:^{108}_{47}Ag\)
b)
Ta có: \(\left\{{}\begin{matrix}p+e+n=95\\p=e\\\dfrac{p+n}{e}=\dfrac{13}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=30\\n=35\end{matrix}\right.\)
\(\Rightarrow A=p+n=30+35=65\left(u\right)\)
\(KHNT:^{65}_{30}Zn\)
c)
Ta có: \(\left\{{}\begin{matrix}p+n=80\\p=e\\n-p=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=35\\n=45\end{matrix}\right.\)
\(\Rightarrow A=p+n=35+45=80\left(u\right)\)
\(KHNT:^{80}_{35}Br\)
d)
Ta có: \(\left\{{}\begin{matrix}p+e+n=52\\p=e\\n-e=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=17\\n=18\end{matrix}\right.\)
\(\Rightarrow A=p+n=17+18=35\left(u\right)\)
\(KHNT:^{35}_{17}Cl\)
a) Ta có: p=e=15
KHNT: \(^{31}_{15}P\)
b) Ta có: \(\left\{{}\begin{matrix}p+n=35\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=18\\p=17\end{matrix}\right.\)
KHNT: \(^{35}_{17}Cl\)
c) Ta có: \(\left\{{}\begin{matrix}e=15\\p=e\\p+n=31\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=15\\n=16\end{matrix}\right.\)
KHNT: \(^{31}_{15}P\)
d) Ta có: \(\left\{{}\begin{matrix}p=19\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=19\\n=20\end{matrix}\right.\)
KHNT: \(^{39}_{19}K\)
e) Ta có: \(\left\{{}\begin{matrix}p+e+n=58\\p=e\\n-e=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=18\\n=22\end{matrix}\right.\)
KHNT: \(^{40}_{18}Ca\)
f) Ta có: \(\left\{{}\begin{matrix}p+e+n=115\\p=e\\p+e-n=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=45\\p=e=35\end{matrix}\right.\)
KHNT: \(^{80}_{35}Br\)
g) Ta có: \(\left\{{}\begin{matrix}p+e+n=46\\p=e\\n=\dfrac{8}{15}\left(p+e\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=15\\n=16\end{matrix}\right.\)
KHNT: \(^{31}_{15}P\)
h) Ta có: \(\left\{{}\begin{matrix}p+e+n=180\\p=e\\\dfrac{n}{p+e}=\dfrac{37}{53}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=74\\p=e=53\end{matrix}\right.\)
KHNT: \(^{127}_{53}I\)
2.
a,Ta có: \(\left\{{}\begin{matrix}p+e+n=28\\p=e\\n=p+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=9\\n=10\end{matrix}\right.\)
b, \(A=p+n=9+10=19\left(đvC\right)\)
c, Đây là flo (F)
Đáp án A
Theo giả thiết ta có:
2 Z X + N X = 23 8 Z Y ( 1 ) 2 Z Y + N Y = 16 5 Z X ( 2 ) N X + N Y = 2 Z Y ( 3 )
⇒ - 6 5 Z X + - 7 8 Z Y + N X + N Y = 0 ( 1 ) + ( 2 ) N X + N Y = 2 Z Y ( 3 )
⇔ 9 8 Z X = 6 5 Z X ⇔ Z X Z Y = 15 16
Ta có: \(\left\{{}\begin{matrix}p+e+n=34\\p=e\\\dfrac{n}{p+n}=\dfrac{12}{23}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=11\\n=12\end{matrix}\right.\)
\(\Rightarrow Z=p=e=11\)
\(KHNT:^{23}_{11}Na\)
Sao từ cái đó mà tương đương dc ra như vậy thế bạn?