a, Gia tốc của xe:

Ta có: \(v=v_0+at\Leftrightarrow a=\dfrac{v-v_0}{t}=\dfrac{5-0}{5}=1\left(m/s^2\right)\)

b, Thời gian để xe đạt vận tốc 20m/s :

Ta có: \(v=v_0+at\Leftrightarrow t=\dfrac{v-v_0}{a}=\dfrac{20-0}{1}=20\left(s\right)\)

c, Quãng đường đi đc sau 10s:

  \(s=v_0t+\dfrac{1}{2}at^2=0.10+\dfrac{1}{2}.1.10^2=50\left(m\right)\)

d, Vận tốc của vật khi đó:

Ta có: \(v^2-v_0^2=2as\Leftrightarrow v=\sqrt{2as+v^2_0}=\sqrt{2.1.200+0}=20\left(m/s\right)\)

\(m_{CuSO_4}=560.20\%=112\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{112}{160}=0,7\left(mol\right)\)

PTHH: Fe + CuSO₄ → FeSO₄ + Cu

Mol:     0,7       0,7           0,7        0,7

\(m_{Fe}=0,7.56=39,2\left(g\right)\)

m_{dd sau pứ} = 39,2 + 560 - 0,7.64 = 554,4 (g)

\(C\%_{ddFeSO_4}=\dfrac{0,7.152.100\%}{554,4}=19,19\%\)

6.

Ta có: \(\left\{{}\begin{matrix}p+e+n=40\\p=e\\p+e-n=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=13\\n=14\end{matrix}\right.\)

   \(\Rightarrow A=z+n=13+14=27\left(u\right)\)

\(KHNT:^{27}_{13}Al\)

7.

Ta có: \(\left\{{}\begin{matrix}p+e+n=82\\p=e\\p+e-n=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=26\\n=30\end{matrix}\right.\)

   \(\Rightarrow A=z+n=26+30=56\left(u\right)\)

\(KHNT:^{56}_{26}Fe\)

8.

Ta có: \(\left\{{}\begin{matrix}p+e+n=40\\p=e\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=13\\n=14\end{matrix}\right.\)

   \(\Rightarrow A=z+n=13+14=27\left(u\right)\)

\(KHNT:^{27}_{13}Al\)

9.

Ta có: \(\left\{{}\begin{matrix}p+e+n=108\\p=e\\n-p=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=33\\n=42\end{matrix}\right.\)

   \(\Rightarrow A=z+n=33+42=75\left(u\right)\)

\(KHNT:^{75}_{33}As\)

10.

Ta có: \(\left\{{}\begin{matrix}p+e+n=34\\p=e\\n-e=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=11\\n=12\end{matrix}\right.\)

   \(\Rightarrow A=z+n=11+12=23\left(u\right)\)

\(KHNT:^{23}_{11}Na\)

anh làm chi tiết câu 2 thôi nhé, tại vì dài quá

2.

Ta có: \(\left\{{}\begin{matrix}p+e+n=92\\p=e\\p+e-n=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2n=68\\p=e\\p+e+n=92\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=34\\p=e=z=29\end{matrix}\right.\)

   \(\Rightarrow A=z+n=29+34=63\left(u\right)\)

\(KHNT:^{63}_{29}Cu\)

3.

Ta có: \(\left\{{}\begin{matrix}p+e+n=155\\p=e\\p+e-n=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=47\\n=61\end{matrix}\right.\)

   \(\Rightarrow A=z+n=47+61=108\left(u\right)\)

\(KHNT:^{108}_{47}Ag\)

4.

Ta có: \(\left\{{}\begin{matrix}p+e+n=58\\p=e\\p+e-n=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=19\\n=20\end{matrix}\right.\)

   \(\Rightarrow A=z+n=19+20=39\left(u\right)\)

\(KHNT:^{39}_{19}K\)

5.

Ta có: \(\left\{{}\begin{matrix}p+e+n=115\\p=e\\p+e-n=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=z=35\\n=45\end{matrix}\right.\)

   \(\Rightarrow A=z+n=35+45=80\left(u\right)\)

\(KHNT:^{80}_{35}Br\)

Câu 3:

\(m_{Be}=9,012u=9,012.1,66.10^{-24}=1,495992.10^{-23}\left(g\right)\)

\(m_{Be}=15,999u=15,999.1,66.10^{-24}=2,655834.10^{-23}\left(g\right)\)

Câu 2:

a, tự làm nha

b, Ta có: \(1u=\dfrac{1}{12}.m_C=\dfrac{1}{12}.\dfrac{12}{6,022.10^{23}}\approx1,66.10^{-24}\left(g\right)\)

c, Vì \(1u=\dfrac{1}{12}.m_C\Rightarrow m_C=12u\)

d, Ta có: \(m_C=11,9059.m_H\Rightarrow m_H=\dfrac{12}{11,9059}=1,0079\left(u\right)\)

Câu 1:

Tỉ số của e so với p là:

\(\dfrac{m_e}{m_p}=\dfrac{9,1094.10^{-31}}{1,6726.10^{-27}}=5,4463.10^{-4}\)

Tỉ số của e so với n là:

\(\dfrac{m_e}{m_n}=\dfrac{9,1094.10^{-31}}{1,6748.10^{-27}}=5,4391.10^{-4}\)

Ta có: \(\overline{M}=\dfrac{24.80+x.\left(100-80\right)}{100}\)

     \(\Leftrightarrow x=\dfrac{100.\overline{M}-24.80}{20}=\dfrac{100.24,2-24.80}{20}=25\)

 ⇒ Đồng vị thứ 2 là ²⁵Mg

Ta có: \(\overline{M}=\dfrac{6.7,5+x.\left(100-7,5\right)}{100}\)

     \(\Leftrightarrow x=\dfrac{100.\overline{M}-6.7,5}{92,5}=\dfrac{100.6,925-6.7,5}{92,5}=7\)

 ⇒ Đồng vị thứ 2 là ⁷Li