\(\Delta t=\overline{\Delta t}+\Delta t_{dc}=0,00456+0,0005=0,00506\left(s\right)\)

\(\overline{v}_{AB}=\dfrac{\overline{s}_{AB}}{\overline{t}_{AB}}=\dfrac{0,05}{1,2946}=38,622\left(m/s\right)\)

\(\delta v=\delta s+\delta t\\ \Leftrightarrow\dfrac{\Delta v}{v}=\dfrac{\Delta s}{s}+\dfrac{\Delta t}{t}\\ \Leftrightarrow\dfrac{\Delta v}{38,622}=\dfrac{0,05}{50}+\dfrac{0,00506}{1,2946}\\ \Leftrightarrow\Delta v=0,19\left(m/s\right)\)

Kết quả đo: \(38,622\pm0,19\left(m/s\right)\)

\(\begin{matrix}\uparrow\downarrow&\uparrow\\1s^2&2s^1\end{matrix}\)

#hoidap247 #Anguyen1127

`X: Fe_3O_4`

`A: H_2`

`B: CO`

`D: HCl`

`E: FeCl_2`

`F: Cl_2`

`G: FeCl_3`

`I: S`

`L: H_2S`

`x: O_2`

\(\xrightarrow[Fe_3O_4+CO\left(2\right)]{Fe_3O_4+H_2\left(1\right)}Fe\left\{{}\begin{matrix}\xrightarrow[\left(3\right)]{HCl}FeCl_2\\\xrightarrow[\left(4\right)]{+Cl_2}FeCl_3\xrightarrow[\left(5\right)]{+Fe}FeCl_2\\\xrightarrow[\left(6\right)]{+S}FeS\xrightarrow[\left(7\right)]{+HCl}H_2S\\\xrightarrow[\left(8\right)]{+O_2,t^o}Fe_3O_4\xrightarrow[\left(9\right)]{+HCl}FeCl_3\end{matrix}\right.\)

#hoidap247 #Anguyen1127

\(^{235}_{92}U+^1_0n\rightarrow^{146}_{57}La+^{87}_{35}Br+3\left(^1_0n\right)\)

Xét một nguyên tử \(^{235}_{92}U:\)

\(\Delta m=235,044+1,00861-145,943-86,912-3.1,00861=0,17178\left(u\right)\\ \Rightarrow\Delta E=\Delta m\left(u\right).931,2\left(MeV/u\right)=0,17178.931,2\approx160\left(MeV\right)=16.10^7\left(eV\right)\)

\(\Rightarrow\Delta E=16.10^7.1,602.10^{-19}=2,5632.10^{-11}\left(J\right)\)

Xét 1 mol \(^{235}_{92}U:\)

\(\Delta E=2,5632.10^{-11}.6,02.10^{23}=1,5430464\left(J\right)\)

#hoidap#Anguyen1127

\(\Delta E=E'-E=h\nu=\dfrac{hc}{\lambda}\left(1\right)\\ \sigma=\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n^2}-\dfrac{1}{n'^2}\right)\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\Delta E=hcR_H\left(\dfrac{1}{n^2}-\dfrac{1}{n'^2}\right)\)

@vtbq2008

`n_{H_2} = (9,4202)/(24,79) = 0,38(mol)`

Gọi `n_{Zn} = a(mol); n_{Al} = b(mol)`

`=> 65a + 27b = 16,24(1)`

`Zn + 2HCl -> ZnCl_2 + H_2`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

Theo PT: `n_{H_2} = n_{Zn} + 3/2 n_{Al}`

`=> a + 1,5b = 0,38(2)`

`(1),(2) => a = 0,2; b = 0,12`

Vậy

`m_{Zn} = 0,2.65 = 13(g)`

`m_{Al} = 0,12.27 = 3,24(g)`

@vtbq2008

Gọi \(\left\{{}\begin{matrix}N_{^{32}P}=a\\N_{^{33}P}=b\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}k_{^{32}P}=\dfrac{ln\left(2\right)}{14,3}\\k_{^{33}P}=\dfrac{ln\left(2\right)}{25,3}\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{ln\left(2\right)}{14,3}.a+\dfrac{ln\left(2\right)}{25,3}.b=9136,2\left(1\right)\)

Sau 14,3 nhày:

\(N_{^{32}P}=a.exp\left(-\dfrac{ln\left(2\right)}{14,3}.14,3\right)=0,5a\\ N_{^{33}P}=b.exp\left(-\dfrac{ln\left(2\right)}{25,3}.14,3\right)=0,6759b\)

\(\Rightarrow A=\dfrac{ln\left(2\right)}{14,3}.0,5a+\dfrac{ln\left(2\right)}{25,3}.0,6759b=4569,7\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=188297\\b=332\end{matrix}\right.\)

Câu 10:

@Anguyen1127

Câu 11:

\(2CO+O_2\xrightarrow[]{t^o}2CO_2\)