Có lẽ đề cho nung kết tủa trong không khí chứ không phải "để" bạn nhỉ?

PT: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)

____0,04_____0,04_____0,04 (mol)

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

0,03_____0,03______0,03 (mol)

⇒ n_{CuSO4 dư} = 0,1 - 0,04 - 0,03 = 0,03 (mol)

- Dung dịch X gồm: ZnSO₄: 0,04 (mol), FeSO₄: 0,03 (mol) và CuSO₄: 0,03 (mol)

PT: \(ZnSO_4+Ba\left(OH\right)_2\rightarrow Zn\left(OH\right)_{2\downarrow}+BaSO_{4\downarrow}\)

______0,04__________________0,04______0,04 (mol)

\(Zn\left(OH\right)_2+Ba\left(OH\right)_2\rightarrow BaZnO_2+2H_2O\)

\(FeSO_4+Ba\left(OH\right)_2\rightarrow Fe\left(OH\right)_{2\downarrow}+BaSO_{4\downarrow}\)

0,03____________________0,03______0,03 (mol)

\(CuSO_4+Ba\left(OH\right)_2\rightarrow Cu\left(OH\right)_{2\downarrow}+BaSO_{4\downarrow}\)

0,03____________________0,03_______0,03 (mol)

- Kết tủa thu được gồm: Fe(OH)₂: 0,03 (mol), Cu(OH)₂: 0,03 (mol) và BaSO₄: 0,1 (mol)

PT: \(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

_________0,03_________0,015 (mol)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

0,03_________0,03 (mol)

⇒ m = m_Fe2O3 + m_CuO + m_BaSO4 = 0,015.160 + 0,03.80 + 0,1.233 = 28,1 (g)

ai giúp mình vs

\(\Delta'=\left(m-1\right)^2+m^3-\left(m+1\right)^2=m^3-4m\ge0\) \(\Rightarrow\left[{}\begin{matrix}m\ge2\\-2\le m\le0\end{matrix}\right.\)

Theo hệ thức Viet:  \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m^3+\left(m+1\right)^2\end{matrix}\right.\)

Do \(x_1+x_2\le4\Rightarrow m-1\le2\Rightarrow m\le3\)

\(\Rightarrow\left[{}\begin{matrix}2\le m\le3\\-2\le m\le0\end{matrix}\right.\)

\(P=x_1^3+x_2^3+3x_1x_2\left(x_1+x_2\right)+8x_1x_2\)

\(=\left(x_1+x_2\right)^3+8x_1x_2\)

\(=8\left(m-1\right)^3+8\left[-m^3+\left(m+1\right)^2\right]\)

\(=8\left(5m-2m^2\right)\)

\(P=8\left(5m-2m^2-2+2\right)=16-8\left(m-2\right)\left(2m-1\right)\le16\)

\(P_{max}=16\) khi \(m=2\)

\(P=8\left(5m-2m^2+18-18\right)=8\left(9-2m\right)\left(m+2\right)-144\ge-144\)

\(P_{min}=-144\) khi \(m=-2\)

\(VT=\dfrac{a+b}{2\sqrt[3]{abc}}+\dfrac{b+c}{2\sqrt[3]{abc}}+\dfrac{c+a}{2\sqrt[3]{abc}}+\dfrac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\) (AM-GM 4 số hạng)

1.C
2.B
3.experiencable
4.presentation
5.powerful
 

1 C

2 C

3 experiences

4 presentation

5 powerful

1, He asks me to watch the video and find more information ..... the given topic

A.in                B.about                 C.with                      D.of

2.We prefer working in groups because it gives us a chance to discuss.....our classmates

A.with            B.about            C.in                         D.to

3.Our class has to.........learning materials at home and discuss the topic in class tomorrow

A.support       B.upload           C.strengthen           D.prepare

4.Our group is.........a lot projects that help us understand the lessons much better

A.making           B.doing             C.taking              D.organising

5.I don't like the traditional methods.Therefore,I wish to.............the new learning activities 

A.review            B.exchange            C.try                      D.encourage 

6.Jacks gets a lot of useful tips on this website and he has......an outline for his presentation 

A.made          B.done             C.taken                  D.read

7.When students have problems,they can ask..........answers or help immediately 

A.on        B.for          C.with           D.about

ai giúp mình với ạ