Câu 16:

PTHH: \(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{Cl_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\n_{NaOH}=\dfrac{600\cdot20\%}{40}=3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) 2 chất p/ứ hết

Mặt khác: \(m_{Cl_2}=1,5\cdot71=106,5\left(g\right)\)

\(\Rightarrow m_{nướcjaven}=m_{Cl_2}+m_{ddNaOH}=706,5\left(g\right)\)

Câu 21:

PTHH: \(H_2SO_{4\left(đ\right)}+CaF_2\rightarrow CaSO_4+2HF\)

Ta có: \(n_{HF}=\dfrac{250\cdot40\%}{20}=5\left(kmol\right)\) 

\(\Rightarrow n_{CaF_2}=2,5\left(kmol\right)\) \(\Rightarrow m_{CaF_2}=2,5\cdot78=195\left(kg\right)\) 

:vvv nhiều quá v bn gửi lẻ ra đi

Câu 8: 

#include <bits/stdc++.h>

using namespace std;

long long a[250],n,i;

int main()

{

cin>>n;

for (i=1; i<=n; i++) cin>>a[i];

sort(a+1,a+n+1);

for (i=1; i<=n; i++) cout<<a[i]<<" ";

return 0;

}

đề đâu e

D. sunny and windy

I

1C

2 D

3 D

4C

5 C

II

1 C

2 A

3 D

4 C

5 D

6 D

7 B

8 B

9 D

10 A

11 C

12 A

13 A

14 C

15 D

16 C

17 D

\(b,\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{\sqrt{15}}=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(d,\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\left(\sqrt{ab}-\sqrt{bc}\right)}=\sqrt{ab}+\sqrt{bc}=\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)\)

\(e,\left(a\sqrt{\dfrac{a}{b}+2\sqrt{ab}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)

\(=a\left(\sqrt{\dfrac{a}{b}+\dfrac{2b.\sqrt{ab}}{b}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)

\(=a\sqrt{a}\sqrt{a+2b\sqrt{ab}}+b\sqrt{a^2}\)

\(=a\sqrt{a^2+2ab\sqrt{ab}}+ab\)

\(=a\left(\sqrt{a^2+2ab\sqrt{ab}}+b\right)\)

\(f,\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(a-\sqrt{a}+1-\sqrt{a}\right)\)

\(=\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2\)

\(=\left(a-1\right)^2=a^2-2a+1\)

Đề bài đâu rồi em?

1, A

2, A

3, B

4, B

5, A

6, C

7, A

8, B

9, A

10, B

11, C

12, D

Câu 7:

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)

c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)

d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)

Câu 8:

a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)

b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)

c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

Bài 6: 

a, \(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)

b, \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaSO_3}=n_{SO_2}=0,05\left(mol\right)\)

\(\Rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{0,05.74}{100}.100\%=3,7\%\)

c, \(m_{CaSO_3}=0,05.120=6\left(g\right)\)