1 A

2 C

3 C

4 D

5 B

6 A

7 B

8 D

9 C

10 C

11 B

12 D

13 B

14 C

15 B

\(a,B=4\sqrt{x+1}-3\sqrt{x+1}+\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\\ b,B=8\Leftrightarrow4\sqrt{x+1}=8\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\left(tm\right)\)

MgO: Mg có điện hóa trị 2+, O có điện hóa trị 2-

FeF₃: Fe có điện hóa trị 3+, F có điện hóa trị 1-

BaCl₂: Ba có điện hóa trị 2+, Cl có điện hóa trị 1-

Ca₃N₂: Ca có điện hóa trị 2+, N có điện hóa trị 3-

A song was sung by her.

Was the letter sent by him?

The bus was stopped.

My car was stolen by a thief.

He was not let go.

The prize wasn't won by her.

Their beds weren't made.

They were not told by me.

Were they told by you?

I was hit by him.

The letter was lost by my mother yesterday.

They key was found by her at home.

Mistakes were made.

That woman was loved by Peter 3 years ago.

The rooms were cleaned.

Câu 1 : 

Gọi $m_{dd} = a(gam)$
$n_{NaOH} = \dfrac{a.20\%}{40} = 0,005a(mol)$
$n_{Ba(OH)_2} = \dfrac{a.8,55\%}{171} = 0,0005a(mol)$

$n_{HNO_3} = \dfrac{224}{1,1.1000}.4,5 = 0,9(mol)$

$NaOH + HNO_3 \to NaNO_3 + H_2O$
$Ba(OH)_2 + 2HNO_3 \to Ba(NO_3)_2 + 2H_2O$
Theo PTHH : 

$0,005a + 0,0005a.2 = 0,9 \Rightarrow a = 150(gam)$

Câu 2 : 

Gọi thể tích dung dịch cần tìm là V(lít)

$n_{H_2SO_4} = 1,1V(mol) ; n_{HCl} = 1,98V(mol)$

$\Rightarrow n_H = 1,1V.2 + 1,98V = 4,18V(mol)$
$n_{NaOH} = 0,19.3 = 0,57(mol) ; n_{Ba(OH)_2} = 0,19.4 = 0,76(mol)$
$\Rightarrow n_{OH} = 0,57 + 0,76.2 = 2,09(mol)$
$H + OH \to H_2O$
$\Rightarrow 4,18V = 2,09 \Rightarrow V = 0,5(lít)$

Câu 3 : 

$n_{ZnO} = \dfrac{48,6}{81} = 0,6(mol) ; n_{HNO_3} = \dfrac{800.6,3\%}{63}= 0,8(mol)$

ZnO + 2HNO₃ → Zn(NO₃)₂ + H₂O

0,4.......0,8...............0,4........................(mol)

Sau phản ứng : 

$m_{dd} = 0,4.81 +800 = 832,4(gam)$
$C\%_{Zn(NO_3)_2} = \dfrac{0,4.189}{832,4}.100\% = 9,1\%$