chiều nay tui giải đc hơm:)

1 was repaired

2 Was - bought

3 are watered

4 wasn't learned

5 was watched

6 Is - read

7 was stolen

8 was broken

9 wasn't cut

10 is cleaned

was repaired

was-bought

are watered

động vật

dong vat

súc vật

\(=\dfrac{1}{2}-\dfrac{1}{2}cos\dfrac{2\pi}{9}+\dfrac{1}{2}-\dfrac{1}{2}cos\dfrac{4\pi}{9}+\dfrac{1}{2}cos\dfrac{\pi}{9}-\dfrac{1}{2}cos\dfrac{3\pi}{9}\)

\(=1-\dfrac{1}{2}\left(cos\dfrac{2\pi}{9}+cos\dfrac{4\pi}{9}\right)+\dfrac{1}{2}cos\dfrac{\pi}{9}-\dfrac{1}{4}\)

\(=\dfrac{3}{4}-cos\dfrac{\pi}{9}.cos\dfrac{3\pi}{9}+\dfrac{1}{2}cos\dfrac{\pi}{9}\)

\(=\dfrac{3}{4}-\dfrac{1}{2}cos\dfrac{\pi}{9}+\dfrac{1}{2}cos\dfrac{\pi}{9}\)

\(=\dfrac{3}{4}\)

\(n_{AgNO_3}=1.0,5=0,5\left(mol\right);n_{HCl}=2.0,3=0,6\left(mol\right)\)

PTHH: AgNO₃ + HCl → AgCl ↓ + HNO₃

Mol:      0,5          0,5                          0,5

Ta có: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\)⇒ AgNO₃ pứ hết, HCl dư

* V_{dd sau pứ} = 0,5+0,6 = 1,1 (l)

\(\Rightarrow C_{M_{ddNO_3}}=\dfrac{0,5}{1,1}=0,4545M\)

\(C_{M_{ddHCldư}}=\dfrac{0,6-0,5}{1,1}=0,0909M\)

\(m_{ddAgNO_3}=500.1,2=600\left(g\right);m_{ddHCl}=300.1,5=450\left(g\right)\)

* m_{dd sau pứ} = 600+450 = 1050 (g)

\(C\%_{ddHNO_3}=\dfrac{0,5.63.100\%}{1050}=3\%\)

\(C\%_{ddHCl}=\dfrac{\left(0,6-0,5\right).36,5.100\%}{1050}=0,348\%\)

D

B

_HT_

B