11. A

12. B

13. C

14. D

15. A

16. A

17. C

18. B

19. A

20. D

21. D

22. C

23. C

24. B

25. B

a) \(\left(-189\right)+135+\left(-111\right)+\left(-135\right)\)

\(=\left(-189-111\right)+\left(135-135\right)\)

\(=-300+0\)

\(=-300\)

b) \(126+345-126+215\)

\(=\left(126-126\right)+\left(345+215\right)\)

\(=0+560\)

\(=560\)

c) \(\left(-213\right)+186+\left|-213\right|-186+100\)

\(=\left(-213\right)+186+213-186+100\)

\(=\left(-213+213\right)+\left(186-186\right)+100\)

\(=0+0+100\)

\(=100\)

d) \(\left(-34\right)+\left|-123\right|+\left|-34\right|-123+200\)

\(=\left(-34\right)+123+34-123+200\)

\(=\left(-34+34\right)+\left(123-123\right)+200\)

\(=0+0+200\)

\(=200\)

e) \(\left(-250\right)+\left(-15\right)-\left(-250\right)+\left|-15\right|+150\)

\(=\left(-250\right)+\left(-15\right)+250+15+150\)

\(=\left(-250+250\right)+\left(-15+15\right)+150\)

\(=0+0+150\)

\(=150\)

\(\left(1\right)4Na+O_2--->2Na_2O\)

\(\left(2\right)Na_2O+H_2O--->2NaOH\)

\(\left(3\right)2NaOH+CO_2--->Na_2CO_3+H_2O\)

\(\left(4\right)Na_2CO_3+H_2SO_4--->Na_2SO_4+H_2O+CO_2\)

\(\left(5\right)Na_2SO_4+BaCl_2--->BaSO_4\downarrow+2NaCl\)

2+2+2=6 

Ok tk mk đi 

====================

2+2+2=6 Tích cho tui nhoa !!!

2+2+2=8

Sao ngày nào bạn cx xin điểm vậy

Bài 22

 Ta có:

∠P + ∠N = 90⁰

⇒ ∠N = 90⁰ - ∠P

= 90⁰ - 58⁰

= 32⁰

sin P = MN/NP

⇒ NP = MN : sin P

= 12 : sin 58⁰

≈ 14,2 (cm)

tan P = MN/MP

⇒ MP = MN : tan P

= 12 : tan 58⁰

≈ 7,5 (cm)

2 D 

3 C 

4 D

33 B 

1 C

2 B 

3 D

4 C

3. \(\left\{{}\begin{matrix}2Z+N=40\\Z-N=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}Z=13\\N=14\end{matrix}\right.\)
A=Z+N=27
Nguyên tử này là : \(\begin{matrix}27\\13\end{matrix}Al\)
4. \(\overline{A}=\dfrac{63x_1+65x_2}{100}=63,54\)
\(\rightarrow63x_1+65x_2=6354\left(1\right)\)
\(x_1+x_2=100\left(2\right)\)
Từ (1), (2) \(\left\{{}\begin{matrix}63x_1+65x_2=6354\\x_1+x_2=100\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=73\\x_2=27\end{matrix}\right.\)
Phần trăm số nguyên tử của : + \(\begin{matrix}63\\29\end{matrix}Cu\) là 73%
                                                +\(\begin{matrix}65\\29\end{matrix}Cu\) là 27%

 

3.

Ta có: \(\left\{{}\begin{matrix}p+e+n=40\\p=e\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=13=z\\n=14\end{matrix}\right.\)

   ⇒ A = n+z = 14+13 = 27

KH: \(^{27}_{13}Al\)

\(x\) ⋮ 48; \(x\) ⋮ 36; ⇒ \(x\) \(\in\) BC(48; 36)

48 = 2⁴.3; 36 = 2².3²

BCNN(48; 36) = 2⁴.3² = 144

\(x\) \(\in\) B(144) = {0; 144; 288; 432; 576;..;}

Vì  100 < \(x\); 576 < \(x\) nên 

⇒  \(x\) = 144.k (k > 5; k \(\in\) n)

A có người cứu rồi .cảm ơn 💕