10 If she is ill, she gets wet.

11 If she knew French, she could apply for the job.

12 Unless he washes his car, it will be dirty.

13 I am looking forward to writing a poem.

14 Because she was rich, she gave me money.

15 Though they are happy, they don't go out.

16 Though he didn't go out, despite the bad result.

17 I walked more slowly than him.

18 Lan is the most curious person in the class.

19 Ba is better than Hoa.

20 She has been planting trees for 20 days.

21 It is his first time baking a cake.

22 She has never published a book before.

23 A song is being sung by them.

24 Money has not been spent by anybody.

Help mình zớiii:>

đề là gì b

Tìm giá trị nhỏ nhất bạn nhé

\(A=\left|x+2\right|+\left|x-3\right|=\left|x+2\right|+\left|3-x\right|\ge\left|x+2+3-x\right|=5\)

Dấu '' = '' xảy ra khi: \(\left(x+2\right)\left(3-x\right)\ge0\Leftrightarrow\left(x+2\right)\left(x-3\right)\le0\Leftrightarrow-2\le x\le3\)

A. MgSO₄ 

- Liên kết ion được hình thành bởi cation kim loại Mg²⁺ và anion gốc axit SO₄^2-

- Liên kết cộng hóa trị giữa S và O.

A: MgSO₄

Liên kết ion giữa Mg²⁺ và SO₄^2-, liên kết cộng hóa trị giữa S và O

Câu 4 : 

\(n_{Fe2O3}=\dfrac{24}{160}=0,15\left(mol\right)\)

Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O|\)

          1               3                    1                 3

       0,15                                 0,15

a) \(n_{Fe2\left(SO4\right)3}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe2\left(SO4\right)3}=0,15.400=60\left(g\right)\)

b) \(C_{M_{Fe2\left(SO4\right)3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

 Chúc bạn học tốt

a,\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

Mol:     0,15         0,45                0,15                   

\(m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)

b,\(C_{M_{ddFe_2\left(SO_4\right)_3}}=\dfrac{0,15}{0,5}=0,3\left(mol\right)\)

Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O

0,15--------------------------------0,15 mol

n Fe2O3=24\160=0,15 mol

=>m Fe2(SO4)3=0,15.400=60g

mà thể tích khôi đổi 

=>CM Fe2(SO4)3=0,15\0,5=0,3 M

Có:

\(m=\dfrac{F_{hd}R^2}{GM}=\dfrac{F'_{hd}\left(R+h\right)^2}{GM}\)

\(\Leftrightarrow F_{hd}R^2=F'_{hd}\left(R+h\right)^2\)

\(\Leftrightarrow45R^2=5\left(R^2+h^2+2Rh\right)\)

\(\Leftrightarrow45R^2=5R^2+5h^2+10Rh\)

\(\Leftrightarrow45R^2-5R^2-5h^2-10Rh=0\)

\(\Leftrightarrow40R^2-5h^2-10Rh=0\)

\(\Leftrightarrow h=12800km\)