a, những

b, chỉ

a.nhung

b.chi 

may mik ko nhan dc ban chiu kho nhin nha 

dung thi like ho mik

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

a) \(7^2-7\left(13-x\right)=14\)

\(7\left(13-x\right)=49-14=35\)

\(13-x=5\)

\(x=13-5=8\)

b) \(5x-5^2=10\)

\(5x=10+25=35\)

\(x=7\)

c) \(4\left(x-5\right)-2^3=2^4.3=48\)

\(4\left(x-5\right)=48+8=56\)

\(x-5=14\)

\(x=19\)

4:

a: \(\Leftrightarrow49+7\left(x-13\right)=14\)

=>7(x-13)=35

=>x-13=5

=>x=18

b: \(5x-5^2=10\)

=>\(5x=10+25=35\)

=>x=7

c: \(4\left(x-5\right)-2^3=2^4\cdot3\)

=>\(4\left(x-5\right)=16\cdot3+8=56\)

=>x-5=14

=>x=19