\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

a. PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PTHH: \(n_{Fe}=n_{H_2}=1,5\left(mol\right)\)

\(\Rightarrow m_{Fe}=56\cdot1,5=84\left(g\right)\)

b. Đổi: \(500ml=0,5l\)

\(CM_{H_2SO_4}=\dfrac{1,5}{0,5}=3M\)

c. \(2H_2+O_2\rightarrow2H_2O\)

Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot1,5=0,75\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,75\cdot22,4=16,8\left(l\right)\)

c, \(2H_2+O_2 \rightarrow2H_2O\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5(mol) \Rightarrow n_{O_2}=0,75(mol)\)

\(V_{O_2}=22,4.0,75=16,8(l)\)

a) Fe+H2SO4---->FeSO4+H2(1)

nH2=33,6:22,4=1,5 mol

Theo PT(1):nFe=nH2=1,5 mol

=>mFe=56.1,5=84g

b)TheoPT(1):nH2SO4=nH2=1,5 mol

=>CM(H2SO4)=1,5:0,5=3 M

c)2H2+O2---->2H2O(2)

TheoPT(2):nO2=1/2nH2=1/2.1,5=0,75mol

=>VO2=22,4.0,75=16,8l

1: Xét ΔOAD và ΔOCB có

OA=OC

\(\widehat{O}\) chung

OD=OB

Do đó: ΔOAD=ΔOCB

Suy ra: AD=CB

a: BC=50cm

AE=25cm

1.theo bất đẳng thức côsi ta có

\(a+b\ge2\sqrt{ab}\\ b+c\ge2\sqrt{ab}\\ c+a\ge2\sqrt{ab}\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{ab.bc.ca}\)

                                       \(\ge8\sqrt{a^2b^2c^2}\\ \ge8abc\)

2.\(a^4+b^2\ge2\sqrt{a^4b^2}=2a^4b^2\)

\(\dfrac{a}{a^4+b^2}\le\dfrac{a}{2a^2b}=\dfrac{1}{2ab}\)

tương tự:\(\dfrac{b}{b^4+a^2}\le\dfrac{1}{2ab}\)

\(\rightarrow\dfrac{a}{a^4+b^2}+\dfrac{b}{b^4+a^2}\le\dfrac{1}{ab}\)

dấu = xảy ra khi \(a^4=b^2\\ b^4=a^2\)\(\rightarrow a^2=b^2=1\)