Ta có:2⁷⁶>2⁷⁵=(2³)²⁵=8²⁵>7²⁵

=>2⁷⁶>7²⁵

Dien tich hinh chu nhat la 2000cm2

n\(_S\)=\(\dfrac{3,2}{32}\)=0,1(mol)

Theo PT ta có: n\(_{SO_2}\)=n\(_S\)\(\Rightarrow\)V\(_{SO_2}\)=22,4 . 0,1= 2,24(l)

PT: 2Cu + O\(_2\) \(\rightarrow\)2CuO

n\(_{Cu}\)=\(\dfrac{48}{64}\)=0,75(mol)

Theo PT ta có:

n\(_{O_2}\)=\(\dfrac{1}{2}\)n\(_{Cu}\)=\(\dfrac{1}{2}\).0,75 = 0.375(mol) \(\Rightarrow\) V\(_{O_2}\)=22,4.0.375=8,4(l)

n\(_{CuO}\)=n\(_{Cu}\)=0,75(mol) \(\Rightarrow\) m\(_{CuO}\)=0,75.80=60(g)

a) \(Fe\) + \(2HCl\) \(\rightarrow\) \(FeCl_2\) + \(H_2\)

b) Fe : HCl=1:2

Fe : FeCl\(_2\)=1:1

Fe : H\(_2\) =1:1

(còn HCl bạn làm tiếp nhé)

c) \(n_{Fe}\)=\(\dfrac{11,2}{56}\)=0,2(mol)

Theo PT ta có: \(n_{H_2}\)=\(n_{Fe}\)=0,2(mol)\(\Rightarrow\)\(V_{H_2}\)=0,2\(\times\)22,4=4,48(l)

d) Theo PT ta có: \(n_{HCl}\)=\(2n_{Fe}\)=2\(\times\)0,2=0,2\(\Rightarrow\)\(m_{HCl}\)=0,4\(\times\)36,5=14,6(g)

e)Theo PT ta có:\(n_{FeCl_2}\)=\(n_{Fe}\)=0,2\(\Rightarrow\)\(m_{FeCl_2}\)=0,2\(\times\)127=25,4(g)

Ta có: \(\left(x_1p-y_1q\right)^{2n}\ge0;\left(x_2p-y_2q\right)^{2n}\ge0;....;\left(x_mp+y_mq\right)^{2n}\ge0\)

=>(x₁p-y₁q)²ⁿ+(x₂p-y₂q)²ⁿ+...+(x_mp-y_mq)²ⁿ > hoặc = 0

Mà theo đề (x₁p-y₁q)²ⁿ+(x₂p-y₂q)²ⁿ+...+(x_mp-y_mq)²ⁿ < hoặc = 0

nên: (x₁p-y₁q)²ⁿ+(x₂p-y₂q)²ⁿ+...+(x_mp-y_mq)²ⁿ=0

=>x₁p-y₁q=0 =>x₁=y₁q/p

    x₂p-y₂q=0 =>x₂=y₂q/p

........

    x_mp-y_mq=0 =>x_m=y_mq/p

Suy ra: \(\frac{x_1+x_2+...+x_n}{y_1+y_2+....+y_n}=\frac{\frac{y_1q}{p}+\frac{y_2q}{p}+...+\frac{y_mq}{p}}{y_1+y_2+...+y_m}=\frac{\frac{q}{p}\left(y_1+y_2+....+y_m\right)}{y_1+y_2+...+y_m}=\frac{q}{p}\)

=>điều phải chứng minh

Đặt tính rồi tính :

d, 18 giờ 36 phút: 6