\(PTK_x=2\cdot56+\left(32+16\cdot4\right)a=400\\ \Rightarrow96a=288\\ \Rightarrow a=3\)

Do đó CTHH \(X:Fe_2\left(SO_4\right)_3\)

Vì \(SO_4\left(II\right)\Rightarrow Fe\left(III\right)\)

tớ cám ơn!

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)

\(=\frac{31}{32}\)

Ta có: \(n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\)

a, PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)

_________1____________________1_________0,5 (mol)

b, V_CO2 = 0,5.22,4 = 11,2 (l)

c, m_CH3COOK = 1.98 = 98 (g)

Bạn tham khảo nhé!

a) PTHH:         2CH₃COOH + K₂CO₃ → 2CH₃COOK + CO₂↑ + H₂O

b) n_CH3COOH = \(\dfrac{m}{M}=\dfrac{60}{60}=1\left(mol\right)\)

Theo PTHH: n_CO2 = \(\dfrac{1}{2}\).n_CH3COOH = 0,5 (mol)

=> V_{CO2 (đktc)} = n.22,4 = 0,5.22,4 = 11,2 (lít)

c) Theo PTHH: n_CH3COOK = n_CH3COOH = 1 (mol)

=> m_CH3COOK = n.M = 1.98 = 98 (g)

Mặc dù hơi muộn nhưng mà chúc bạn thi đạt kết quả tốt nha ^_^

Tin học à

Lớp mấy vậy 

\(5^{11}và7^7=\left(5.7\right)^4và\left(7.5\right)^2=35^4>35^2\)

5 mũ 11 > 7 mũ 7

a: Bảng tần số:

Lớp 8A có 18 bạn nam

Mốt của dấu hiệu là 20

b: thời gian trung bình là;

\(\dfrac{15\cdot1+17\cdot3+18\cdot2+19\cdot1+20\cdot6+23\cdot5}{18}\simeq19,78\)
d: Thời gian chạy của bạn đó là;

20*19-19,78*18=23,96(giây)

\(=\left(\dfrac{\sqrt{5}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}+\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}\right):\sqrt{\dfrac{5}{2}}\)

\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(=2\sqrt{5}.\dfrac{\sqrt{2}}{\sqrt{5}}=2\sqrt{2}\)

a) Ta có: \(\left(\dfrac{\sqrt{15}-\sqrt{20}}{\sqrt{3}-2}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{6}+\sqrt{5}}\right):\sqrt{\dfrac{5}{2}}\)

\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{10}}{2}\)

\(=2\sqrt{5}\cdot\dfrac{2}{\sqrt{10}}=2\sqrt{2}\)

`#040911`

`b)`

\(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{19\times21}\)

`=`\(\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{19\times21}\right)\)

`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{21}\right)\)

\(=\dfrac{1}{2}\times\dfrac{20}{21}\\ =\dfrac{10}{21}\\ \text{ Vậy, A = }\dfrac{10}{21}\)

Bài 3:

a: \(M=\left(6x-2\right)^2+\left(5x-2\right)^2-\left(4-12x\right)\left(2-5x\right)\)

\(=\left(6x-2\right)^2-2\left(6x-2\right)\left(5x-2\right)+\left(5x-2\right)^2\)

\(=\left(6x-2-5x+2\right)^2=x^2\)

b: \(N=\left(x^2+9x+1\right)^2+9\left(1-6x+9x^2\right)-6\left(3x-1\right)\left(x^2+9x+1\right)\)

\(=\left(x^2+9x+1\right)^2-2\left(x^2+9x+1\right)\left(9x-3\right)+\left(9x-3\right)^2\)

\(=\left(x^2+9x+1-9x+3\right)^2=\left(x^2+4\right)^2=x^4+8x^2+16\)

c: \(P=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)

\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)

\(=-9x^2\)

d: \(Q=\left(a^2+4a+4\right)^2-\left\lbrack2\left(a^2-4\right)^{}\right\rbrack^2+4\left(a-2\right)^4\)

\(=\left(a^2+4a+4\right)^2-4\left(a^2+4a+4\right)\left(a^2-4a+4\right)+4\left(a^2-4a+4\right)^2\)

\(=\left(a^2+4a+4-2a^2+8a-8\right)^2=\left(-a^2+12a-4\right)^2\)

Bài 5:

a: \(A=-2x^2+3x-5\)

\(=-2\left(x^2-\frac32x+\frac52\right)\)

\(=-2\left(x^2-2\cdot x\cdot\frac34+\frac{9}{16}+\frac{31}{16}\right)=-2\left(x-\frac34\right)^2-\frac{31}{8}\le-\frac{31}{8}\forall x\)

Dấu '=' xảy ra khi \(x-\frac34=0\)

=>x=3/4

b: \(B=\frac12x^2+x-1\)

\(=\frac12\left(x^2+2x-2\right)\)

\(=\frac12\left(x^2+2x+1-3\right)=\frac12\left(x+1\right)^2-\frac32\ge-\frac32\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

d: D=(x-1)(x+2)(x+3)(x+6)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi \(x^2+5x=0\)

=>x(x+5)=0

=>x∈{0;-5}