$n_{CO_2} = \dfrac{48,4}{44} = 1,1(kmol)$
$n_{O_2} = n_{CO_2} = 1,1(mol)$

Suy ra $V_{O_2} = 1,1.22,4.1000 = 24640(lít)$

Ta có : \(M_{kk}=\dfrac{20.32+80.28}{20+80}=28,8\) ( đvc )

\(\Rightarrow d_{\dfrac{KK}{H2}}=\dfrac{M_{kk}}{M_{h2}}=\dfrac{28,8}{2}=14,4\)

Vậy ...

c) \(=\left(4x-3\right)^2-\left(9x^2-4\right)\)

\(=16x^2-24x+9-9x^2+4=7x^2-24x+13\)

d) \(=\left(x^2-3x+2\right)\left(x+3\right)-\left(x^3-5x^2\right)\)

\(=x^3+3x^2-3x^2-9x+2x+6-x^3+5x^2\)

\(=5x^2-7x+6\)

c. (4x - 3)(4x - 3) - (3x + 2)(3x - 2)

= (4x - 3)² - (9x² - 4)

= 16x² - 24x + 9 - 9x² + 4

= 16x² - 9x² - 24x + 9 + 4

= 7x² - 24x + 13

d. (x - 2)(x - 1)(x + 3) - x²(x - 5)

= (x² - 1 - 2x + 2)(x + 3) - x²(x - 5)

= x³ + 3x² - x - 3 - 2x² - 6x + 2x + 6 - x³ + 5

= x³ - x³ + 3x² - 2x² - x - 6x + 2x + 6 + 5 - 3

= x² - 5x + 8

Ta có:-(3-0,2.x)-80%=7,5

-3+0,2x-0,8=7,5

0,2x=7,5+3+0,8

X=11,3:0,2

X=56,5

Vậy x=56,5

Ta có:-(3-0,2.x)-80%=7,5

-3+0,2x-0,8=7,5

0,2x=7,5+3+0,8

X=11,3:0,2

X=56,5

Vậy x=56,5

Gọi biểu thức trên là A, ta có:

\(A=\frac{1}{2\cdot15}+\frac{1}{15\cdot3}+\frac{1}{3\cdot21}+\frac{1}{21\cdot4}+...+\frac{1}{87\cdot90}\)

\(13A=\frac{13}{2\cdot15}+\frac{13}{15\cdot3}+\frac{13}{3\cdot21}+\frac{13}{21\cdot4}+...+\frac{13}{87\cdot90}\)

\(13A=\frac{1}{2}-\frac{1}{15}+\frac{1}{15}-\frac{1}{3}+\frac{1}{3}-\frac{1}{21}+\frac{1}{21}-\frac{1}{4}+...+\frac{1}{87}-\frac{1}{90}\)

\(13A=\frac{1}{2}-\frac{1}{90}\)

\(13A=\frac{22}{45}\)

\(A=\frac{22}{45\text{x}13}=\frac{22}{585}\)

a) Ta có: \(M_{XCO_3}=4\cdot25=100\) \(\Rightarrow M_X=100-12-16\cdot3=40\left(đvC\right)\)

\(\Rightarrow\) X là Canxi (Ca)

b) \(\%O=\dfrac{16\cdot3}{100}\cdot100\%=48\%\)

Bài 3:

a) M(XCO3)=25. M(He)= 25.4=100(đ.v.C)

Mặt khác: M(XCO3)=M(X)+ 60

=> M(X)+60=100

<=>M(X)=40(đ.v.C) 

=> X là Canxi (Ca=40)

b) %mO=[(3.16)/100].100=48%