a) 2KClO₃ \(\underrightarrow{t^o}\) 2KCl + 3O₂

mol \(\dfrac{a}{122,5}\rightarrow\dfrac{a}{122,5}\dfrac{3a}{245}\)

2KMnO₄ \(\underrightarrow{t^o}\) K₂MnO₄ + MnO₂ + O ₂

mol \(\dfrac{b}{158}\rightarrow\dfrac{b}{316}\dfrac{b}{316}\dfrac{b}{316}\)

\(74,5.\dfrac{a}{122,5}=197.\dfrac{b}{316}+87.\dfrac{b}{316}\)

⇔ \(\dfrac{74,5a}{122,5}=\dfrac{71b}{79}\)

⇒ \(\dfrac{a}{b}=\dfrac{71.122,5}{74,5.79}\approx1,478\)

b) \(\dfrac{3a}{245}:\dfrac{b}{316}=\dfrac{3a.316}{245.b}=\dfrac{948}{245}.\dfrac{a}{b}=\dfrac{948}{245}.1,478\approx5,72\)

\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

PT: CuSO₄ + 2NaOH → Cu(OH)₂ + Na₂SO₄

mol 0,02 0,04 ← 0,02 → 0,02

2Cu(OH)₂ + O₂ \(\underrightarrow{t^o}\) 2CuO + 2H₂O

mol 0,02 0,01 ← 0,02 → 0,02

a) \(m_{Cu\left(OH\right)_2}=0,02.98=1,96\left(g\right)\)

b) V_{dung dịch sau phản ứng} = 200 + 300 = 500 (ml) = 0,5 (l)

\(C_{MNa_2SO_4}=\dfrac{0,02}{0,5}=0,04M\)

vo: 39000

sach: 52000

a) x³ - 3x² + 3x - 1 = 0

⇔ (x - 1)³ = 0

⇔ x - 1 = 0

⇔ x = 1

Vậy tập nghiệm phương trình là S = \(\left\{1\right\}\)

b) (2x - 5)² - (x + 2)² = 0

⇔ (2x - 5 - x - 2)(2x - 5 + x + 2) = 0

⇔ (x - 7)(3x - 3) = 0

⇔ \(\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=7\\3x=3\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy tập nghiệm phương trình là S = \(\left\{1;7\right\}\)

Vì Cu đứng sau H nên không phản ứng với dung dịch HCl

⇒ chất rắn không tan là Cu

m_Al = 10 - 4,6 = 5,4 (g)

n_Al = \(\dfrac{5,4}{27}=0,2\left(mol\right)\)

a) % m Al = \(\dfrac{5,4}{10}.100\%=54\%\)

% m Cu = 100% - 54% = 46%

b) PT: 2Al + 6HCl → 2AlCl₃ + 3H₂

mol 0,2 → 0,6 0,2 0,3

\(V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) m_HCl = 0,6.36,5 = 21,9 (g)

m_{dung dịch sau phản ứng} = 10 + 600 = 610 (g)

\(C\%_{HCl}=\dfrac{21,9}{610}.100\%=3,59\%\)

d) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{26,7}{610}.100\%=4,38\%\)

|x + 2| - 5 = 2018

⇔ |x + 2| = 2018 + 5

⇔ |x + 2| = 2023

⇔ \(\left[{}\begin{matrix}x+2=2023\\x+2=-2023\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=2023-2\\x=-2023-2\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=2021\\x=-2025\end{matrix}\right.\)

Vậy x ∈ \(\left\{2021;-2025\right\}\)

n_{hỗn hợp} = \(\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(n_{H_2O}=\dfrac{16,2}{18}=0,9\left(mol\right)\)

Gọi \(n_{CH_4}\) = a (mol); \(n_{H_2}\) = b (mol)

PT: CH₄ + 2O₂ \(\underrightarrow{t^o}\) CO₂ + 2H₂O

mol a → 2a a 2a

2H₂ + O₂ \(\underrightarrow{t^o}\) 2H₂O

mol b → 0,5b b

\(\left\{{}\begin{matrix}a+b=0,5\\2a+b=0,9\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}a=0,4\\b=0,1\end{matrix}\right.\)

a) Vì tỉ lệ V = tỉ lệ n nên:

% V CH₄ = \(\dfrac{0,4}{0,5}.100\%=80\%\)

% V H₂ = 100% - 80% = 20%

b) \(V_{CO_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\)

2Fe + 3Cl₂ → 2FeCl₃

-23.63 + 23.21 - 58.23

= 23.(-63 + 21 - 58)

= 23.(-100)

= -2300

1) 35.18 - 5.7.28

= 35.18 - 35.28

= 35.(18 - 28)

= 35.(-10)

= -350

2) 45 - 5.(12 + 9)

= 45 - 5.12 - 5.9

= 45 - 60 - 45

= (45 - 45) - 60

= 0 - 60

= -60

3) 24.(16 - 5) - 16.(24 - 5)

= 24.16 - 24.5 - 16.24 + 16.5

= (24.16 - 16.24) - (24.5 + 16.5)

= 0 - (24.5 + 16.5)

= -24.5 + 16.5

= 5.(-24 + 16)

= 5.(-8)

= -40

4) 29.(19 - 13) - 19.(29 - 13)

= 29.19 - 29.13 - 19.29 + 19.13

= (29.19 - 19.29) + (19.13 - 29.13)

= 0 + (19.13 - 29.13)

= 19.13 - 29.13

= 13.(19 - 29)

= 13.(-10)

= -130

5) 31.(-18) + 31.(-81) - 31

= 31.(-18 - 81 - 1)

= 31.(-100)

= -3100

6) (-12).47 + (-12).52 + (-12)

= (-12).(47 + 52 + 1)

= (-12).100

= -1200

7) 13.(23 + 22) - 3.(17 + 28)

= 13.45 - 3.45

= 45.(13 - 3)

= 45.10

= 450

8) -48 + 48.(-78) + 48.(-21)

= 48.(-1) + 48.(-78) + 48.(-21)

= 48.(-1 - 78 - 21)

= 48.(-100)

= -4800