bạn đăng tách ra tầm 10 câu mỗi lần đăng nha, chứ dài ntnay ngại làm lắm~

e có tách 3 bài ra rồi ạ, phiền anh/chị/bạn giúp e với ạ, e cảm ơn ạ

a, \(n_{CaCO_3}=\dfrac{41,2}{100}=0,412\left(mol\right)\)

PTHH: CaO + H₂O → Ca(OH)₂

Mol:      0,412                 0,412                        

PTHH: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Mol:       0,412        0,412          0,412

\(m_{CaO}=0,412.56=23,072\left(g\right)\)

b, \(V_{CO_2}=0,412.22,4=9,2288\left(l\right)\)

\(m_{Na_2CO_3}=100.16,96\%=16,96\left(g\right)\Rightarrow n_{Na_2CO_3}=\dfrac{16,96}{106}=0,16\left(mol\right)\)

\(m_{BaCl_2}=200.10,4\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + BaCl₂ → BaCO₃ + 2NaCl

Mol:      0,1             0,1                           0,2

Ta có: \(\dfrac{0,16}{1}>\dfrac{0,1}{1}\) ⇒ Na₂CO₃ dư, BaCl₂ hết

m_{dd sau pứ} = 100 + 200 = 300 (g)

\(C\%_{ddNaCl}=\dfrac{0,1.58,5.100\%}{300}=1,95\%\)

\(C\%_{ddNa_2CO_3}=\dfrac{\left(0,16-0,1\right).106.100\%}{300}=2,12\%\)

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I will organize free tutoring classes, donate books and school supplies, and raise funds for students in need.

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Because education is very important, and I want to help students have better opportunities to learn and succeed.

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Yes, they will. They can improve their knowledge, have enough materials to study, and feel more confident about their future.

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I can share information on social media, invite friends to join, and cooperate with schools or local organizations to spread the message.

5.

1. going

2.lying/reading

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9. will travel

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6.

1.beautiful

2. boring

3. decoration

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8

Hình mờ lắm ak bn!

Vì dd có 3 axit có thể tích bằng nhau nên: n_H⁺ = 0,1.0,1.2 + 0,1.0,2 + 0,1.0,3 = 0,07 mol
nOH^- = 0,2V + 2.0,1V = 0,4V mol.
Ta có: H⁺ + OH^- → H2O
Dung dịch C có pH = 1 => H⁺ dư
[H⁺]dư = \(\dfrac{0,07-0,4V}{0,3+C}\)= 10^-1

=> V = 0,08 lít

=> Chọn B