Câu 4:

a) nC2H6O=0,3(mol)

PTHH: C2H6O + 3 O2 -to-> 2 CO2 + 3 H2O

0,3___________0,9_____0,6(mol)

=>V(CO2,đktc)=0,6 x 22,4= 13,44(l)

b) V(kk,dktc)=V(O2,dktc) . 100/20 = (0,9.22,4).5=100,8(l)

Câu 5:

C2H6O + 3 O2 -to-> 2 CO2 + 3 H2O

nH2O=0,9(mol)

=> nCO2= 2/3. 0,9=0,6(mol)

a) V(CO2,đktc)=0,6.22,4=13,44(l)

b) Vkk=5.V(O2,dktc)= 5.(0,9.22,4)= 100,8(l)

Em cần hỗ trợ cụ thể bài nào em?

(1)  S + O₂ --t^o--> SO₂

(2) 2SO₂ + O₂ --t^o, V₂O₅--> 2SO₃

(3) SO₃ + H₂O --> H₂SO₄

(4) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

(1)\(S+O_2\rightarrow\left(t^o\right)SO_2\)

(2)\(2SO_2+O_2\rightarrow\left(t^o,V_2O_5\right)2SO_3\)

(3)\(SO_3+H_2O\rightarrow H_2SO_4\)

(4)\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

S+O₂----t^o---> SO₂

SO₂+2NO---->SO₃₊N₂O

SO₃+H₂O-----> H₂SO₄

2Al+3H₂SO₄---->Al₂(SO₄)₃+3H₂

Kẻ AH⊥BC

ta có: \(VP=AB^2+BC^2-2.AB.BC.cosB=AB^2+BC^2-2.AB.BC.\dfrac{BH}{AB}=AB^2+BC^2-2.BH.BC=AB^2-BH^2+BC^2-2.BH.BC+BH^2=AH^2+\left(BC-BH\right)^2=AH^2+CH^2=AC^2=VT\)

She does not go to school at 8 o'clock
He does not have dinner at 9 o'clock
She does not go to bed at 7 o'clock
He does not go home at 4 o'clock
He does not go to bed at 9 o'clock