Ta có: \(\widehat{DIB}=\widehat{IBC}\)

\(\widehat{DBT}=\widehat{TBC}\)

Do đó: \(\widehat{DIB}=\widehat{DBT}\)

\(n_{Fe_2O_3}=\dfrac{4}{160}=0,025\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

Mol:     0,025        0,075           0,025

\(m_{ddH_2SO_4}=\dfrac{0,075.98.100}{9,8}=75\left(g\right)\)

m_{dd sau pứ} = 4 + 75 = 79 (g)

\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,025.400.100\%}{79}=12,66\%\)

giúp mình với ạ

mình không thấy rõ bạn chụp lại được không

II.

1.enjoyable

2.permissble

3.information

4.sickness

5.occupiable(ko chắc)

6.anxiously

7.mountainous

8.convenience

1.enjoyable

2.permissble

3.information

4.stickness

5.??????

6.axiously

7.mountainous

8.convenience

Lộn giúp em 4 câu gấp đc ko mn oi=_)

4D

D

a) Ta có: 36-x=44

nên x=36-44=-8

Vậy: x=-8

b) Ta có: \(\dfrac{5}{6}-x=\dfrac{7}{4}-0.75-3\)

\(\Leftrightarrow\dfrac{5}{6}-x=\dfrac{7}{4}-\dfrac{3}{4}-3=1-3=-2\)

hay \(x=\dfrac{5}{6}+2=\dfrac{17}{6}\)

a) 36-x=44                                  b) 5/6-x=7/4-0,75-3

=> x= 36-44                                => 5/6-x= -2

=> x= -8                                      => x= 5/6-(-2)

                                                    => x=17/6

1.

\(y'=\left(cos^2\left(2x+3\right)\right)'=2cos\left(2x+3\right).\left(cos\left(2x+3\right)\right)'\)

\(=2cos\left(2x+3\right).\left(-sin\left(2x+3\right)\right).\left(2x+3\right)'\)

\(=-4sin\left(2x+3\right).cos\left(2x+3\right)\)

\(=-4sin\left(4x+6\right)\)

2.

\(f'\left(x\right)=-x^2+\left(3m-2\right)x-\left(2m^2-5m-2\right)\)

Để \(f'\left(x\right)< 0;\forall x\in R\)

\(\Leftrightarrow\Delta=\left(3m-2\right)^2-4\left(2m^2-5m-2\right)< 0\)

\(\Leftrightarrow m^2+8m+12< 0\)

\(\Rightarrow-6< m< -2\)