1.

\(V=22,4.\left(0,2+0,15\right)=7,84\left(l\right)\)

2.

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

3.

a, \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\)

\(\Rightarrow V_{CO_2}=22,4.0,1=2,24\left(l\right)\)

b, \(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\)

\(\Rightarrow V_{O_2}=22,4.0,2=4,48\left(l\right)\)

Câu 1: D

Câu 2: C

Câu 3: C

Câu 4: D

Câu 5: A

 1: D

 2: C

 3: C

 4: D

 5: A

C

1, B
2, A
3, C
4, B
5, C
6, B

Bài 4:

Vì  \(\left\{{}\begin{matrix}AB=AC\\BM=MC\\AM.chung\end{matrix}\right.\) nên \(\Delta AMB=\Delta AMC\left(c.c.c\right)\)

Do đó \(\left\{{}\begin{matrix}\widehat{AMB}=\widehat{AMC}\\\widehat{MAB}=\widehat{MAC}\end{matrix}\right.\)

Mà \(\widehat{AMB}+\widehat{AMC}=180^0\) nên \(\widehat{AMB}=\widehat{AMC}=90^0\) hay AM⊥BC

\(\widehat{MAB}=\widehat{MAC}\) (cm trên) nên AM là pg góc BAC

a) \(\left|x-3\right|\ge0\)

\(\Rightarrow\left|x-3\right|+2021\ge2021\)

\(\Rightarrow A\ge\dfrac{2020}{2021}\)

Dấu = xảy ra khi x = 3

Vậy:............

b) Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{3}{5}\right)^2\ge0\\\left|y+1\right|\ge0\end{matrix}\right.\)

PT => \(\left\{{}\begin{matrix}x=\dfrac{3}{5}\\y=-1\end{matrix}\right.\)

Chụp thêm câu b,c với d đi

B