Gọi cthc: XO₃

Theo gt: \(\dfrac{M_X}{M_X+16.3}=40\%\)

\(\Rightarrow M_X=32\)

X là lưu huỳnh

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Mg + 2HCl -> MgCl2 + H2 (1)

2Al + 6HCl -> 2AlCl3 + 3H2 (2)

Fe + 2HCl -> FeCl2 + H2 (3)

(1) ( 2)(3) => \(n_{HCl}=2n_{H_2}=0,6.2=1,2\left(mol\right)\)

\(m=25,12+1,2.36,5-0,6.2=67,72\left(g\right)\)

Đáp án A.

Cho hỗn hợp K và Na vào nước thu được dd A và có khí bay ra:

ta có : \(n_{H_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)

\(H_2O\rightarrow OH^-+H^+\)

0,375--- 0,375 ---- 0,375

Cho dung dịch A vào AlCl₃

Pt: \(3OH^-+Al^{3+}\rightarrow Al\left(OH\right)_3\)

0,375 ------------------ 0,125

\(m_{ktua}=0,125.78=9,75\left(g\right)\)

Lần đầu giải ion không biết đúng không ' '

í lộn friendly

friendliness

\(\dfrac{3\left(2x+1\right)}{4}-3=\dfrac{2\left(3x+1\right)}{5}+\dfrac{3x+2}{10}\)

\(\Leftrightarrow15\left(2x+1\right)-60=8\left(3x+1\right)+2\left(3x+2\right)\)

\(\Leftrightarrow30x+15-60=24x+8+6x+4\)

\(\Leftrightarrow0x=57\)

=> x vô nghiệm

9. Please turn ___on___ the gas, I want to cook my lunch. (over / back / on / in)

10. I like bananas, ___but___ my brother doesn’t. (because / but / and / even though)

11. I shall go __because____ you stay here. (and / because / however / although)

12. Jane will be admitted to the university __even though____ she has bad grades. (because / even though / because of / but)

13. _Despite_____ his physica l handicap, he has become a successful businessman. (Although / Despite / Because of / Because)

14. She was absent _because_____ her cold was worse. (because / or / because of / so)

15. The examination was long _and_____ difficult. (or / and / but / however)

16. Joh n was ill, __so____ he went to bed early. (but / so / because of / however)

17. I’d love to play the table tennis ___but___ I have to complete an assignment. (and / so / but / or)

CMR: Với mọi số tự nhiên n thì n^2 chia 5 dư 0 hoặc 1 hoặc 4.

a) \(\left|x+\dfrac{1}{4}\right|-4=-2\)

\(\Rightarrow\left|x+\dfrac{1}{4}\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=2\\x+\dfrac{1}{4}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

b) \(-\dfrac{15}{12}x+\dfrac{3}{7}=\dfrac{6}{5}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{-49}{20}x=\dfrac{-13}{14}\)

\(\Rightarrow x=\dfrac{130}{343}\)