1C 2D 3B

có 4 hinhg thang cân

Hình thang cân:4

Hình chữ nhật:5

Hình vuông:4

a) Zn + 2HCl $\to$ ZnCl2 + H2

b) 

n H2 = n Zn = 9,75/65 = 0,15(mol)

V H2 = 0,15.22,4 = 3,36(lít)

n HCl = 2n H2 = 0,3 mol

=> m HCl = 0,3.36,5 = 10,95 gam

c) n CuO = 24/80 = 0,3(mol)

CuO + H2 $\xrightarrow{t^o}$ Cu + H2O

n CuO = 0,3 > n H2 = 0,15 => CuO dư

n Cu = n H2 = 0,15 mol

=> m Cu = 0,15.64 = 9,6 gam

Câu 3 : 

\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.15.......0.3..................0.15\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(n_{CuO}=\dfrac{24}{80}=0.3\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(1..........1\)

\(0.3........0.15\)

\(LTL:\dfrac{0.3}{1}>\dfrac{0.15}{1}\Rightarrow CuOdư\)

\(n_{Cu}=n_{H_2}=0.15\left(mol\right)\)

\(m_{Cu}=0.15\cdot64=9.6\left(g\right)\)

a: A=[-5;4]

B=[7;14)

C=(2;+\(\infty\))

2. C

3. D

4. A

b: Ta có: \(\sqrt{x-2}=4-x\)

\(\Leftrightarrow x^2-8x+16-x+2=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-6\right)=0\)

hay x=3

6, \(\Rightarrow-2\left(x-3\right)-8=5\left(x+2\right)\Leftrightarrow-2x-2=5x+10\)

\(\Leftrightarrow7x=-12\Leftrightarrow x=-\dfrac{12}{7}\)

7, \(\Rightarrow6\left(2x+1\right)-5\left(x+6\right)=5-4x\Leftrightarrow7x-24=5-4x\)

\(\Leftrightarrow11x=29\Leftrightarrow x=\dfrac{29}{11}\)

8, \(\Rightarrow35-15x-10-2x=10\Leftrightarrow25-17x=10\Leftrightarrow x=\dfrac{15}{17}\)

6: \(\Leftrightarrow-2\left(x-3\right)-8=5x+10\)

=>5x+10=-2x+6-8

=>5x+10=-2x-2

=>7x=-12

hay x=-12/7

7: \(\Leftrightarrow6\left(2x+1\right)-5\left(x+6\right)=5-4x\)

=>12x+6-5x-30-5+4x=0

=>11x-29=0

hay x=29/11

8: \(\Leftrightarrow5\left(7-3x\right)-2\left(x+5\right)=10\)

=>35-15x-2x-10=10

=>-17x=-15

hay x=15/17

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

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

\(-4x-4=10x+20\)

\(-14x=24\)

\(x=\dfrac{24}{-14}=-\dfrac{12}{7}\)