a) PTHH: Fe + H₂SO₄ \(\rightarrow\) FeSO₄ + H₂\(\uparrow\)

Thep pt: 1mol....1mol........1mol......1mol

Theo đề: 0,2mol...0,2mol......0,2mol....0,2mol

b) ⁿFe = \(\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(\Rightarrow\)^mFeSO4 = n.M = 0,2.152 = 30,4 (g)

c) ^mH2SO4 = n.M = 0,2.98 = 19,6 (g)

\(\Rightarrow\)^C%ddH₂SO_4(l) = \(\dfrac{19,6}{100}.100\%=19,6\%\)

d) ^VH_2(đktc) = n.22,4 = 0,2.22,4 = 4,48 (l)

e) ^mH₂ = n.M = 0,2.2 = 0,4 (g)

^mddspứ = ^mFe + ^mddH₂SO₄ + ^mH₂ = 11,2 + 100 + 0,4 = 111,6 (g)

\(\Rightarrow\)Dung dịch sau phản ứng chỉ gồm FeSO₄

_{\(\Rightarrow\)}C% = \(\dfrac{30,4}{111,6}.100\%=27,24\%\)

10. My sister speaks French fluently.

=> My sister is a fluent French speaker.

11. The black car is cheaper than the red car.

=> The red car is more expensive than the black car.

12. My house is the oldest house on the streer.

=> No house on the street is older than my house.

13. Having a vacation abroad is very interesting.

=> It's very interesting to have a vacation abroad.

14. How much does it cost to mail a letter to the USA?

=> What is the price of mailing a letter to the USA?

15. It takes us about half an hour to walk to school.

=> We spend about half an hour walking to school.

16. Do you have a cheaper computer than this?

=> Is this the cheapest computer you have?

17. There are over eight hundred stamps in Tim's collection.

=> Tim's collection has over eight hundred stamps.

18. What is the price of this bag?

=> How much is this bag?/ does this bag cost?

19. How heavy is your son?

=> What is the heaviness of your son?

20. You shouldn't eat too much candies.

=> You ought not to eat too much candies.

Viết lại câu nghĩa không đổi

1. They are frequent church goers.

=> They goes to the church frequently.

2. She is a fluent speaker.

=> She speaks fluently.

3. Nam is a hard worker.

=> Nam works hard.

4. Lan is a good singer.

=> Lan sings well.

5. They are fast runners.

=> They run fast.

Câu 6.

a) \(6x-3=4x+5\)

\(\Leftrightarrow6x-4x=5+3\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\)

Vậy \(S=\left\{4\right\}\)

b) \(\dfrac{2x+3}{x+1}-\dfrac{6}{x}=2\) (ĐK: \(x\ne-1\) )

\(\Leftrightarrow\dfrac{x\left(2x+3\right)-6\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)

\(\Leftrightarrow x\left(2x+3\right)-6\left(x+1\right)=2x\left(x+1\right)\)

\(\Leftrightarrow2x^2+3x-6x-6=2x^2+2x\)

\(\Leftrightarrow2x^2-2x^2+3x-6x-2x=6\)

\(\Leftrightarrow-5x=6\)

\(\Leftrightarrow x=-\dfrac{6}{5}\) (thỏa mãn)

Vậy \(S=\left\{-\dfrac{6}{5}\right\}\)

c) \(\left|3x-1\right|=3x\) (1)

TH1: \(3x-1\ge0\Leftrightarrow x\ge\dfrac{1}{3}\)

Khi đó: \(\left|3x-1\right|=3x-1\)

(1) \(\Leftrightarrow3x-1=3x\)

\(\Leftrightarrow3x-3x=1\)

\(\Leftrightarrow0=1\) (vô lí => loại)

TH2: \(3x-1< 0\Leftrightarrow x< \dfrac{1}{3}\)

Khi đó: \(\left|3x-1\right|=-3x+1\)

(1) \(\Leftrightarrow-3x+1=3x\)

\(\Leftrightarrow-3x-3x=-1\)

\(\Leftrightarrow-6x=-1\)

\(\Leftrightarrow x=\dfrac{1}{6}\) (thỏa mãn => nhận)

Vậy \(S=\left\{\dfrac{1}{6}\right\}\)

Câu 7.

a) \(\dfrac{x-4}{6}+\dfrac{1}{2}>\dfrac{2x-5}{3}\)

\(\Leftrightarrow\dfrac{x-4+3}{6}>\dfrac{2\left(2x-5\right)}{6}\)

\(\Leftrightarrow x-4+3>2\left(2x-5\right)\)

\(\Leftrightarrow x-1>4x-10\)

\(\Leftrightarrow x-4x>-10+1\)

\(\Leftrightarrow-3x>-9\)

\(\Leftrightarrow x< 3\)

Vậy \(S=\left\{x|x< 3\right\}\)

b) \(\dfrac{2}{3-x}< 0\) (ĐK: \(x\ne3\) )

\(\Leftrightarrow2.\dfrac{1}{3-x}< 0\)

\(\Leftrightarrow\dfrac{1}{3-x}< 0\)

\(\Leftrightarrow\dfrac{1.\left(3-x\right)}{\left(3-x\right)^2}< \dfrac{0}{\left(3-x\right)^2}\)

\(\Leftrightarrow3-x< 0\)

\(\Leftrightarrow-x< -3\)

\(\Leftrightarrow x>3\) (nhận)

Vậy \(S=\left\{x|x>3\right\}\)

Câu 8.

Gọi độ dài quãng đường AB là x (km) \(\left(x\ne0\right)\)

Thời gian người đó đi xe máy từ A đến B là \(\dfrac{x}{30}\left(h\right)\)

Thời gian người đó đi xe máy từ B về A là \(\dfrac{x}{35}\left(h\right)\)

Đổi: 30 phút = \(\dfrac{1}{2}h\)

Ta có phương trình: \(\dfrac{x}{30}-\dfrac{x}{35}=\dfrac{1}{2}\)

Giải phương trình, ta được x = 105

Vậy độ dài quãng đường AB là 105km.

Bài 1:

a) PTHH: CuO + H₂SO₄ \(\rightarrow\) CuSO₄ + H₂O

Theo pt: ..... 1 ......... 1 .............. 1 ......... 1 ... (mol)

Theo đề: .. 0,04 .... 0,04 ......... 0,04 .... 0,04 . (mol)

b) \(n_{CuO}=\dfrac{m}{M}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{m_{ddH_2SO_4}.C\%}{100\%.M}=\dfrac{200.20\%}{100\%.98}=\dfrac{20}{49}\left(mol\right)\)

So sánh \(\dfrac{n_{CuO_{đề}}}{n_{CuO_{pt}}}< \dfrac{n_{H_2SO_{4_{đề}}}}{n_{H_2SO_{4_{pt}}}}\left(0,04< \dfrac{20}{49}\right)\)

=> H₂SO₄ dư, tính theo ⁿCuO

=> Dung dịch thu được sau phản ứng gồm H₂SO₄ dư và CuSO₄.

c) \(m_{CuSO_4}=n.M=0,04.160=6,4\left(g\right)\)

d) \(m_{H_2SO_{4_{dư}}}=n.M=\left(\dfrac{20}{49}-0,04\right).98=36,08\left(g\right)\)

\(m_{ddspứ}=m_{CuO}+m_{ddH_2SO_4}=3,2+200=203,2\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{m_{CuSO_4}}{m_{ddspứ}}.100\%=\dfrac{6,4}{203,2}.100\%\approx3,15\%\)

\(C\%_{H_2SO_{4_{dư}}}=\dfrac{m_{H_2SO_{4_{dư}}}}{m_{ddspứ}}.100\%=\dfrac{36,8}{203,2}.100\%\approx18,11\%\)

= 48 học sinh tick nha.

PTHH: CaCO₃ + 2HCl \(\rightarrow\) CaCl₂ + H₂O + CO₂\(\uparrow\) (1)

Theo pt: .... 1 ......... 2 ............ 1 ......... 1 ....... 1 .... (mol)

Theo đề: .. 0,25 .... 0,5 ........................................... (mol)

PTHH: 2HCl_dư + MgO \(\rightarrow\) MgCl₂ + H₂O (2)

Theo pt: ... 2 .......... 1 ............ 1 .......... 1 .. (mol)

Theo đề: .. 0,5 ..... 0,25 ............................. (mol)

\(n_{MgO}=\dfrac{m}{M}=\dfrac{10}{40}=0,25\left(mol\right)\)

\(m_{HCl_{dư}}=n.M=0,5.36,5=18,25\left(g\right)\)

\(m_{HCl\left(1\right)}=m_{HCl_{bđ}}-m_{HCl_{dư}}=36,5-18,25=18,25\left(g\right)\)

\(n_{HCl\left(1\right)}=\dfrac{m}{M}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

\(m_{CaCO_3}=n.M=0,25.100=25\left(g\right)\)

Vậy m = 25g

x-12 chia hết cho cái j

Bài 2: Tìm MIN

\(A=x^2-6x+5\)

\(A=x^2-2x.3+3^2-3^2+5\)

\(A=\left(x-3\right)^2-4\ge-4\)

Dấu "=" xảy ra khi \(x-3=0\Rightarrow x=3\)

Vậy MIN_A = -4 khi x = 3

\(B=x^2-x+1\)

\(B=x^2-2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)2-\left(\dfrac{1}{2}\right)^2+1\)

\(B=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)