2) \(n_{CO_2}=\dfrac{0,45.4}{3}=0,6\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=\dfrac{80}{74}\approx1,08\left(mol\right)\)

CO₂ + Ca(OH)₂ \(\rightarrow\) CaCO₃ + H₂O

Ta co: \(\dfrac{0,6}{1}< \dfrac{1,08}{1}\Rightarrow Ca\left(OH\right)_2dư\)

CO₂ + Ca(OH)₂ \(\rightarrow\) CaCO₃ + H₂O

de: 0,6 1,08 0 0

pu: 0,6 0,6 0,6 0,6

sau pu: 0 0,48 0,6 0,6

\(m_{Ca\left(OH\right)_2dư}=0,48.74=35,52g\)

\(m_{CaCO_3}=0,6.100=60g\)

goi x la so mol cua Cu

y la so mol cua CuO

\(m_{H_2SO_4}=\dfrac{70.112}{100}=78,4g\)

\(n_{H_2SO_4}=\dfrac{78,4}{98}=0,8\left(mol\right)\)

Cu+2H₂SO₄(d,n)\(\underrightarrow{t^o}\)CuSO₂+2H₂O+SO₂

de: x 2x x 2x x

CuO + H₂SO₄\(\rightarrow\) CuSO₄ +H₂O

de: y y y y

Ta co: 64x + 80y = 28

2x + y = 0,8

\(\Rightarrow\left\{{}\begin{matrix}x=0,375\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(m_{Cu}=0,375.64=24g\)

\(m_{CuO}=0,05.80=4g\)

\(\%m_{Cu}=\dfrac{24}{28}.100\%\approx85,71\%\)

\(\%m_{CuO}=\dfrac{4}{28}.100\%\approx14,29\%\)

mk chỉ lam ý thu 2 thoi con ý 1 mk k hieu bn muon tinh cai j

mk sẽ sửa lại cau a

Ta co: \(\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)

\(\sqrt{11}-\sqrt{10}=\dfrac{1}{\sqrt{11}+\sqrt{10}}\)

Ta thấy: \(\sqrt{12}+\sqrt{11}>\sqrt{11}+\sqrt{10}\)

\(\Rightarrow\dfrac{1}{\sqrt{12}+\sqrt{11}}< \dfrac{1}{\sqrt{11}+\sqrt{10}}\)

\(\Rightarrow\sqrt{12}-\sqrt{11}< \sqrt{11}-\sqrt{10}\)

2,van toc cano di xuoi dong:84:7=12km/h

van toc cano di nguoc dong:63:7=9km/h

goi x(km/h) la van toc rieng cua cano

y(km/h) la van toc dong nc

DK: 9 < x < 12 , y < x , y>0

Ta co: van toc xuoi dong: x + y = 12

\(\Rightarrow x=12-y\) (1)

van toc nguoc dog: x - y = 9\(\Rightarrow x=9+y\)(2)

(1) (2) \(\Rightarrow12-y=9+y\)

\(\Leftrightarrow-2y=-3\Leftrightarrow y=1,5\)(km/h)

\(\Rightarrow x=12-1,5=10,5\)(km/h)

vậy........

(bn tự kết luận nha)

chiều dài hình chữ nhật là

 232:8=29 m

chu vi hình chữ nhật là

(29+8)x2=74 m

tick minh nha

a, \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

\(=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}-y\sqrt{y}+y\sqrt{x}+x\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{y\sqrt{x}+x\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

b, \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

Bai 1:

\(n_{Al_2O_3}=\dfrac{5,1}{102}=0,05\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

Ta co:\(\dfrac{0,05}{1}< \dfrac{0,25}{3}\Rightarrow H_2SO_4dư\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

de:0,05 0,25

PU:0,05 0,15 0,05 0,05

sau: 0 0,1 0,05 0,05

\(m_{H_2SO_4dư}=0,1.98=9,8g\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\)

nếu cho Fe vào các chất thu đc sau PU :

* 3Fe + Al₂(SO₄)₃ \(\rightarrow\) 3FeSO₄ + 2Al

(PU này k sinh ra H₂O)

* Fe + H₂SO₄(loãng)\(\rightarrow\) H₂ + FeSO₄

(PU này k sinh ra H₂O)

* 2Fe+ 6H₂SO₄(đặc,nóng)\(\rightarrow\)Fe₂(SO₄)₃ + 6H₂O + 3SO₂

\(n_{H_2O}=n_{H_2SO_4}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2O}=22,4.0,1=2,24l\)

phần thêm Fe mk k chắc là nó đúng, nên có j sai mấy bn góp ý cho mk nha

Ta co: \(2\sqrt{5}=\sqrt{20}\)

\(5=\sqrt{5^2}=\sqrt{25}\)

vì 20 < 25 \(\Rightarrow\sqrt{20}< \sqrt{25}\Rightarrow2\sqrt{5}< 5\)

cái đề bn viết k rõ ràng j hết. Mk viết lại đề nha: Phân tích thành nhân tử

a) \(3x^2-7x+2\)

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

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

b) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)​\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-24\)\(=\left(x^2+7x+11\right)^2-1-24\)​

\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

c) \(x^4+4=\left(x^2\right)^2+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-4x^2\)

\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)

\(\sqrt{3}x^2-\sqrt{12}=0\)

\(\Leftrightarrow\sqrt{3}x^2-2\sqrt{3}=0\)

\(\Leftrightarrow\sqrt{3}\left(x^2-2\right)=0\)

\(\Leftrightarrow x^2=2\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\sqrt{2};-\sqrt{2}\right\}\)