1)goi A là kim loai trong oxit

x la hóa tri cua A

CTC: A₂O_x doi: 800ml = 0,8l

\(n_{HNO_{3_{ }}}=0,8.3=2,4\left(mol\right)\)

\(m_{HNO_3}=2,4.63=151,2g\)

A₂O_x + 2xHNO₃ \(\rightarrow\) 2A(NO₃)_x + xH₂O

pt:2A+16x 126x (g)

de: 64 151,2 (g)

ta co: 151,2.(2A+16x) = 126x.64

\(\Leftrightarrow302,4A+2419,2x=8064x\)

\(\Leftrightarrow302,4A=5644,8x\)

\(\Leftrightarrow A=\dfrac{5644,8x}{302,4}=\dfrac{56x}{3}\)

bien luan:

\(+x=1\Rightarrow A=\dfrac{56}{3}\left(loai\right)\)

\(+x=2\Rightarrow A=\dfrac{112}{3}\left(loai\right)\)

\(+x=3\Rightarrow A=56\left(lay\right)\)

vậy CT oxit la: Fe₂O₃

1. \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)

\(=\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}\)

\(=\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}\)

\(=\sqrt{11}+1-\sqrt{11}+1=2\)

2.a)\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{5-2.\sqrt{5}.2+2^2}-\sqrt{5}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}=-2\)

b)\(\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{2+2\sqrt{2}+1}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)

\(=\sqrt{2}-1-\sqrt{2}-1=-2\)

c)\(\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)

\(=\sqrt{9-2\sqrt{2}.3+2}+3+\sqrt{2}\)

\(=\sqrt{\left(3-\sqrt{2}\right)^2}+3+\sqrt{2}\)

\(=3-\sqrt{2}+3+\sqrt{2}=6\)

d)\(\sqrt{7-2\sqrt{6}}+\sqrt{7+2\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}+1\right)^2}\)

\(=\sqrt{6}-1+\sqrt{6}+1=2\sqrt{6}\)

gọi x la so mol cua HCl

y la so mol cua HNO₃

n_CaO= \(\dfrac{11,2}{56}=0,2\left(mol\right)\)

CaO + 2HCl \(\rightarrow\) CaCl₂ + H₂O (1)

de: \(\dfrac{x}{2}\) \(\leftarrow\) x \(\rightarrow\) \(\dfrac{x}{2}\rightarrow\) \(\dfrac{x}{2}\)

CaO + 2HNO₃ \(\rightarrow\) Ca(NO₃)₂ + H₂O (2)

de: \(\dfrac{y}{2}\leftarrow\) y \(\rightarrow\dfrac{y}{2}\) \(\rightarrow\dfrac{y}{2}\)

Ta co: C_{M HCl} = \(\dfrac{n_{HCl}}{V}=0,2M\)

C_{M HNO3} = \(\dfrac{n_{HNO_3}}{V}=0,2M\)

\(\Rightarrow\dfrac{n_{HCl}}{V}=\dfrac{n_{HNO_3}}{V}\Rightarrow n_{HCl}=n_{HNO_3}\)

\(\Rightarrow x=y\)

Ta co: \(\dfrac{x}{2}+\dfrac{y}{2}=0,2\Rightarrow\dfrac{2x}{2}=0,2\)

\(\Rightarrow x=y=0,2\left(mol\right)\)

Ta co: \(\Rightarrow\dfrac{n_{HCl}}{V}+\dfrac{n_{HNO_3}}{V}=0,4\)

\(\Rightarrow V=\dfrac{2x}{0,4}=1l\)

\(m_{CaCl_2}=0,2.111=22,2g\)

\(m_{Ca\left(NO_3\right)_2}=0,2.160=32g\)

a.18cm²

đề của bn bị thiếu vận tốc lúc về

goi x la thu nhat. DK: 0<x<120

khi do: so thu hai la: 120 - x

theo de bai ta co pt:120 - x = 3x

\(\Leftrightarrow4x=120\Leftrightarrow x=30\left(TMDK\right)\)

so thu hai la: 120-30 = 90

vay so thu nhat la:30

so thu hai la:90

a, xet \(\Delta BFC\) vuong tai F va \(\Delta CEB\) vuong tai E

BC chung; \(\widehat{FBC}=\widehat{ECB}\left(\Delta ABCcan\right)\)

\(\Rightarrow\Delta BFC=\Delta CEB\left(ch-gn\right)\)

\(\Rightarrow BF=CE\)

b, Ta co: AF= AB- BF va AE= AC- EC

ma AB=AC (ABC can); BF=EC(cmt)

\(\Rightarrow AF=AE\Rightarrow\dfrac{AF}{AB}=\dfrac{AE}{AC}\)

=> EF//BC(d/li Talet dao)

c, ke duong cao AD, vi tg ABC can tai A nen AD vua la duong cao vua la duong trung truc

\(BD=DC=\dfrac{1}{2}BC=3cm\)

xet \(\Delta BEC\) va \(\Delta ADC\)

\(\widehat{C}chung;\widehat{ADC}=\widehat{BEC}=90\)

\(\Rightarrow\Delta BEC\infty\Delta ADC\Rightarrow\dfrac{AC}{BC}=\dfrac{DC}{EC}\)

\(\Rightarrow EC=\dfrac{BC.DC}{AC}=2cm\)

Ta co: AF=AE= AC-EC = 7cm

xet \(\Delta AFE\) va \(\Delta ABC\)

\(\dfrac{AF}{AB}=\dfrac{AE}{AC};\widehat{A}chung\)

\(\Rightarrow\Delta AFE\infty\Delta ABC\Rightarrow\dfrac{AF}{AB}=\dfrac{EF}{BC}\)

\(\Rightarrow EF=\dfrac{AF.BC}{AB}=\dfrac{14}{3}\approx4,67cm\)