Vì Om là phân giác của \(\widehat{xOy}\)

\(\Rightarrow\widehat{IOE}=\widehat{IOF}=\dfrac{1}{2}\widehat{EOF}\)

Vì \(\left\{{}\begin{matrix}IE\perp Ox\\IF\perp Oy\end{matrix}\right.\left(gt\right)\Rightarrow\widehat{IEO}=\widehat{IFO}=90^o\)

Xét \(\Delta IOE\) và \(\Delta IOF\) có: \(\left\{{}\begin{matrix}\widehat{IEO}=\widehat{IFO}\left(=90^o\right)\\OI:chung\\\widehat{IOE}=\widehat{IOF}\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta IOE=\Delta IOF\left(\text{cạnh huyền - góc nhọn}\right)\)

b) Vì \(\Delta IOE=\Delta IOF\left(cmt\right)\Rightarrow OE=OF\left(\text{2 cạnh tương ứng}\right)\)

Xét \(\Delta EOF\) có: \(OE=OF\left(cmt\right)\)

\(\Rightarrow\Delta EOF\) cân ở O

\(\Rightarrow\widehat{OEF}=\widehat{OFE}\)

Xét \(\Delta EOF\) có:

\(\widehat{EOF}+\widehat{OFE}+\widehat{OEF}=180^o\)

\(\Rightarrow2\widehat{EOI}+2\widehat{OEF}=180^o\\ \Rightarrow\widehat{EOI}+\widehat{OEF}=90^o\)

Gọi \(EF\cap OI\equiv M\)

Xét \(\Delta OME\) có: 

\(\widehat{OEF}+\widehat{EOI}+\widehat{OME}=180^o\\ \Rightarrow90^o+\widehat{OME}=180^o\\ \Rightarrow\widehat{OME}=180^o-90^o=90^o\\ \Rightarrow EF\perp Om\left(\text{đpcm}\right)\)

a)

Xét \(\Delta AOD\) và \(\Delta COB\) có: \(\left\{{}\begin{matrix}OA=OC\left(gt\right)\\\widehat{O}:chung\\OB=OD\left(gt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta AOD=\Delta COB\left(c.g.c\right)\)

\(\Rightarrow AD=BC\left(\text{2 cạnh tương ứng}\right)\left(\text{đpcm}\right)\)

b) 

Nối A với C

Ta có: \(\left\{{}\begin{matrix}OA=OC\\OB=OD\end{matrix}\right.\left(gt\right)\Rightarrow OA-OB=OC-OD\)

Hay \(AB=CD\)

Xét \(\Delta ABC\) và \(\Delta CDA\) có: \(\left\{{}\begin{matrix}AB=CD\left(cmt\right)\\AC:chung\\AD=BC\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta ABC=\Delta DCA\left(c.c.c\right)\)

\(\Rightarrow\widehat{ABC}=\widehat{CDA}\left(\text{2 góc tương ứng}\right)\)

Vì \(\Delta AOD=\Delta COB\left(cmt\right)\Rightarrow\widehat{A}=\widehat{C}\left(\text{2 góc tương ứng}\right)\)

Xét \(\Delta ABE\) và \(\Delta CDE\) có: \(\left\{{}\begin{matrix}\widehat{ABC}=\widehat{CDA}\left(cmt\right)\\AB=CD\left(cmt\right)\\\widehat{A}=\widehat{C}\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta ABE=\Delta CDE\left(g.c.g\right)\left(\text{đpcm}\right)\)

c) Vì \(\Delta ABE=\Delta CDE\left(cmt\right)\Rightarrow AE=CE\left(\text{2 cạnh tương ứng}\right)\)

Xét \(\Delta AOE\) và \(\Delta COE\) có: \(\left\{{}\begin{matrix}OA=OC\left(gt\right)\\\widehat{A}=\widehat{C}\left(cmt\right)\\AE=CE\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta AOE=\Delta COE\left(c.g.c\right)\\ \Rightarrow\widehat{AOE}=\widehat{COE}\left(\text{2 góc tương ứng}\right)\)

`=> OE` là phân giác của \(\widehat{xOy}\) (đpcm)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH:
\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\\ 2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=2.\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)

a) \(n_{C_2H_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)

           0,3---->0,9------>0,6

\(\Rightarrow V_{kk}=\dfrac{0,9.22,4}{20\%}=100,8\left(l\right)\)

b) \(V_{CO_2}=0,6.22,4=13,44\left(l\right)\)

c) \(n_P=\dfrac{15,5}{31}=0,5\left(mol\right)\)

PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)

Xét tỉ lệ: \(\dfrac{0,5}{4}< \dfrac{0,9}{5}\Rightarrow\) O₂ dư, P cháy hết

a) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

            0,3--->0,2----->0,1

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

b) \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)

c) \(n_{O_2\left(hao,h\text{ụt}\right)}=0,2.10\%=0,02\left(mol\right)\)

\(\Rightarrow n_{O_2\left(t\text{ổng}\right)}=0,2+0,02=0,22\left(mol\right)\)

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

              0,44<------------------------------------0,22

\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)

\(V_{C_2H_5OH}=\dfrac{600.50}{100}=300\left(ml\right)\\ \Rightarrow V_{ddC_2H_5OH\left(25^o\right)}=\dfrac{300.100}{25}=1200\left(ml\right)\)