Fe+S-to>FeS

n Fe=0,1 mol

n S=0,025 mol

=>Fe dư

Fe+2HCl->Fecl2+H2

0,075-------------------0,075

FeS+2HCl->FeCl2+H2S

0,025-----------------------0,025

=>A là H2, Blaf H2S

20A 21C 22B

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2. She drives cautiously

1. handy

2. She drives cautiously

Chúc bạn học tốt ^^ !!!

21.

\(\left\{{}\begin{matrix}SA\perp AB\\AC\perp AB\end{matrix}\right.\) \(\Rightarrow AB\perp\left(SAC\right)\)

E là trung điểm SA, F là trung điểm SB \(\Rightarrow\) EF là đường trung bình tam giác SAB

\(\Rightarrow EF||AB\Rightarrow EF\perp\left(SAC\right)\)

\(\Rightarrow EF=d\left(F;\left(SEK\right)\right)\)

\(SE=\dfrac{1}{2}SA=\dfrac{3a}{2}\) ; \(EF=\dfrac{1}{2}AB=a\)

 \(SC=\sqrt{SA^2+AC^2}=a\sqrt{13}\Rightarrow SK=\dfrac{2}{3}SC=\dfrac{2a\sqrt{13}}{3}\)

\(\Rightarrow S_{SEK}=\dfrac{1}{2}SE.SK.sin\widehat{ASC}=\dfrac{1}{2}.\dfrac{3a}{2}.\dfrac{2a\sqrt{13}}{3}.\dfrac{2a}{a\sqrt{13}}=a^2\)

\(\Rightarrow V_{S.EFK}=\dfrac{1}{3}EF.S_{SEK}=\dfrac{1}{3}.a.a^2=\dfrac{a^3}{3}\)

\(AB\perp\left(SAC\right)\Rightarrow AB\perp\left(SEK\right)\Rightarrow AB=d\left(B;\left(SEK\right)\right)\)

\(\Rightarrow V_{S.EBK}=\dfrac{1}{3}AB.S_{SEK}=\dfrac{1}{3}.2a.a^2=\dfrac{2a^3}{3}\)

22.

Gọi D là trung điểm AB

Do tam giác ABC đều \(\Rightarrow CD\perp AB\Rightarrow CD\perp\left(SAB\right)\)

\(\Rightarrow CD=d\left(C;\left(SAB\right)\right)\)

\(CD=\dfrac{AB\sqrt{3}}{2}=a\sqrt{3}\) (trung tuyến tam giác đều)

N là trung điểm SC \(\Rightarrow d\left(N;\left(SAB\right)\right)=\dfrac{1}{2}d\left(C;\left(SAB\right)\right)=\dfrac{a\sqrt{3}}{2}\)

\(S_{SAB}=\dfrac{1}{2}SA.AB=a^2\sqrt{3}\) \(\Rightarrow S_{SAM}=\dfrac{1}{2}S_{SAB}=\dfrac{a^2\sqrt{3}}{2}\)

\(\Rightarrow V_{SAMN}=\dfrac{1}{3}.\dfrac{a\sqrt{3}}{2}.\dfrac{a^2\sqrt{3}}{2}=\dfrac{a^3}{4}\)

Lại có:

\(V_{SABC}=\dfrac{1}{3}SA.S_{ABC}=\dfrac{1}{3}.a\sqrt{3}.\dfrac{\left(2a\right)^2\sqrt{3}}{4}=a^3\)

\(\Rightarrow V_{A.BCMN}=V_{SABC}-V_{SANM}=\dfrac{3a^3}{4}\)

Hình vẽ câu 21:

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{x+2017-\left(2015-x\right)}{\sqrt[3]{\left(x+2017\right)^2}+\sqrt[3]{\left(x+2017\right)\left(2015-x\right)}+\sqrt[3]{\left(2015-x\right)^2}}}{\dfrac{2000+x-\left(1998-x\right)}{\sqrt{2000+x}+\sqrt{1998-x}}}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{2000+x}+\sqrt{1998-x}}{\sqrt[3]{\left(x+2017\right)^2}+\sqrt[3]{\left(x+2017\right)\left(2015-x\right)}+\sqrt[3]{\left(2015-x\right)^2}}\)

\(=\dfrac{\sqrt{1999}+\sqrt{1999}}{\sqrt[3]{2016^2}+\sqrt[3]{2016^2}+\sqrt[3]{2016^2}}=\dfrac{2\sqrt{1999}}{3.24\sqrt[3]{294}}=\dfrac{\sqrt{1999}}{36\sqrt[3]{294}}\)

\(\Rightarrow a+b=1999+294\)