CH4+Cl2-as->CH3Cl+HCl

CaC2+2H2O->Ca(OH)2+C2H2

C2H2+5\2O2-to>2CO2+H2O

C2H2+2H2-xt->C2H6

bài 5

C2H5OH+3O2-to>2CO2+3H2O

0,5------------1,5--------1 mol

n C2H5OH=0,5 mol

=>VCo2=1,5.22,4=33,6l

b)Vkk=1,5.22,4.5=168l

:))))

Bạn nên chịu khó gõ đề ra khả năng được giúp sẽ cao hơn.

Câu h của em đây nhé

h, ( 1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1 - \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\))

= \(\dfrac{\sqrt{3}-1+3-\sqrt{3}}{\sqrt{3}-1}\).\(\dfrac{\sqrt{3}+1-3-\sqrt{3}}{\sqrt{3}+1}\)

= \(\dfrac{2}{\sqrt{3}-1}\).\(\dfrac{-2}{\sqrt{3}+1}\)

= \(\dfrac{-4}{2}\)

= -2

\(\text{#}HaimeeOkk\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)

\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)

\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)

\(A=1-\dfrac{1}{2020}\)

\(A=\dfrac{2019}{2020}\)

Vậy \(A=\dfrac{2019}{2020}\)

nNa2SO4= 9,94/142=0,07(mol);

mBa(OH)2= 20,52(g) -> nBa(OH)2=0,12(mol)

PTHH: Na2SO4 + Ba(OH)2 -> BaSO4 + 2 NaOH

Ta cps: 0,07/1 < 0,12/1

=> Ba(OH)2 dư, Na2SO4  hết, tính theo nNa2SO4.

-> nBaSO4=nNa2SO4= 0,07(mol)

=> m(kết tủa)=mBaSO4=0,07.233=16,31(g)

=>m=16,31(g)

b) Dung dịch A thu được bao gồm NaOH và Ba(OH)2 dư.

nNaOH=2.0,07=0,14(mol) => mNaOH= 0,14.40=5,6(g)

nBa(OH)2 (dư)=0,12-0,07=0,05(mol)

=> mBa(OH)2 (dư)= 0,05.171=8,55(g)

=> mddA=Na2SO4 + mddBa(OH)2 - mBaSO4 = 9,94+ 100 - 16,31= 93,63(g)

=> C%ddBa(OH)2 (dư)= (8,55/93,63).100=9,132%

C%ddNaOH= (5,6/93,63).100=5,981%

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