giải giúp tớ với ạ :((
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Ta có: \(n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\)
a, PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)
_________1____________________1_________0,5 (mol)
b, VCO2 = 0,5.22,4 = 11,2 (l)
c, mCH3COOK = 1.98 = 98 (g)
Bạn tham khảo nhé!
a) PTHH: 2CH3COOH + K2CO3 → 2CH3COOK + CO2↑ + H2O
b) nCH3COOH = \(\dfrac{m}{M}=\dfrac{60}{60}=1\left(mol\right)\)
Theo PTHH: nCO2 = \(\dfrac{1}{2}\).nCH3COOH = 0,5 (mol)
=> VCO2 (đktc) = n.22,4 = 0,5.22,4 = 11,2 (lít)
c) Theo PTHH: nCH3COOK = nCH3COOH = 1 (mol)
=> mCH3COOK = n.M = 1.98 = 98 (g)
Mặc dù hơi muộn nhưng mà chúc bạn thi đạt kết quả tốt nha ^_^
\(=\left(\dfrac{\sqrt{5}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}+\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}\right):\sqrt{\dfrac{5}{2}}\)
\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{5}}{\sqrt{2}}\)
\(=2\sqrt{5}.\dfrac{\sqrt{2}}{\sqrt{5}}=2\sqrt{2}\)
a) Ta có: \(\left(\dfrac{\sqrt{15}-\sqrt{20}}{\sqrt{3}-2}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{6}+\sqrt{5}}\right):\sqrt{\dfrac{5}{2}}\)
\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{10}}{2}\)
\(=2\sqrt{5}\cdot\dfrac{2}{\sqrt{10}}=2\sqrt{2}\)
`#040911`
`b)`
\(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{19\times21}\)
`=`\(\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{19\times21}\right)\)
`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{21}\right)\)
\(=\dfrac{1}{2}\times\dfrac{20}{21}\\ =\dfrac{10}{21}\\ \text{ Vậy, A = }\dfrac{10}{21}\)
Lời giải:
$4x-6=2x+4$
$\Leftrightarrow (4x-6)-(2x+4)=0$
$\Leftrightarrow 2x-10=0$
$\Leftrightarrow 2x=10$
$\Leftrightarrow x=5$
Bài 1:
a: \(x-\dfrac{3}{7}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{2}{3}+\dfrac{3}{7}=\dfrac{14+9}{21}=\dfrac{23}{21}\)
Ta có: \(B=\sin^254^0+\sin^249^0+\sin^241^0+\sin^236^0-\dfrac{\cos23^0}{\sin67^0}-\tan18^0\cdot\tan72^0\)
\(=1+1-1-1\)
=0
VI
1 If you play sports, you'll feel better
2 If students don't take notes during the lesson, they won't understand what the teacher says
3 If people had wings, they would be able fly
4 If I were you, I wouldn't do that
5 If you try your best, you'll win
6 If there isn't sun, the flowers won't open
VII
1 that he couldn't do this test
3 that she would go to China the following week
4 if he like to played this game
5 if I had to do this work
6 if they would go to the stadium the following day