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, V_CO2 = 0,5.22,4 = 11,2 (l)

c, m_CH3COOK = 1.98 = 98 (g)

Bạn tham khảo nhé!

a) PTHH:         2CH₃COOH + K₂CO₃ → 2CH₃COOK + CO₂↑ + H₂O

b) n_CH3COOH = \(\dfrac{m}{M}=\dfrac{60}{60}=1\left(mol\right)\)

Theo PTHH: n_CO2 = \(\dfrac{1}{2}\).n_CH3COOH = 0,5 (mol)

=> V_{CO2 (đktc)} = n.22,4 = 0,5.22,4 = 11,2 (lít)

c) Theo PTHH: n_CH3COOK = n_CH3COOH = 1 (mol)

=> m_CH3COOK = 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:

$4x-6=2x+4$

$\Leftrightarrow (4x-6)-(2x+4)=0$

$\Leftrightarrow 2x-10=0$

$\Leftrightarrow 2x=10$

$\Leftrightarrow x=5$

\(4x-6=2x+4\\ \Leftrightarrow4x-2x=4+6\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\)

Vậy phương trình có nghiệm là: \(x=5\)

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