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\(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(10x=21\)
\(x=2,1\)


1 haven't cleaned
2 will be
3 is sleeping - gets
4 is teaching
5 does - isn't going - is watching
6 have written
7 hadn't disturbed
8 have you worked
9 get - get
6 studies - are trying
7 had stood
8 drove
9 have you studied
10 have waited - hasn't come
11 have lived


/ -2 / = 2x + 3
\(=>2=2x+3\)
\(=>-1=2x\)
\(=>-2x=1\)
\(=>x=\frac{-1}{2}\)
Ủng hộ mik nha

-12.(x-5)+7.(3-x)=5
-12x-60+21-7x=5
(-12x-7x)-(60+21)=5
-19x-81=5
-19x=5-81
-19x=-76
x=-76:(-19)
x=4
x=
x=44:(-19)
\(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(60-12x+21-7x=5\)
\(81-19x=5\)
\(19x=81-5\)
\(19x=76\)
\(x=4\)

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 ^_^