a. PTHH: CuO + H₂SO₄ ---> CuSO₄ + H₂O

b. Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\)

=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{9,8}{m_{dd_{H_2SO_4}}}.100\%=4,9\%\)

=> \(m_{dd_{H_2SO_4}}=200\left(g\right)\)

c. Ta có: \(m_{dd_{CuSO_4}}=200+8=208\left(g\right)\)

Theo PT: \(n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

=> \(C_{\%_{CuSO_4}}=\dfrac{16}{208}.100\%=7,69\%\)

a: \(=5x^2-10x-5x^2+7x=-3x\)

b: \(=2x^3+3xy^2-4y-3xy^2=2x^3-4y\)

a.\(C\%_{KCl}=\dfrac{20}{600}.100=3,33\%\)

b.2,5kg = 2500g

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{2500}.100=1,368\%\)

`(a): C%_[KCl] = 20 / 600 . 100 ~~ 3,33 %`

________________________________________

`(b):` Đổi `2,5 kg = 2500 g`

`C%_[Al_2 (SO_4)_3] = [ 34,2 ] / 2500 . 100 = 1,368%`

a. \(C_{\%_{KCl}}=\dfrac{20}{600}.100\%=3,33\%\)

b. \(2,5kg=2500g\)

\(C_{\%_{Al_2\left(SO_4\right)_3}}=\dfrac{34,2}{2500}.100\%=1,368\%\)

1have to

2 had to

3shouldn't

4has to

Giúp mik vs mọi người đâu hết rùi😄😄😄