giup e phan btvn a

B5:

\(n_{CuSO_4}=0,2.0,1=0,02\left(mol\right)\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ Cu\left(OH\right)_2\rightarrow\left(t^o\right)CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,02\left(mol\right)\\ m_D=m_{CuO}=0,02.80=1,6\left(g\right)\\ m_{Na_2SO_4}=142.0,02=2,84\left(g\right)\)

câu hỏi đâu nhỉ em?

K có câu hỏi sao​

câu hỏi đâu bn???

Theo đầu bài ta có:
\(\frac{1}{5}\cdot a+2+\frac{1}{2}\cdot a+7=a\)
\(\Rightarrow2+7=a-\frac{1}{2}\cdot a-\frac{1}{5}\cdot a\)
\(\Rightarrow a\cdot\frac{3}{10}=9\)
\(\Rightarrow a=30\)

\(\frac{1}{5}a+2+\frac{1}{2}a+7=a\left(\frac{1}{5}+\frac{1}{2}\right)+2+7=\frac{7}{10}a+10=\frac{7a}{10}+10\)

tìm x hả ?

|x| ko  thể bằng -3,5

=> x ko  tồn tại

easy

x(4+1) : x(x+1)=0

(x:x) 5(x+1)=0

5x+5=0

5x=-5

x=-1

$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$n_{H_2} = n_{Zn} = \dfrac{19,5}{65} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH : 

$n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)$
$m_{Al} = 0,2.27 = 5,4(gam)$

? 

\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)

\(\Rightarrow3x^2-3^2-3x^2+15x=1\)

\(\Rightarrow3x^2-9-3x^2+15x=1\)

\(\Rightarrow-9+15x=1\)

\(\Rightarrow15x=-8\)

\(\Rightarrow x=\frac{-8}{15}\)