bó tay chấm com

bó tay chấm com chấm campuchia!!! 

Bài 1:
\(m_{H_2SO_4.40\%}=200\times40\%=80\left(g\right)\)

\(\Rightarrow m_{H_2O}=200-80=120\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4.19,6\%}=\frac{120}{100\%-19,6\%}=149,25\left(g\right)\)

\(\Rightarrow m_{H_2SO_4.19,6\%}=149,25\times19,6\%=29,253\left(g\right)\)

\(\Rightarrow m_{H_2SO_4}=80-29,253=50,747\left(g\right)\)

Bài 2:

Gọi \(m_{ddNaCl.20\%}=x\left(g\right)\Rightarrow m_{NaCl.20\%}=20\%x=0,2x\left(g\right)\)

\(m_{ddNaCl.30\%}=y\left(g\right)\Rightarrow m_{NaCl.30\%}=30\%y=0,3y\left(g\right)\)

\(m_{NaCl.26\%}=300\times26\%=78\left(g\right)\)

Ta có hệ: \(\left\{{}\begin{matrix}x+y=300\\0,2x+0,3y=78\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=120\\y=180\end{matrix}\right.\)

Vậy \(m_{ddNaCl.20\%}=120\left(g\right)\)

\(m_{ddNaCl.30\%}=180\left(g\right)\)

mik dc k

Câu 1:

\(m_{NaCl.8\%}=500\times8\%=40\left(g\right)\)

\(\Rightarrow m_{H_2O}=500-40=460\left(g\right)\)

\(\Rightarrow m_{ddNaCl.12\%}=\frac{460}{100\%-12\%}=522,73\left(g\right)\)

\(\Rightarrow m_{NaCl.12\%}=522,73\times12\%=62,7276\left(g\right)\)

\(\Rightarrow m_{NaCl}thêm=62,7276-40=22,7276\left(g\right)\)

Câu 2:

\(m_{NaCl.12\%}=500\times12\%=60\left(g\right)\)

\(\Rightarrow m_{ddNaCl.8\%}=\frac{60}{8\%}=750\left(g\right)\)

\(\Rightarrow m_{H_2O}thêm=750-500=250\left(g\right)\)

Câu 1:

\(m_{NaCl}=500\times10\%=50\left(g\right)\)

Câu 2:

\(m_{CuSO_4}=500\times8\%=40\left(g\right)\)

\(\Rightarrow n_{CuSO_4}=\frac{40}{160}=0,25\left(mol\right)\)

Ta có: \(n_{CuSO_4.5H_2O}=n_{CuSO_4}=0,25\left(mol\right)\)

\(\Rightarrow m_{CuSO_4.5H_2O}=0,25\times250=62,5\left(g\right)\)

\(\Rightarrow m_{H_2O}=500-62,5=437,5\left(g\right)\)

;V

bn nào bt cho mik xin ạ

có lẽ là 4

Lời giải:

$x^2\geq 0, \forall x\in\mathbb{R}$

$\Rightarrow Q(x)=x^2+\sqrt{3}\geq \sqrt{3}>0$ với mọi $x\in\mathbb{R}$

Do đó đa thức $Q(x)$ vô nghiệm.

a, pthh: 4Na+O2--->2Na2O

Theo ĐLBTKL: mNa+mO2=mNa2O

=> mO2= 6,2-4,6= 1,6 (g)

=> nO2= \(\dfrac{1,6}{32}=0,05\) mol

=> VO2= 0,05.22,4= 1,12 (l)

b, C1:nO2= \(\dfrac{0,448}{22,4}=0,02\) mol

Theo pt: nNa= 4.nO2= 4.0,02= 0,08 mol

=> mNa= 0,08.23= 1,84 (g)

C2: nO2= 0,02 mol

=> mO2= 0,02.32= 0,64 (g)

Theo pt: nNa2O= 2.nO2= 2.0,02= 0,04 mol

=> mNa2O= 0,04.62= 2,48 (g)

Theo ĐLBTKL: mNa+mO2=mNa2O

=> mNa= 2,48-0,64= 1,84 (g)