Nó cầm dao và đấm vào ngực nó (đười ươi hay làm thế).

Bài 1:

a) \(n_{CO}=x;n_{H_2S}=y\)

\(M_X=32.0,95=30,4\rightarrow\dfrac{28x+34y}{x+y}=30,4\rightarrow2,4x=3,6y\rightarrow\dfrac{x}{y}=\dfrac{3,6}{2,4}=\dfrac{3}{2}\)

%V_CO=\(\dfrac{3}{3+2}.100=60\%\)

%\(V_{H_2S}=\dfrac{2}{3+2}.100=40\%\)

b) \(V_{CO}=12,32.\dfrac{3}{5}=7,392l\)

\(V_{H_2S}=12,32.\dfrac{2}{5}=4,928l\)

-Gọi \(V_{CO_2}=x\left(lít\right)\)

\(\overline{M}=\dfrac{44x+28.7,392+34.4,928}{x+12,32}=9.2=18\)

Giải ra x\(\approx5,9\left(lít\right)\)

a) \(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(n_{Cl}=n_{HCl}=2n_{H_2}=2.0,06=0,12\left(mol\right)\)

m=m_{kim loại}+m_Cl=1,965+0,12.35,5=6,225(gam)

b) \(H_2+CuO\overset{t^0}{\rightarrow}Cu+H_2O\left(1\right)\)

\(yH_2+Fe_xO_y\overset{t^0}{\rightarrow}xFe+yH_2O\left(2\right)\)

Fe+H₂SO₄\(\rightarrow FeSO_4+H_2\left(3\right)\)

\(m_{Cu}=1,28\left(g\right)\rightarrow n_{Cu}=\dfrac{1,28}{64}=0,02\left(mol\right)\)

-Theo(1): \(n_{H_2\left(1\right)}=n_{CuO}=n_{Cu}=0,02\left(mol\right)\)

\(\rightarrow m_{CuO}=0,02.80=1,6\left(g\right)\rightarrow m_{Fe_xO_y}=3,92-1,6=2,32\left(g\right)\)

\(\rightarrow n_{H_2\left(2\right)}=n_{H_2}-n_{H_2\left(1\right)}=0,06-0,02=0,04\left(mol\right)\)

-Theo(2): \(n_{Fe_xO_y}=\dfrac{1}{y}n_{H_2}=\dfrac{0,04}{y}mol\rightarrow M_{Fe_xO_y}=\dfrac{2,32}{\dfrac{0,04}{y}}=58y\)

\(\rightarrow56x+16y=58y\rightarrow56x=42y\rightarrow\dfrac{x}{y}=\dfrac{42}{56}=\dfrac{3}{4}\)

\(\rightarrow Fe_3O_4\)

%CuO=\(\dfrac{1,6}{3,92}.100\approx40,8\%\)

%Fe₃O₄=100%-40,8%=59,2%

33 bạn chuẩn luôn

ko phải lớp 1 đâu nha bạn

Bài 1: Gọi oxit là R₂O_n

\(n_{HCl}=\dfrac{54,75.20}{36,5.100}=0,3mol\)

R₂O_n+2nHCl\(\rightarrow2RCl_n+nH_2O\)

\(n_{R_2O_n}=\dfrac{1}{2n}n_{HCl}=\dfrac{0,3}{2n}mol\)

\(M_{R_2O_n}=\dfrac{5,1}{\dfrac{0,3}{2n}}=34n\rightarrow2R+16n=34n\)

\(\rightarrow2R=18n\text{​​}\rightarrow R=9n\rightarrow\)nghiệm phù hợp n=3 và R=27(Al)\(\rightarrow oxit:Al_2O_3\)

10Fe+36HNO₃\(\rightarrow\)10Fe(NO₃)₃+3N₂+18H₂O

3Mg+8HNO₃\(\rightarrow\)3Mg(NO₃)₂+2NO+4H₂O

Fe_xO_y+(y-x)H₂\(\rightarrow\)xFeO+(y-x)H₂O

C_xH_yO_z+(\(x+\dfrac{y}{4}-\dfrac{z}{2}\))O₂\(\rightarrow\)xCO₂+\(\dfrac{y}{2}H_2O\)

a)

- Gọi A là C_xO_y: %O=100%-42,6%=57,4%

Ta có: x:y=\(\dfrac{42,6}{12}:\dfrac{57,4}{16}=3,55:3,5875\approx1:1\)

- Gọi B là C_aO_b: %O=100%-27,3%=72,7%

Ta có: x:y=\(\dfrac{27,3}{12}:\dfrac{72,7}{16}=2,275:4,54375\approx1:2\)

b)

-Nếu A chỉ có 1C thì công thức A là CO\(\rightarrow M_{CO}=12+16=28\left(đvC\right)\)

-Nếu B có 1C thì công thức B là CO₂\(\rightarrow M_{CO_2}=12+16.2=44\left(đvC\right)\)

a) Nguyên tử khối của R:

\(M_R=\dfrac{66,4.10^{-24}}{1,6605.10^{-24}}\approx40\left(đvC\right)\rightarrow R\) là Ca

b) CaX₂

\(M_{CaX_2}=2,775.M_{Ca}=2,755.40=111\left(đvC\right)\)

\(\rightarrow40+2M_X=111\rightarrow X=35,5\left(Cl\right)\)

\(\left\{{}\begin{matrix}2P+N=95\\N-P=5\end{matrix}\right.\)

Giải ra P=30(Zn), N=35

-Vậy: P=E=30;N=35

Số khối A=P+N=30+35=65