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Gọi CTHH là: XH3
Theo đề, ta có: \(d_{\dfrac{XH_3}{H_2}}=\dfrac{M_{XH_3}}{M_{H_2}}=\dfrac{M_{XH_3}}{2}=8,5\left(lần\right)\)
=> \(M_{XH_3}=17\left(g\right)\)
Ta có: \(M_{XH_3}=M_X+1.3=17\left(g\right)\)
=> MX = 14(g)
Dựa vào bảng hóa trị, suy ra:
X là nitơ (N)
=> CTHH của hợp chất là NH3
Chọn B
\(n_P=\dfrac{9.3}{31}=0.3\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
\(0.3.....0.375.....0.15\)
\(V_{O_2}=0.375\cdot22.4=8.4\left(l\right)\)
\(m_{P_2O_5}=0.15\cdot142=21.3\left(g\right)\)
PT: 4P + 5O2 → 2P2O5.
Ta có: nP= 9,3/31=0,3(mol)
Theo PT: nO2= 5/4 . nP=5/4 . 0,3=0,375(mol)
=> VO2=0,375.22,4=8,4(lít)
Theo PT: nP2O5=1/2 . nP=1/2 . 0,3=0,15(mol)
=> mP2O5= 0,15.142=21,3(g)
BT4: Hiệu suất phản ứng:
\(H=\dfrac{m_{tt}}{m_{lt}}.100\%=\dfrac{36,48}{48}.100\%=76\%\)
BT5 Khối lượng đồng thu được:
\(H=\dfrac{m_{tt}}{m_{lt}}.100\%\Rightarrow m_{tt}=\dfrac{m_{lt}.H}{100\%}=\dfrac{48.95}{100\%}=45,6\left(g\right)\)
Số hạt mang điện là:
34 : (11 + 6) . 11 = 22 (hạt)
Số proton của X là:
22 : 2 = 11 (hạt)
Số hạt mang điện là:
34 : (11 + 6) . 11 = 22 (hạt)
Số proton của X là:
22 : 2 = 11 (hạt)
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11/ 2Fe(OH)3 --> Fe2O3 + 3H2O
12/ Fe(OH)3 + 3HCl --> FeCl3 + 3H2O
13/ CaCl2 + 2AgNO3 --> Ca(NO3)2 + 2AgCl
14/ 4P + 5O2 --> 2P2O5
15/ N2O5 + H2O --> 2HNO3
16/ Zn + 2HCl --> ZnCl2 + H2
17/ 2Al + 3CuCl2 --> 2AlCl3 + 3Cu
18/ CO2 + Ca(OH)2 --> CaCO3 + H2O
19/ SO2 + Ba(OH)2 --> BaSO3 + H2O
20/ 2KMnO4 --> K2MnO4 + MnO2 + O2
Mình cân bằng pt rồi bạn tự chọn tỉ lệ theo cặp chất nhé! :))
1. H2SO4 + K2CO3 \(\rightarrow\) K2SO4 + H2O + CO2\(\uparrow\)
2. Fe + 2HCl \(\rightarrow\) FeCl2 + H2
3, 2Al + 3PbCl2 \(\rightarrow\) 2AlCl3 + 3Pb