\(n_{KMnO_4}=\frac{632}{158}=4(mol)\\ a/ 2KMnO_4 \buildrel{{t^o}}\over\longrightarrow MnO_2+O_2+K_2MnO_4\\ b/\\ n_{MnO_2}=\frac{1}{2}.n_{KMnO_4}=\frac{1}{2}.4=2(mol)\\ m_{MnO_2}=87.2=174(g)\\ c// n_{O_2}=\frac{1}{2}.n_{KMnO_4}=\frac{1}{2}.4=2(mol)\\ V_{O_2}=2.22,4=44,8(l) \)

\((1) 2HCl + CuO \to CuCl_2+H_2O\\ (2)CuCl_2+2NaOH \to Cu(OH)_2 +2NaCl\\ (3)Cu(OH)_2+H_2SO_4 \to CuSO_4+2H_2O\\ (4)CuSO_4+Ba(OH)_2 \to BaSO_4+Cu(OH)_2\)

\(0,05<0,08\\ \)

\(\text{Ta có: } S_xO_y\\ x:y=\frac{2}{32}: \frac{3}{16}\\ \Leftrightarrow x:y= 1:3\\ \Rightarrow: SO_3\)

B,Động năng của vật càng lớn khi khối lượng càng nhỏ.

Tính không ra được

\(n_{CaO}=\frac{6,5}{56}=\frac{13}{112}(mol)\\ CaO+H_2O \to Ca(OH)_2\\ a/ \\ \text{Chất tạo thành làm quỳ hoá xanh, đó là: }Ca(OH)_2\\ b/\\ n_{Ca(OH)_2}=n_{CaO}=\frac{13}{112}(mol)\\ C\%_{Ca(OH)_2}=\frac{\frac{13}{112}.74\\}{6,5+500.1}.100\%=1,7\%\)

\(\text{Hoà tan đc:} Ag_2SO_4\\ Ag_2SO_4+2NaOH \to Na_2SO_4+2AgOH\)

\(a/ 2Mg+O_2 \to 2MgO\\ b/ \\ n_{Mg}=\frac{3,6}{24}=0,15(mol)\\ n_{Mg}=n_{MgO}=0,15(mol)\\ m_{MgO}=0,15.40=6(g)\)

\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ Zn+H_2SO_4 \to ZnSO_4+H_2\\ n_{H_2}=\frac{2,24}{22,4}=0,1(mol)\\ n_{Al}=a(mol)\\ n_{Zn}=b(mol)\\ m_{hh}=27a+65b=3,68(1)\\ n_{H_2}=1,5a+b=0,1(2)\\ (1)(2)\\ a=b=0,04(mol)\\ n_{H_2SO_4}=1,5a+b=1,5.0,04+0,04=0,11mol\\ m_{dd H_2SO_4}=\frac{98.0,1.100\%}{10\%}=98(g)\\ m_{dd}=3,68+98-0,1.2=101,48(g)\)