CTHH:Ca3(PO4)2

PTK=40x3+(31+16x4)x2=310

\(a,CTTQ:Al_x^{III}O_y^{II}\Rightarrow x.III=y.II\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow x=2;y=3\\ \Rightarrow Al_2O_3\\ PTK_{Al_2O_3}=27.2+16.3=102\left(đvC\right)\\ b,CTTQ:Al_x^{III}\left(SO_4\right)_y^{II}\Rightarrow x.III=y.II\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow x=2;y=3\\ \Rightarrow Al_2\left(SO_4\right)_3\\ PTK_{Al_2O_3}=27.2+\left(32+16.4\right).3=342\left(đvC\right)\)

\(c,CTTQ:N_x^{IV}O_y^{II}\Rightarrow x.IV=y.II\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow NO_2\\ PTK_{NO_2}=14+16.2=46\left(đvC\right)\\ d,CTTQ:Mg_x^{II}\left(OH\right)_y^I\Rightarrow x.II=y.I\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow Mg\left(OH\right)_2\\ PTK_{Mg\left(OH\right)_2}=24+17.2=58\left(đvC\right)\)

\(-PTK_{BaSO_4}=137+32+16.4=233\left(đvC\right)\)

\(-PTK_{Fe\left(OH\right)_3}=56+\left(16+1\right).3=107\left(đvC\right)\)

\(-PTK_{Na_2SO_3}=23.2+32+16.3=126\left(đvC\right)\)

\(-PTK_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342\left(đvC\right)\)

\(-PTK_{C_{12}H_{22}O_{11}}=12.12+1.22+16.11=342\left(đvC\right)\)

\(-PTK_{Ca\left(NO_3\right)_2}=40+\left(14+16.3\right).2=164\left(đvC\right)\)

\(a,CTHH:Al_2O_3\\ PTK:27.2+16.3=102\left(đvC\right)\\ b,CTHH:Al_2\left(SO_4\right)_3\\ PTK:27.2+\left(32+16.4\right).3=342\left(đvC\right)\\ c,CTHH:NO_2\\ PTK:14+16.2=46\left(đvC\right)\\ d,CTHH:Mg\left(OH\right)_2\\ PTK:24+17.2=58\left(đvC\right)\)

\(a\)) \(CTHH\) giữa \(Al\left(III\right)\) và  \(O\left(II\right)\) là: \(Al_2O_3\)

\(PTK\) \(_{Al_2O_3}\) \(=27.2+16.3=102\left(đvC\right)\)

\(b\)) \(CTHH\) giữa \(Al\left(III\right)\) và \(SO_4\left(II\right)\) là : \(Al_2\left(SO_4\right)_3\)

\(PTK_{Al_2\left(SO_4\right)_3}\) \(=27.2+\left(32+16.4\right).3=342\left(đvC\right)\)

\(c\)) \(CTHH\) giữa \(N\left(IV\right)\) và \(O\) là : \(NO_2\)

\(PTK_{NO_2}\) \(=14+16.2=46\left(đvC\right)\)

\(d\) ) \(CTHH\) giữa \(Mg\left(II\right)\) và \(\left(OH\right)\left(I\right)\) là : \(Mg\left(OH\right)_2\)

\(PTK_{Mg\left(OH\right)_2}\)\(=24+\left(16+1\right).2=58\left(đvC\right)\)

kiểm tra ko giúp dc nhé

hmmmmmm dạng bài tập hả 
cân bằng pthh 
tính theo pthh 
cách đọc tên phân loại axit bazo muối 

a) CTHH:S₂O₅

PTK: 64+ 80 = 144 DvC

b)CTHH:FeCl₃

PTK: 56+106,5=162,5 DvC

còn lại tự làm nha

3. CO₂, BaO, Al₂O₃, Zn, P₂O₅, Na₂O, Cu, K, S (phân loại chắc bạn học rồi cũng tự phân loại được :) )

a, CO₂, BaO, P₂O₅, Na₂O, K

\(CO_2+H_2O\rightarrow H_2CO_3\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ Na_2O+H_2O\rightarrow2NaOH\\ 2K+2H_2O\rightarrow2KOH+H_2\)

b, Zn, Cu, K, S

\(2Zn+O_2\underrightarrow{t^o}2ZnO\\ 2Cu+O_2\underrightarrow{t^o}2CuO\\ 4K+O_2\underrightarrow{t^o}2K_2O\\ S+O_2\underrightarrow{t^o}SO_2\)

c, Zn, K

\(Zn+2HCl\rightarrow ZnCl_2+H_2\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ 2K+2HCl\rightarrow2KCl+H_2\\ 2K+H_2SO_4\rightarrow K_2SO_4+H_2\)

a) mình viết theo stt nhé 
 

S  : đơn chất phi kim 
a) \(CO_2+H_2O\rightarrow H_2CO_3\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ Na_2O+H_2O\rightarrow2NaOH\\ 2K+2H_2O\rightarrow2KOH+H_2\) 
 \(2CO_2+O_2\underrightarrow{t^o}2CO_3\\ 2Zn+O_2\underrightarrow{t^o}2ZnO\\ 2Cu+O_2\underrightarrow{t^o}2CuO\\ 4K+O_2\underrightarrow{t^o}2K_2O\\ S+O_2\underrightarrow{t^o}SO_2\) 
\(Zn+2HCl\rightarrow ZnCl_2+H_2\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(K:potassium\\ O_2:Oxygen\\ H_2:Hydrogen\\ O_3:Ozone\\ N_2:Nitrogen\\ I_2:iodine\\ Cl_2:Chlorine\\ Na:Sodium\\ Mg:Magnesium\\ Fe:Iron\)