Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Mình lập nhanh nhé!
SO2 hoặc SO3
Al2O3
CaO
Al2(PO4)3
Fe(OH)2 hoặc Fe(OH)3
MgSO4
Ca3(PO4)2
`(1)`
Gọi ct chung: \(\text{Al}_{\text{x}}\text{O}_{\text{y}}\)
`@` Theo quy tắc hóa trị: `III*x=y*II -> x/y=(II)/(III)`
`-> \text {x=2, y=3}`
`->`\(\text{CTHH: Al}_2\text{O}_3\)
\(\text{KLPT = }27\cdot2+16\cdot3=102\text{ }< \text{amu}>\)
`(2)`
Gọi ct chung: \(\text{Mg}_{\text{x}}\text{(OH)}_{\text{y}}\)
`@` Theo quy tắc hóa trị: `II*x=I*y -> x/y=I/(II)`
`-> \text {x = 1, y = 2}`
`->`\(\text{CTHH: Mg(OH)}_2\)
\(\text{KLNT = }24+\left(16+1\right)\cdot2=58\text{ }< \text{amu}>.\)
\(a,CTTQ:Al_x^{III}O_y^{II}\Rightarrow x\cdot III=y\cdot II\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow x=2;y=3\\ \Rightarrow Al_2O_3\\ b,CTTQ:Fe_x^{II}\left(NO_3\right)_y^I\Rightarrow x\cdot II=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow Fe\left(NO_3\right)_2\\ c,CTTQ:Mg_x^{II}\left(OH\right)_y^I\Rightarrow x\cdot II=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow Mg\left(OH\right)_2\\ d,CTTQ:N_x^{III}H_y^I\Rightarrow x\cdot III=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{3}\Rightarrow x=1;y=3\\ \Rightarrow NH_3\)
\(Đặt:Al_a^{III}O_b^{II}\left(a,b:nguyên,dương\right)\\ QT.hoá.trị:a.III=II.b\Rightarrow\dfrac{a}{b}=\dfrac{II}{III}=\dfrac{2}{3}\Rightarrow a=2;b=3\\ CTTQ:Al_2O_3\\ m_{Al_2O_3}=2.27+3.16=102\left(đ.v.C\right)\)
a)CuO
b)SO4
c)K2SO4
d)BA3(PO4)2
e)FeCl3
f)Al(NO3)3
g)PO5
h)Zn(OH)2
k)MgSO4
l)FeSO3
3.1:
- Hợp chất: \(Al_2O_3\)
- \(PTK_{Al_2O_3}\) \(= \) \(2.27 + 3.16 = 102\) (đvC)
3.2:
- Hợp chất: \(NH_3\)
- \(PTK_{NH_3}\)\( = 14 + 3.1 = 17\) (đvC)
Gọi CTHH là: \(\overset{\left(III\right)}{Al_x}\overset{\left(I\right)}{\left(OH\right)_y}\)
Ta có: \(III.x=I.y\)
\(\Leftrightarrow\dfrac{x}{y}=\dfrac{I}{III}=\dfrac{1}{3}\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy CTHH là Al(OH)3
\(\Rightarrow PTK_{Al\left(OH\right)_3}=27+\left(16+1\right).3=78\left(đvC\right)\)
b. Gọi CTHH là: \(\overset{\left(II\right)}{Ca_a}\overset{\left(III\right)}{\left(PO_4\right)_b}\)
Ta có: \(II.a=III.b\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{III}{II}=\dfrac{3}{2}\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\)
Vậy CTHH là Ca3(PO4)2
\(\Rightarrow PTK_{Ca_3\left(PO_4\right)_2}=40.3+\left(31+16.4\right).2=310\left(đvC\right)\)
a: \(Al_2O_3\)
M=27*2+16*3=48+54=102
b: MgO
M=24+16=40
c: \(Al\left(OH\right)_3\)
\(M=27+\left(16+1\right)\cdot3=27+3\cdot17=27+51=78\)