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`a,`
\(K.L.P.T_{CuSO_{\text{4}}}=64+32+16.4=160< amu>.\)
\(\%Cu=\dfrac{64.100}{160}=40\%\)
\(\%S=\dfrac{32.100}{160}=20\%\)
\(\%O=100\%-40\%-20\%=40\%\)
`b,` \(K.L.P.T_{Ca\left(NO_3\right)}=40+14+16.3=102< amu>.\)
\(\%Ca=\dfrac{40.100}{102}\approx39,22\%\)
\(\%N=\dfrac{14.100}{102}\approx13,73\%\)
\(\%O=100\%-39,22\%-13,73\%=47,05\%\)
`c,` (bạn sửa lại đề: CTHH là `Al_2(SO_4)_3`)
\(K.L.P.T_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342< amu>.\)
\(\%Al=\dfrac{27.2.100}{342}\approx15,79\%\)
\(\%S=\dfrac{32.3.100}{342}\approx28,07\%\)
\(\%O=100\%-15,79\%-28,07\%=56,04\%\)
\(a,CTHH:KCl\) , \(\text{K.L.P.T}=39+35,5=74,5< amu>.\)
\(CTHH:BaS\) , \(\text{K.L.P.T}=137+32=169< amu>.\)
\(CTHH:Al_2O_3\) , \(\text{K.L.P.T}=27.2+16.3=102< amu>.\)
\(b,CTHH:K_2SO_4\) , \(\text{K.L.P.T}=39.2+32+16.4=174< amu>.\)
\(CTHH:Al_2\left(SO_4\right)_3\), \(\text{K.L.P.T}=27.2+\left(32+16.4\right).3=342< amu>.\)
\(CTHH:MgCO_3\), \(\text{K.L.P.T}=24+12+16.3=84< amu>.\)
`(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}`
`->`\(\text{CTHH: Mg(OH)}_2\)
\(\text{KLNT = }24+\left(16+1\right)\cdot2=58\text{ }< \text{amu}>.\)
e)
Đáp án: MCa(HCO3)2= 162g
%Ca= (40/162)* 100%= 24,7%
%H= (2/162)*100%= 1,23%
%C= (24/162)*100%= 14,81%
%O2=(96/162)*100%= 59,26%