a) K2CrO4(aq) with AgNO3(aq)
b) K2CrO4(aq) with Pb(NO3)2(aq)
c) K2CrO4(aq) with Hg2(NO3)2(aq)
d) AgNO3(aq) with HCl(aq)
e) Pb(NO3)2(aq) with HCl(aq)
f) Hg2(NO3)2(aq) with HCl(aq)
g) AgCl(s) with NH3(aq)
Net Ionic Equations?
2014-06-16 10:13 pm
回答 (2)
2014-06-16 11:46 pm
I'll show you the steps and hopefully you learn from it!
1) First we do double displacement reaction
K2CrO4 + AgNO3 ---> KNO3 + Ag2CrO4
Then we look for compound that remains a solid. Theres usually a chart for this, Ag2CrO4 is insoluable so it doesnt dissociates
Now we split every single compound to its basic components except for Ag2CrO4
2K + CrO4^-2 + Ag + NO3 ---> K + NO3 + Ag2CrO4
Now we match and cancel
CrO4^-2 + Ag ----> Ag2CrO4
I hope you understand :)
1) First we do double displacement reaction
K2CrO4 + AgNO3 ---> KNO3 + Ag2CrO4
Then we look for compound that remains a solid. Theres usually a chart for this, Ag2CrO4 is insoluable so it doesnt dissociates
Now we split every single compound to its basic components except for Ag2CrO4
2K + CrO4^-2 + Ag + NO3 ---> K + NO3 + Ag2CrO4
Now we match and cancel
CrO4^-2 + Ag ----> Ag2CrO4
I hope you understand :)
2014-06-16 10:30 pm
a)
CrO4{2-} + Ag{+} → Ag2CrO4
b)
CrO4{2-} + Pb{2+} → PbCrO4
c)
CrO4{2-} + Hg(2+} → HgCrO4
d)
Ag(+) + Cl{-} → AgCl
e)
Pb{2+} + Cl{-} → PbCl2
f)
Hg(+) + Cl{-} → HgCl
g)
Ag{+} + 2 NH3 → [Ag(NH3)2]{+}
CrO4{2-} + Ag{+} → Ag2CrO4
b)
CrO4{2-} + Pb{2+} → PbCrO4
c)
CrO4{2-} + Hg(2+} → HgCrO4
d)
Ag(+) + Cl{-} → AgCl
e)
Pb{2+} + Cl{-} → PbCl2
f)
Hg(+) + Cl{-} → HgCl
g)
Ag{+} + 2 NH3 → [Ag(NH3)2]{+}
收錄日期: 2021-04-21 00:08:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140616141303AAUmyzY