Mn(OH)2 , according to a homework problem I did
this equation is wrong, it should be Al2(SO4)3 + Ca(OH)2 --> Al(OH)3 +CaSO4 and when balanced, it is Al2(SO4)3 + 3Ca(OH)2 --> 2Al(OH)3 +3CaSO4
-2 for O, +3 for Bi
Im assuming you mean Mg(OH)2 and not MgOHThe reaction between Mg(OH)2 and HCl is as follows: Mg(OH)2 (aq) + 2HCl (aq) -> MgCl2 (aq) + 2H2O (l)First the number of mole is found: n(HCl) = c × v = 0.100M × 0.200L = 0.0200mol (to 3 significant figures)Next we find the number of mole of Mg(OH)2:n(Mg(OH)2) ÷ n(HCl) = Coefficient of Mg(OH)2 ÷ Coefficient of HCl n(Mg(OH)2) ÷ n(HCl) = 1 ÷ 2 therefore:n(Mg(OH)2) = (1 ÷ 2) × n(HCl) n(Mg(OH)2) = (1 ÷ 2) × 0.0200moln(Mg(OH)2) = 0.0100mol (to 3 significant figures)Finally we calculate the volume of Mg(OH)2 reacted:v(Mg(OH)2) = n ÷ cv(Mg(OH)2) = 0.0100mol ÷ 0.500MThereforev(Mg(OH)2) = 0.0200L (to 3 significant figures) = 20.0ml (to 3 significant figures)
H2O-----------------H+ + OH-Al2(SO4)3----------2 Al(3+) + 3 (SO4)2-NaOH---------------Na+ + OH-
"Oh si, eres bi?"
Yes it can by the following balance equationBiCl3 + 3 NaOH = 3 NaCl + Bi(OH)3
NaOH and NaClAl(OH)3 and NaOHBa(NO3)2 and Al(OH)3Ba(OH)2 and Ba(NO3)2
-2 for O, +3 for Bi
This Life - 1996 The Bi Who Came in from the Cold 2-3 is rated/received certificates of: Australia:M
Mn(OH)2 , according to a homework problem I did
11
oh MI-oh ba-BI-no CA-roh
The subscript for cobalt II hydroxide is Co(OH)2
this equation is wrong, it should be Al2(SO4)3 + Ca(OH)2 --> Al(OH)3 +CaSO4 and when balanced, it is Al2(SO4)3 + 3Ca(OH)2 --> 2Al(OH)3 +3CaSO4
Amphoteric nature of Ferric hydroxide:Basic nature; It forms Iron(III) ions:Fe(OH)3 + 3 H+ ----> Fe3+ + 3 H2OLikeFe(OH)3 + CH3COOH ----> (CH3COO)3Fe + H2OAcidic nature; It forms Ferrate(VI) ions (with oxidizers):2 Fe(OH)3 + OH- + 5 [O] ----> 2 FeO4-2 + 3 H2OLike2 Fe(OH)3 + OH- + 5 ClO- ----> 2FeO4-2 + 3 H2O + 5 Cl-
1 3/4 cups (US), 2 cups (BI), or 1 3/5 cup (1.25 BI)