HIO4(aq) stands for periodic acid, which is a strong oxidizing agent commonly used in organic chemistry for oxidation reactions. It is a colorless, crystalline solid when in pure form and is soluble in water to form a clear solution.
HCN(aq) ==> H^+(aq) + CN^-(aq)Ka = [H+][CN-]/[HCN] and the value can be looked up in a book or on line.
Barium chloride and sodium sulfate Molecular equation: BaCl2(aq) + Na2SO4(aq)--> BaSO4(s) + 2NaCl(aq) ionic equation: Ba+2(aq) + 2Cl-(aq) + 2Na+(aq) + SO4-2(aq) --> BaSO4(s) + 2Na+(aq) + 2Cl-(aq) Net ionic equation is Ba+2(aq) + SO4-2(aq) --> BaSO4(s)
A precipitate
aq bhoola
Mg(s) + H2SO4(aq) ==> MgSO4(aq) + H2(g) ... molecular equationMg(s) + 2H+(aq) + SO4^2-(aq) ==> Mg^2+(aq) + SO4^2-(aq) + H2(g) ... ionic equationMg(s) + 2H^+(aq) ==> Mg^2+(aq) + H2(g) ... net ionic equationSpectator is SO4^2- (sulfate ion).
The conjugate base of HIO4 is IO4-. This is because when HIO4 donates a proton (H+), it forms IO4- by losing the hydrogen ion.
Yes, HIO3 is more acidic than HIO4. This is because the acidity of oxoacids generally increases with the number of oxygen atoms attached to the central atom. Since HIO3 has one more oxygen atom compared to HIO4, it is more acidic.
Periodic Acid. Pronounce Per - I-oh-dic. not periodic like the table of the elements.
In HIO4 (periodic acid), the oxidation number of hydrogen (H) is +1, of iodine (I) is +7, and of oxygen (O) is -2. Therefore, the overall oxidation number of HIO4 is zero since the sum of the oxidation numbers in a neutral molecule must equal zero.
HIO4 (periodic acid) is commonly used in organic synthesis to cleave vicinal diols to aldehydes and ketones, a process known as periodate cleavage. This reaction is useful for structural elucidation and for creating synthons in organic transformations. Additionally, HIO4 can be used for oxidative cleavage of double bonds in olefins.
HIO4 is known as periodic acid. It is a strong oxidizing agent and is used in organic chemistry for reactions involving oxidation of alcohols to carbonyl compounds. Periodic acid is a colorless solid and is highly soluble in water.
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) -> H2O(l) + Na+(aq) + Cl-(aq)
Molecular Eq HC2H3O2(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq) Ionic Eq H+(aq) + C2H3O2-(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq) Net Ionic Eq H+(aq) + NH3(aq) -> NH4+(aq)
Mg(NO3)2(aq) + H2SO4(aq) ==> MgSO4(aq) + 2HNO3(aq) Complete molecular equationTotal ionic equation:Mg^2+(aq) + 2NO3^-(aq) + 2H^+(aq) + SO4^2-(aq) => Mg^2+(aq) + SO4^2-(aq) + 2H^+(aq) + 2NO3^-(aq)RESULT - NO REACTION
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) KOH(aq) + HCl(aq) → KCl(aq) + H2O(l) Pb(NO3)2(aq) + KI(aq) → PbI2(s) + KNO3(aq) BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
2Na+(aq) + SO42-(aq) + Ca2+(aq) + 2Cl-(aq) CaSO4(s) + 2Na+(aq) + 2Cl-(aq)
HIO4 cleaves C-C bonds between vicinal -OH groups so you get the malaprade reaction: HOCH2(CHOH)4CHO + 4 HIO4 ------> H2C=O + 4 HCOOH + 4 HIO3 Only 4 HIO4 molecules required to cleave..........