2Na2S2O3 + I2 → 2NaI + Na2S4O6 K2Cr2O7 + 6KI + 7H2SO4 → Cr2(SO4)3 + 4K2SO4 + 7H2O + 3I2
The way to calculate a primary standard is with an equation. This equation is M^pdt = (N*V*meq^pb)/1000. For example, for K2Cr2O7, the number is 294,187 grams.
K2Cr2O7 + Na2C2O4 --> K2C2O4 + Na2Cr2O7
k2CrO7 +6FeSO4 +7H2SO4=Cr2(SO4)3 + 3Fe(SO4)3 + 7H2O + K2SO4
2KI+MnO_2+3H_2 SO_4→2I+2KHSO_4+MnSO_4+2H_2 O
the reaction between bleach anb potassium iodide is KI+NaCl2--->KCl2+NaI
potassium bromide + fluorine --> potassium fluoride + bromide
ethanal
ethanol is converted into acetic acid....
K2Cr2O7 + Na2C2O4 --> K2C2O4 + Na2Cr2O7
Cr2O72- + 6Fe2++ 14H+ --> 2Cr3+ + 6Fe3+ + 7H2O
k2CrO7 +6FeSO4 +7H2SO4=Cr2(SO4)3 + 3Fe(SO4)3 + 7H2O + K2SO4
I think the reaction equation can be written as follows: K2Cr2O7 (aq) + BaCl2(aq) ------ BaCr2O7(s) + 2KCl(aq).
The only known and possible reaction is the following redox (reduction-oxidation) reaction between I3- (Iodine-Iodide complex) and S2O32- (thiosulfate)I3- + 2 S2O32- --> 3 I- + S4O62-ox. + red.So the reaction between potassium iodide (KI) and potassium thiosulfate (K2S2O3) is NOT possible because they both are reductors (electron donors).Iodine-Iodide complex is essentially Iodine is an oxidator, bound to a non-reacting I- ion (Iodide)
One would expect that the reaction between an alkene and cold, dilute potassium dichromate is an oxidation to a bifunctional alcohol at the carbons in the double bond. The result is also called a vincinal diol or a glycol. The reaction should be similar to the reaction of an alkene with cold, dilute potassium permangante, however, dichromate is a milder oxidizing agent and may not be as effective. The mechanism for this reaction involves the formation of an intermediete 'ester' with the metal at the carbons of the double bond, breaking the double bond. The it can be shown that the resonance structure of the intermediete complex transfers electron density to the ester linkage and protonates from water at both of the ester sites completing the oxidation.
No reaction.
2KI+MnO_2+3H_2 SO_4→2I+2KHSO_4+MnSO_4+2H_2 O
the reaction between bleach anb potassium iodide is KI+NaCl2--->KCl2+NaI
potassium bromide + fluorine --> potassium fluoride + bromide