There is no real "net ionic" equation, as KI is a catalyst that decomposes H2O2.
The "slow step" is H2O2 + I- --> OI- + H2O
The "fast step" is H2O2 + OI- --> H2O + I- O2
The overall is: 2 H2O2 --> 2H2O + O2
The chemical reaction is:2 K + I2 = 2 KI
Yes, a reaction will occur between copper(II) sulfate (copersulphate) and potassium iodide (KI) to form copper(II) iodide (CuI₂) and potassium sulfate (K₂SO₄). This is a double displacement reaction where the cations and anions switch partners.
The balanced equation for KI + BaS is 2Kl + BaS -> BaI2 + K2S.
Acetic acid is added in the titration reaction to provide the acidic conditions necessary for the reaction between KI and N-bromosuccinimide to occur effectively. The acidic medium helps to convert KI to iodine, which can then react with N-bromosuccinimide. This reaction is commonly used to determine the vitamin C content in a solution.
Yes, when Barium chloride (BaCl2) and Potassium iodide (KI) are mixed, a reaction will occur. BaCl2 and KI will undergo a double displacement reaction to form Barium iodide (BaI2) and Potassium chloride (KCl).
The balanced equation for the reaction between KI and Cl2 to form KCl and I2 is: 2KI + Cl2 -> 2KCl + I2 This equation is already balanced.
The balanced chemical equation for the reaction between sodium chloride (NaCl) and potassium iodide (KI) is: 2NaCl + KI → NaI + KCl. This equation ensures that there is the same number of each type of atom on both sides of the reaction.
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The reaction between potassium iodide (KI) and dilute sulfuric acid (H2SO4) can be represented by the following equation: 2KI + H2SO4 -> 2KHSO4 + HI
KI would be potassium iodine, but you asked KL, and there is no L element.
The balanced chemical equation for the reaction between HI and KOH is: HI + KOH --> KI + H2O. In this reaction, hydrogen iodide (HI) reacts with potassium hydroxide (KOH) to form potassium iodide (KI) and water (H2O). The equation is balanced in terms of atoms and charge.
The net ionic equation for the reaction between Pb(NO3)2 and KI is: Pb2+ (aq) + 2I- (aq) -> PbI2 (s) This represents the formation of solid lead(II) iodide.
The balanced chemical equation for the reaction between lead (II) nitrate (Pb(NO3)2) and potassium iodide (KI) is: Pb(NO3)2 + 2KI -> PbI2 + 2KNO3 This equation is already balanced as it has the same number of atoms of each element on both sides of the reaction arrow.
There is no reaction, because all product ( if reaction goes) would be soluble. so : H+ + Cl- + K+ + I- = H+ + Cl- + K+ + I- and than all of them dare present as a ions only.
If you were to try and react potassium iodide with hydrochloride you would have the following reaction: KI + HCl - > KCl + HI Hydroiodic acid is actually quite a strong acid, so this reaction does not occur in real life.
The chemical equation for the reaction between lead nitrate (Pb(NO3)2) and potassium iodide (KI) to form lead iodide (PbI2) and potassium nitrate (KNO3) is: Pb(NO3)2 + 2KI → 2KNO3 + PbI2
The chemical equation for the reaction between ethyl iodide and aqueous potassium hydroxide is: C2H5I + KOH → C2H5OH + KI This reaction involves the substitution of the iodine in ethyl iodide with hydroxide from KOH, resulting in the formation of ethanol and potassium iodide.