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yep as uv heard coz Cl is more reactive than iodine it displaces it leaving iodine molecules - thats the brown though typically Iodine is blue-black/purple soo something probably has gone wrong..
the equation is
Cl2(aq) + 2I-(aq) -------> 2Cl-(aq) + I2 (aq)
soz bout no subscript
originally sodium iodide is colourless as the iodine is not "free" - it is bonded to the sodium = and in organic solution it goes its true purple
the equation is
Cl2(aq) + 2I-(aq) -------> 2Cl-(aq) + I2 (aq)
soz bout no subscript
originally sodium iodide is colourless as the iodine is not "free" - it is bonded to the sodium = and in organic solution it goes its true purple
Ammar21
yep as uv heard coz Cl is more reactive than iodine it displaces it leaving iodine molecules - thats the brown though typically Iodine is blue-black/purple soo something probably has gone wrong..
the equation is
Cl2(aq) + 2I-(aq) -------> 2Cl-(aq) + I2 (aq)
soz bout no subscript
originally sodium iodide is colourless as the iodine is not "free" - it is bonded to the sodium = and in organic solution it goes its true purple
the equation is
Cl2(aq) + 2I-(aq) -------> 2Cl-(aq) + I2 (aq)
soz bout no subscript
originally sodium iodide is colourless as the iodine is not "free" - it is bonded to the sodium = and in organic solution it goes its true purple
Originally the iodine is not 'bonded ' to the sodium. It is in the form of ions and iodide ions are colourless.
Iodine in aqueous solution is brown. When mixed with excess iodide ions iodine forms the triiodide ion, which is red.
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