CE level chemistry

the colour of ammonium hydroxide in (1) is colourless
the colour of sodium dichromate in (2) is red-orange colour
I hope this can help with your understanding. =)

2007-11-27 03:42:13 補充：
In reality, for (1), the reaction will be NH4Cl + NaOH ---> Na+　+　Cl-　+　NH4+　+　OH-as NH4OH will never form ppt, the resulant is just a mixture of four ions which is colourless.

2007-11-27 03:43:38 補充：
In reality, for (2), colour of Na2Cr2O7 is red-orange, or dichromate ion is red-orange in colourcolour of Na2CrO4 is yellow, or chromate ion is yellow in colourH2O + (Cr2O7)2- <=> 2 (CrO4)2- + 2 H+

2007-11-27 03:46:34 補充：
H2O + (Cr2O7)2- ＜＝＞ 2 (CrO4)2- + 2 H+ Na2Cr2O7 + KOH the colour of solution shifts from orange to yellow as OH- increasesthe reaction will beK2Cr2O7 + 2 NaOH ---＞ 2 K+　+　2 Na+　+　2 (CrO4)2-　+　H2O

2007-11-27 03:52:48 補充：
The net effect is that chromate is produced, dichromate is used up, and the orange solution turns yellow. Moreover, no ppt will be formed as both K2Cr2O7, Na2Cr2O7, KOH, NaOH, K2CrO4, and Na2CrO4 are soluble in water.

2007-11-27 03:54:48 補充：
The observations made during this experiment may be explained through Le Chatelier's principle: if a stress (such as a change in concentration, temperature, or pressure) is applied to a chemical system at equilibrium, the equilibrium shifts in such a way as to undo the applied stress.

1)The colour is colourless as ammonium ions and hydroxide ions are colourless.
2)Thr colour is orange as dichromate ions are orange in colour though sodium ions are colourless.