chem 送分題~

1. There is no observable change.
Cl2(aq) + H2O(l) ⇌ HOCl (aq) + H+(aq) + Cl-(aq)
This exists as an equilibrium state.
By having more Cl-(aq), the reverse reaction is more apparent.
However, all of these liquids/solutions are colourless (given that the conc. of Cl2 is low).

2. Hydrogen has a very low density that it normally will not be present in the atmosphere. Liquefied air also does not contain hydrogen.

3. Carbon dioxide dissolves in rainwater to form a weak acid, carbonic acid, which has a pH of 5.6-5.7, i.e. CO2(g) + H2O(l) ⇌ H2CO3(aq). However, this is the common pH of rainwater. If carbon dioxide causes acid rain, common precipitation will also be considered as acid rain, which does not make sense. Acid rain normally has a pH between 4.5 and 5.6.

4. Glacial ethanoic acid is hygroscopic, i.e. it absorbs water vapour in air; its mass cannot be accurately determined. Thus it is not suitable to dissolve glacial ethanoic acid to prepare 1M ethanoic acid.

On the other hand, we can dilute another standard solution of ethanoic acid of a higher concentration, say 10.0M.
Firstly, pipette 25.0 cm^3 of the solution into a 250.0 cm^3 volumetric flask. Then, add in water by using a washbottle, until the meniscus reaches 3 to 4 cm below the graduation mark. After that, use a dropper to add more water until the meniscus reaches the graduation mark. It is reminded that the reading should be taken at eye level. Stopper the volumetric flask and invert it for a few times for more complete mixing. The resulting solution is 1M ethanoic acid

1.
There is NO observable change when chlorine water is added to sodium chloride solution. There is NO reaction between chlorine water and chloride/sodium ion.

2.
There is NO hydrogen in normal air. Therefore, there would be NO hydrogen in liquefied air.

3.
Acid rain is defined as the rain water having a pH value lower than 5.6.
Carbon dioxide is a weak acid. When rain water is saturated with carbon dioxide, the pH is 5.6. Therefore, carbon dioxide does not cause acid rain.

4.
Dissolve 15.0 g of glacial ethanoic acid (pure ethanoic acid) in distilled water. Then dilute the solution to the mark in a 250 cm3 volumetric flask.
Mass of CH3COOH = 15.0 g
Volume of the solution = 250 cm3 = 0.250 dm3
Relative molecular mass of CH3COOH = 12.0x2 + 1.0x4 + 16.0x2 = 60.0 g mol-1
No. of moles of CH2COOH = 15.0 / 60.0 = 0.250 mol
Molarity = 0.250 / 0.250 = 1.00 M