1. Draw me a beaker for each of the following 3 solutions. Show (label) the particles that exist when 1 particle of each dissolves. You don't have to show the water molecules!
H3P04 C12H22011 HCHOO
2. Balance the following redox reaction that occurs in basic solution.
NO2- (aq) + Al (s) ---> NH3 (g) + Al02- (aq)
3. Describe, in detail, how you'd make 100 mL of a 2.0 M HC1 solution from a 6.0 M HC1 solution. How much solvent do you expect to add? For this essay, you may answer in list form (ie, Step 1 ...etc). Support your answer with calculations to indicate amounts of all liquids.
Equipment available: graduated cylinder, pipet, volumetric flask, beaker, 100 niL of 6.OM HC1, 100 mL of distilled water, 100 mL of tap water
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H3PO4 (phosphoric acid) is a strong acid. Strong acids ionize completely. When it dissolves, it will form 1 PO43- ion and 3 H+ ions (if we are talking about one molecule).
Reaction: H3PO4 ----> 3H+ + PO43-
C12H22O11 (sucrose or table sugar) is what is known as a molecular solid. When it dissolves, it does not split up into ions.
HCHOO (formic acid) is a weak acid. If there is a large amount, just a few molecules will dissolve (split into ions). If we are talking about one of these molecule that splits into ions...
Reaction: HCHOO ----> H+ + HCOO-
1. Split into half reactions (group molecules containing the same atom).
NO2-(aq) ----> NH3(g)
Al(s) ----> AlO2-(aq)
2. Add enough H2O to balance O on both sides.
NO2-(aq) ----> NH3(g) + 2 H2O
2 H2O + Al(s) ----> AlO2-(aq)
3. Add H+ to balance H.
7 H+ + NO2-(aq) ----> NH3(g) + 2 H2O
2 H2O + Al(s) ----> AlO2-(aq) + 4 H+
4. Add electrons to balance the charge (electrons have a negative charge).
6 e + 7 H+ + NO2-(aq) ----> NH3(g) + 2 H2O (-6 + 7 + -1 = 0 ----> 0 + 0 = 0 ...
Detailed explanations are presented for dilution, redox and titration problems.