My Dream Essay

1376 Words Apr 19th, 2011 6 Pages
Shinhye Lee
Chem101
4.16 Specify what ions are present upon dissolving each of the following substances in water:
a) MgI2(aq) -> Mg 2+ (aq) +2OH-(aq)
b) Al(NO3)3(aq) -> Al3+(aq)+3NO3- (aq)
c) HCLO4(aq) -> H+ (aq) + ClO4- (aq)
d) NaCH3COO(aq) ->Na + (aq) + CH3COO-(aq)
4.22) In each reaction, the precipitate is in bold type.
a) Ni(NO3)2 (aq) +2NaOH (aq) ->+2NaNO3(aq) + NiOH2 (s) b )No precipate, therefore, no reaction.
c) Na2S(aq) + CuCH3COO (aq) -> CuS(s) + 2NaCH3COO (aq)
4.24) Write balanced net ionic equations & Identify spectator ion or ions in each reaction.
a) 2Cr3+ (aq) +3CO32-(aq) -> Cr2(CO3)3(s) ,spectators: NH4,SO42-
b) Ba2+(aq) + SO42-(aq) ->BaSO4(s), spectators: K+, NO3-
c) Fe 2+
…show more content…
Total volume does not affect cocentration
с) 0ю050 Ь РСд= 0ю050 Ь Сд-ж 0ю020 Ь СвСд2 – 0ю040 Ь Сд-
0ю0
4.76 a) The amount of AgNo3 needed is:
0.150 M X 0.1750 L = 0.02625 = 0.263 mol AgNo3
0.02625 mol AgNo3 * 169.88 g Agno3 / 1mol AgNo3 = 4.4594 = 4.46 AgNo3
Add this amount of solid to a 175 mL volumetric container , dissolve in a small amount of wáter , bring the total volumen to exactly 175 ml, and agitate well.
D) Dilute the 3.6 M HNO3 to prepare 100 mL of 0.50 M HNO3. To determine the volume of 3.6 M HNO3 needed, calculate the moles HNO3 present in 100ml of 0.50 M HNo3 and then the volume of 6.-0 Msolution that contains this number of moles.
4.88
a) HNO3(aq) + Na OH(s) -> NanO3 (aq) + H2 O (l)
b) determine the limiting reactant, then the identitiy and concentration of ions remainging in solution. Assume that the H2O (l) produced by the reaction does hnot increase the total solution volume.
12.0g NaOH * 1mol NaOH/ 40.00 g NaOh = 0.300 mol NaOH
0.200M HNO3 * HNO3 * 0.0750 L HNO3 = 0.0150 mol HNO3.
The mol ratio is 1:1, so HNO3 is the limiting reactant No excess h +, remains in solution. The remaining ions are OH- (excess reactant), Na +, and NO3- (spectators)
OH-: 0.300 mol OH – initial – 0.0150 mol OH – react

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