Chemistry

Topics: Ion, Chemistry, Ammonia Pages: 5 (1009 words) Published: October 3, 2013
Unit 1 – Lesson 1
Chemistry Key Questions
1.
Element
Standard Notation
# Of Electrons
# Of Protons
# Of Neutrons
Sodium
23 11 Na
11
11
12
Chromium
24 Ca
26
26
27 ?
Phosphorus
32 P
15
15
16

2. Fireworks; Which metals burn to produce these colours?
Red – Lithium (Sr)
Blue – Cesium (Cs)
Pink – Potassium (K)
Yellow/Green – Copper (Cu)
Orange – Carbon (Ca)
3. Let’s say that the three bottles contain Fe(NO3)3, BaCl, and AgNO3. These are Iron (III) Nitrate, Barium Chloride, and Silver Nitrate. All of these are white solids that dissolve in water. We know that silver chloride is highly insoluble in water. So if we reacted an aqueous solution of each of the chemicals with a dilute solution of hydrochloric acid (HCI), the one with Ag+ in it will form AgCl and precipitate as a solid. The other two samples will give no chemical reaction. We have now identified BaCl. We know that Barium Sulfate is highly insoluble in water. So if we reacted an aqueous solution of each of the chemicals with a dilute solution of sulfuric acid (HS204), the one with the Ba2+ in it will for BaS04 and precipitate as a solid. The other two samples will give no chemical reaction. We have now identified which powder is BaCl. We know that Iron (III) hydroxide is insoluble in water. So if we reacted an aqueous solution of each of the samples with a dilute solution of sodium hydroxide (NaOH), the one with the Fe3+ in it will form Fe(OH)3 and precipitate as a solid. The other two samples will give no chemical reaction. So we have now identified which powder is Fe(NO3)3. We have now designed a qualitative analysis experiment in which we have matched up the identities of the three starting solids.

Unit 1- Lesson 2
Chemistry Key Questions
4.
Name
Formula
Cobalt (II) Nitrate
Co(NO3)2
Calcium Phosphate
Ca 3 (PO 4) 2
Disphosphorus trioxide
P203
Ammonium Sulfate
(NH 4) 2 (SO 4)
Gold (III) hydroxide
AuH3O3
Platinum (II) Oxide
PtO
Gallium carbonate
Ga2(CO3)3
Magnesium Chloride
MgCl 2
Aluminum nitrate
Al(NO3)3
Sulfur Dioxide
SO 2

5. a) CaCl 2 + Al 2 (SO 4) 3 -> AlCl3 + 3CaS04(s) double displacement b) Al 2 O 3 -> 4 Al+3 02 decomposition
c) CuO + Cl 2 -> 2CuC;2+ O2 single replacement
d) Ca + O 2 -> CaO composition

6. a) 2Fe203+ 3C -> 3 CO2+ 4e
b) This is a single replacement reaction because carbon is taking the place of iron in the product. c) C replaces Fe
d) C is a nonmetal and Fe is a metal
Unit 1 – Lesson 3
Chemistry Key Questions
7. a) Sodium Chloride and Silver Nitrate – NaCl(aq) + AgNO 3 (aq) -> AgCl(s) + NaNo 3 (aq) White precipitate pf AgCl forms.

b) Coppe (II) Chloride and Sodium Nitrate – CuCl 3 (aq) + 2NaNo 3 (aq) -> Cu(No3)2 (aq) + 2NaCl(aq) No precipitate, therefore no reactions occurs.

c) Potassium hydroxide and Ammonium Chloride – KOH(aq) + NH 4 Cl(aq) -> KCl (aq) + NH4OH(aq) No reaction.

8. Iron is removed from municipal water supplies by a process known as oxidation. Soluble iron ions in water are in the ferrous oxidation state (Fe2+). When the iron is oxidized, it is converted to the ferric oxidation state (Fe3+). The ferric iron readily combines with hydroxide ions in the water to form iron hydroxide:

Fe3 + 3(OH) -> Fe(OH)3

9. Bottles 1 and 3 contain cations and bottles 2 and 4 contain anions. Our choices are Pb2+, Cl-, Ag+ and (S4)2-. We are mixing bottles 1 and 2, 1 and 4, and 2 and 3. The results for all three are white insoluble precipitates. We know that we are mixing cations with anions to get new inorganic salts. We know that lead (Pb2+) forms many inorganic salts that are not water soluble. Mixing Pb2+ with Cl- will make PbCl4 and mixing Pb2+ with (SO4)2- will make Pb(SO4). Both of these are white solids that are not water soluble. So we could label bottle 1 Pb2+, bottle 2 Cl, and bottle 4 (SO4)2-. By process of elimination, this would make bottle 3 Ag+. We must also mix bottles 2 and 3 which would be mixing Cl- and Ag+ to give AgCl. This is a white...
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