Today we started class off by reviewing Lindsey’s blog from the previous day. Her blog can be used as a reliable reference if you need help with the five different reaction types. mr H also mentioned referencing packet pages 5 and 6 for pictures and definitions of the reaction types. as a review, the five reaction types are: synthesis, decomposition, combustion(REMEMBER O2!!!), single replacement, and double replacement. these reaction types were then used to determine the type of reaction on packet pages 7 and 8. the answers are as followed:
[page 7]
1. DECOMPOSITION
2. COMBUSTION/SYNTHESIS
3. SYNTHESIS
4. DECOMPOSITION
5. DECOMPOSITION
6. DOUBLE REPLACEMENT
7. SINGLE REPLACEMENT
8. SINGLE REPLACEMENT
9. SKIP!!!
10. DOUBLE REPLACEMENT
[page 8]
11. DOUBLE REPLACEMENT
12. SKIP!!!
13. COMBUSTION
14. DECOMPOSITION
15. SINGLE REPLACEMENT
16. SINGLE REPLACEMENT
17. DOUBLE REPLACEMENT
18. DECOMPOSITION
19. COMBUSTION
20. SYNTHESIS
21. SINGLE REPLACEMENT

After determining type of reaction the chemical equations were, we moved on to the bottom of packet page 9. On this page, there were two questions giving the verbal chemical equation. Our task was to put the verbal chemical equation into symbols. mr. H told us to find the reactants, find the products, and then balance equation. Problem number 31 states: Mercury(II) oxide decomposes into its elements. By using our knowledge from previous units, we were able to determine that the formula for mercury oxide is HgO, because they both have a charge of two which cancels out. next, we knew that decomposes means that the two elements (Hg and O) would separate. Keeping mr. H’s UNCLE HONClBrIF in mind, specifically his cousin olive, we were able to determine that the oxygen would be O2 since all of mr. H’s uncle’s children have two of arms, legs, noses, mouths, etc. since the element oxygen was labeled O2, we would have to balance the equation out by putting a 2 in front of HgO and a 2 in front of Hg for the product. the final answer for question number 31 is: 2Hg0–>2Hg+02. the next problem (#32)follows the same criteria as #31. the problem states: Iron(II) oxide is synthesized from its elements. The reactants would both be the separated elements, because the equation is synthesized. with cousin Olive in mind again, the reactants would be Fe+O2 with products of FeO. Because there is two O elements to begin with, you would make the FeO into 2FeO. with changing the product, you would also have to change the reactant Fe to 2Fe. This makes the final answer: 2Fe+O2–>2FeO.

After these brief packet pages, we worked on the rest of our Classifying Chemical Reactions Lab (CR1). we were able to finish the eight reactions and afterwards, we shortly discussed the reactants and products for each station.
They were:
Reaction #1 R: Mg2 and O2 (synthesis)
Reaction #2 R: 2HCl and 2Mg P:H2 and 2MgCl (single replacement)
Reaction #3 R:NH4CO3 (decomposition)
Reaction #4 R: CaCO3 and 2HCl P: CaCl2 and CO2 and H2O (decomposition)
Reaction #5 R:CuCl2 and Zn P:ZnCl2 and Cu (single replacement)
Reaction #6 R:CuCl2 and Na3PO4 P:NaCl and Cu3(PO4)2 (double replacement)
Reaction #7 R:NaHO and HCl P:NaCl and H2O (double replacement)
Reaction #8 (my group didn’t get there yet)

This concluded our day!
HOMEWORK:
due friday–4.2 Reading sheet W.A.
due monday–Web Assign (one in class)
due tuesday–Web Assign
due wednesday–web assign and test prep

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