February 4 - Lab 5B

So today we did lab 5b, which had to do with chemical reactions.

The 4 main chemical reactions that were observed in this lab were synthesis, decomposition, single replacement, and double replacement.

The purpose of the lab was to :

1. observe a variety of chemical reactions
2. interpret and explain observations with balanced chemical equations
3. to classify each reaction as one of the four main types

So in this lab, we did seven different experiments, and each one would be classified as a synthesis, decomposition, single replacement, or double replacement reaction.  

Reaction 1: burning a copper wire.

Reaction 2: putting an iron nail in copper (II) sulfate solution.

Reaction 3: heating up copper (II) sulfate pentahydrate in a test tube.

Reaction 4: add water to product of reaction 3.

Reaction 5: combine calcium chloride with sodium carbonate.

Reaction 6: put mossy zinc into hydrochoric acid solution.

Reaction 7: mix hydrogen peroxide solution with manganese (IV) oxide.  Then put glowing splint of wood into mouth of test tube.

February 2 - Chemical Reactions Cont'd

We learned about the rest of the chemical reactions today which were: Double Replacement, Combustion, and Neutralization.


Double displacement: This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These reactions are in the general form: AB + CD ---> AD + CB

One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium iodide to form lead (II) iodide and potassium nitrate:

Pb(NO₃)₂ + 2 KI ---> PbI₂ + 2 KNO₃

Combustion: A combustion reaction is when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic, meaning they produce heat:
AB + O2  --> AO + BO

An example of this kind of reaction is the burning of napthalene:

C₁₀H₈ + 12 O₂ ---> 10 CO₂ + 4 H₂O

Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H⁺ ion in the acid reacts with the OH^- ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water:            HA + BOH ---> H₂O + BA

One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide:HBr + NaOH ---> NaBr + H₂O


1)  Does your reaction have oxygen as one of it's reactants and carbon dioxide and water as products? If yes, then it's a combustion reaction
2) Does your reaction have water as one of the products? If yes, then it's an acid-base reaction
3) If you haven't answered "yes" to any of the questions, then you've got a double displacement reaction

Vids have low volume so turn it up a bit

Links that might help http://www.visionlearning.com/library/module_viewer.php?mid=54

January 27 - Types of Chemical Reactions

So today we learned about 3 types of chemical reactions : Synthesis, Decomposition, and Single Replacement.

Synthesis:  A synthesis reaction is when two or more simple compounds combine to form a more complicated one. These reactions come in the general form of: A + B --> AB


Eg. 8 Fe + S₈ ---> 8 FeS


Decomposition:A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions come in the general 
form: AB --> A + B


Eg.2 H₂O ---> 2 H₂ + O₂




Single Replacement:This is when one element trades places with another element in a compound. These reactions come in the general form of: A + BC ---> AC + B


Eg. Mg + 2 H₂O ---> Mg(OH)₂ + H₂


Ask these questions to figure out which type of chemical reaction it is:

1. Does your reaction have two (or more) chemicals combining to form one chemical? If yes, then it's a synthesis reaction.

2. Does your reaction have one large molecule falling apart to make several small ones? If yes, then it's a decomposition reaction.

3. Does your reaction have any molecules that contain only one element? If yes, then it's a single replacement reaction.

January 25 - Balancing Equations

So today we learned how to balance equations. In case you forgot how to, here are the steps to do it.


1. Write the "skeletal" equation. This is just what will react; forget the coefficients for now. However, do remember to balance the charges. For example, if you were reacting copper(II) nitrate with Sodium, you would write:
Cu(NO₃)₂ + Na --> NaNO₃ + Cu

2. Determine how much of each atom or ion you have on both sides of the equation. In the example, there is 1 copper atom, 2 nitrate ions, and 1 sodium atom being reacted. Meanwhile, there is 1 sodium ion, 1 nitrate ion, and 1 copper atom being produced.

3.Figure out if what is reacting is the same as what is produced. If it is the same, your equation is already balanced, and you don't need to do anything else. In the example problem, however, there is 1 more nitrate atom that is reacted than produced, which means coefficients will have to be added.

4. Add coefficients to the element or ion that needs to be increased. For example, since you needed another nitrate ion, you'll need to add a 2 to NaNO₃ so you'll have 2 nitrate ions. You'll now have the following:

Cu(NO₃)₂ + Na --> 2NaNO₃ + Cu

5. Continue adding coefficients until you have the same amount of atoms on each side of the equation. For example, by placing a 2 by NaNO₃, there are now 2 sodium atoms, but only 1 on the reactant side. A 2 would have to be added to Na in the reactants so that there are 2 Na's on each side of the equation. The final equation would be this:

Cu(NO₃)₂ + 2Na --> 2NaNO₃ + Cu


A balancing equations game!!!!!<----click

January 11

January 8

January 5 2011---Molarity

Molarity is a measure of concentration of a solute in a solution or of any chemical species in terms of amount of substance in a given volume.  The unit for molarity is mol/L.  A solution of concentration 1 mol/L is also denoted as 1 molar (1 M). 


To calculate molarity you use this forumula : Molar Concentration (M) = moles of solute(mol)/Volume of solution (L)

Eg.


1. How many grams of solute are present in 175 mL of 0.67 M glucose (C₆H₁₂O₆)?

0.67M/L * 0.175L * 180g/mol = 21.11g

2. How many grams of MgBr₂ should be dissolved in 275 mL of solution to produce a 0.32 M solution?

0.32M/L * 0.275L * 180g/mol = 15.84g

3. What volume of 12.4 M HCl would you need to make 500.0 mL of 3.50 M HCl?

(12.4M) * V = (0.500L)(3.50M)
V = .141 L = 14.1 mL

4. How many mL of 0.24 M KCl solution contains 0.56 g of KCl?

0.24 M * 0.56 g * 1 mol/74g = .0018 mL


Video!!!!