Lab+3+Chemistry


 * LABORATORY OVERVIEW**
 * Reading the Periodic Table**
 * Element || Proton || AMU ||
 * Nitrogen || 7 || 14.00674 ||
 * Oxygen || 8 || 15.9984032 ||
 * Aluminum || 13 || 26.981538 ||
 * Chlorine || 17 || 35.4527 ||
 * Copper || 29 || 63.546 ||

Molar Weight (AMU) of Na = __22.989770__ Molar Weight (AMU) of Cl = __35.4527__ MW of NaCl = __58.4425__ If you want 0.5M of NaCl, you would add 1/2 a mole of salt or __29.2212 g__ to 1 liter of water. If you want 0.1M of NaCl, you would add 1/10 a mole of salt or __5.84425 g__ to1 liter of water. MW of Cu = __63.546__ MW of N x 2 = __28.0135__ MW of O x 6 = __95.9904__ MW of Cu(NO3)2 = __187.55__ 0.2 or 1/5 grams of Cu(NO3)2 is = __37.51__ MW of H = __1.00794__ MW of Cl = __35.4527__ MW of HCl = __36.4606 g__ MW of Na = __22.989770__ MW of O = __15.9984032__ MW of H = __1.00794__ MW of NaOH = __39.9961__ Covalent Molecular Weight Calculator **Here**
 * Calculating Molar Solutions**
 * a) NaCl**
 * b) Cu(NO3)2**
 * c) HCl**
 * d) NaOH**

2 Hydrogen 1 Oxygen 1 Mole of Na reacts with 1 Mole of Cl. 2 Moles of Hydrogen for 1 Mole of Oxygen.
 * Balancing Chemical Equations**
 * a)** How many Hydrogen and Oxygen are there on each side?
 * b)** How many Moles of Na sodium reacts with how many Moles of Cl?
 * c)** How Moles of Hydrogen react with how many Moles of Oxygen?

//Chemical reaction being observed// How do you know? What do you observe? Both sides of the equation are equal. atoms on left ||  || # atoms on right ||  || c)What is a mole? A __MOLE__ is 6.022 x 10*23 atoms or molecules of something. This number is called **__AVOGADRO'S NUMBER__**. moles ||  || # moles ||   ||
 * 1. Cu(NO3)2 +2NaOH -> Cu(OH)2 + 2NaNO3**
 * a)**Which of the two product is the blue precipitate?
 * Cu(OH)2** is the blue precipitat because it is insouble. This is what creates the precipitate in the first place. This type of reaction is called precipitation reaction. For more info click here.
 * b)**Compare the number of each type of atom on the left and right sides of the equation.
 * 1 || Cu || 1 || Cu ||
 * 2 || N || 2 || N ||
 * 8 || O || 8 || O ||
 * 2 || Na || 2 || Na ||
 * 2 || H || 2 || H ||
 * 1:9 || Cu(NO3)2 || 1:5 || Cu(OH)2 ||
 * 1:3 || NaOH || 1:5 || NaNO3 ||

//Chemical reaction being observed// atoms on left ||  || # atoms on left ||  || The equation is balance on both sides. The light blue because H2O is the water that is seen at the top which is less dense then **CuO**.
 * 2. Cu(OH)2 -> CuO + H2O**
 * 1 || Cu || 1 || Cu ||
 * 2 || O || 2 || O ||
 * 2 || H || 2 || H ||
 * a)** Is the equation balanced?
 * b)** What color is **CuO**?

//Chemical reaction being observed// The precipitate mixes back into the solution. The precipitate disappears. The solution changes into a clear blue substance. atoms on left ||  || # atoms on right ||  || NO New balanced equation.
 * 3. CuO + HCl -> CuCl2 + H2O**
 * a)**What happens to the precipitate?
 * b)**What color do you observe?
 * 1 || Cu || 1 || Cu ||
 * 1 || O || 1 || O ||
 * 1 || H || 2 || H ||
 * 1 || Cl || 2 || Cl ||
 * c)**Is the equation balanced?
 * CuO + 2HCl --> CuCl2 + H2O **

//Chemical reaction being observed// The foil began to turn a reddish brown color with a little black in there. Black atoms on left ||  || # atoms on right ||  || No New balanced equation.
 * 4. CuCl2 + Al -> AlCl3 + Cu**
 * a)**What changes occur to the aluminum foil?
 * b)**What color is the new compound?
 * 1 || Cu || 1 || Cu ||
 * 2 || Cl || 3 || Cl ||
 * 1 || Al || 1 || Al ||
 * c)**Is the equation balanced?
 * 3CuCl2 + Al2 -> 2AlCl3 + Cu3 **

What chemical reactions occurred and how do you know? The first reaction is a precipitation reac because the Sodium Nitrate forms a think substance with in the solution. The second reaction is a polymerization reaction because of the removal of water on the right side of the equation, and in that case reaction three can be considered polymerization as well. The last reaction is an oxidation or redox reaction because one can see that the aluminum in tarnishing or being coded by the copper in the formula.
 * Discussion**