#### Determination of an Equilibrium Constant

Determination of an Equilibrium Constant & Analysis of an Iron Tablet Khan Salinder Results and Analysis Part One: Prepare and Test Standard Solutions Table One Beaker Number 0.0020 M SCN- (L) 0.200 M Fe(NO3)3 (L) [FeSCN2+] in Equilibrium Absorbance 0.000 0.005 0.00M 0.000 2 0.002 0.005 8.0 x 10-5 M 0.273 3 0.003 0.005 .2 x 10-4 M 0.396 4 0.004 0.005 .6 x 10-4 M 0.527 5 0.005 0.005 2.0 x 10-4 M 0.652 Sample Calculation (Beaker 2): Calculating [FeSCN2+] when Reaction "Complete" Fe3+ (aq) + SCN-(aq) ?FeSCN2+(aq) One- one ratio between SCN- & FeSCN2+ Moles of FeSCN2+ n=MV = (0.002) (0.0020) n= 0.000004 mol SCN-= 0.000004 mol FeSCN2+ Molarity of FeSCN2+ n=MV 0.000004= (M) (0.05) M=8.0 x 10-5 M Equation for Graph: y=3256.1x + 0.0049 (y=Absorption, x= [FeSCN2+]) Part Two: Prepare and Test Equilibrium Systems Table Two Test Tube Number Absorbance 0.026 2 0.111 3 0.185 4 0.235 5 0.285 Sample Calculation (Tube 2): Calculating Kc Value Calculation of [ FeSCN2+] In Equilibrium y=3256.1x + 0.0049 0.111= 3256.1x + 0.0049 0.1061=3256.1x x = 3.26 x 10-5 [FeSCN2+] = 3.26 x 10-5 Calculation of [Fe3+] & [SCN-] In Equilibrium Fe3+ + SCN-?FeSCN2+ Kc = [FeSCN2+] [Fe3+] [SCN-] [ ] Fe3+ + SCN- --> FeSCN2+ Calculation of Kc Value Kc = [FeSCN2+] = 3.26 x 10-5 .

• Word count: 474
• Level: International Baccalaureate
• Subject: Chemistry

#### Graham's law. In this experiment the relationship between the molar mass of a gas and the speed with which it diffuses was to be assessed.

Abstract In this experiment the relationship between the molar mass of a gas and the speed with which it diffuses was to be assessed. The idea we had behind finding this relationship was to first form a participate and measure the distance from this participate ring. In Graham's Law a gas's rate of effusion is inversely proportional to the square root of its molecular weight. Because of this, by finding the molar mass and distance from the participate ring the relationship can be experimentally analyzed. Based on this fact that no outside factors will influence the experiment we can directly plug our values into Graham's Law. Errors could have happened from gas escaping and can be prevented by more immediate actions. Purpose To demonstrate the relationship between the molar mass of a gas and the speed with which it diffuses. Materials In the hood: NH3 17 M, HCL 12M Cotton swabs Acetone Glass tube(60 cm in length) Meter stick Procedure First, three glass tubes were placed on a level table. Next, in the fume hood, an end of the cotton swab was soaked into the 12 M HCl and another in the 17 M NH3. After being soaked for a brief period of time each were inserted simultaneously into opposite ends of one of tubes and covered with paper towels to prevent excess air flow. This was repeated twice and a white ring participate was formed. The distance of this ring to both

• Word count: 472
• Level: International Baccalaureate
• Subject: Chemistry

#### Energy output of Bioethanol and Industrial Ethanol

Ken Chen Energy output of Bioethanol and Industrial Ethanol Aim The aim of this investigation is to determine which of the two derivatives of ethanol releases more energy, Bioethanol or Industrial Ethanol. Hypothesis The main compound in both alcohols has the same chemical formulae C2H5OH, and the other compounds are unknown therefore the Energy output cannot be calculated exactly by using standard bond enthalpies. However, as the main compound is known and the same for both ethanol's the energy output should be very similar. Therefore the difference in energy output only depends on the compounds which are blended to the alcohols. Apparatus - Thermometer - Metal can x2 - Draught shield x4 - Gloves - Safety spectacles - Heat-resistant mat - Insulation card - 100cm3 measuring cylinder x2 - Spirit burner - Clamp stand - Balance - Matches Substances - E100 Bioethanol - Industrial Ethanol - Water Method The idea to do the experiment the way described is due to knowledge from past experiments and the IB textbook. Follow these steps 10 times, five times for each alcohol. Step 1: Measure 50cm3 of alcohol in a measuring cylinder Step 2: Weigh and record the empty spirit burner Step 3: Fill the 50cm3 of alcohol in the spirit burner Step 4: Weigh and record the filled spirit burner Step 5: Measure 100cm3 of Water in a measuring cylinder Step 6: Fill the

• Word count: 471
• Level: International Baccalaureate
• Subject: Chemistry

#### Determining the Percentage Yield of Sodium Chloride produced from Sodium Bicarbonate and Hydrochloric Acid

Determining the Percentage Yield of Sodium Chloride produced from Sodium Bicarbonate Hydrochloric Acid Introduction/Purpose: Determination of the amount of sodium chloride produced from known quantities of sodium bicarbonate and hydrochloric acid. In the experiment, three trials were undertaken to determine the viability of the results. The purpose of the lab was to compare the experimental value with the theoretical value and calculate the percent yield to determine how much product was actually made compared with the amount of product that was expected. Hypothesis: The percent yield will be high, because there would be little opportunity for mass to escape, providing the expected result of the product. Material: . Sodium bicarbonate 2. Test tube rack 3. Bunsen burner 4. Electronic balance 5. Test tubes x 3 6. Sodium Chloride 7. Test tube clamp Method: At the beginning, a table has been made in order to record all the data. After the test tubes were numbered from one to three, they were weighted to 0.001 grams. Later, sodium bicarbonate was added to each test tube and they were again separately weighted. After that, the mass of sodium bicarbonate was measured in each tube. To begin the reaction, a drop of hydrochloric acid was added to each test tube separately. Then, the fluid was evaporated from each test tube by heating it above a Bunsen burner flame

• Word count: 468
• Level: International Baccalaureate
• Subject: Chemistry

#### Chemistry experiment

Chemistry Experiment ------ Industrial Discharges My experiment is about Industrial Discharges. The main idea of my experiment is to show the disadvantages of Industrial Discharges. According to my research, I found that factories letting out the polluted water to the sea, oceans and rivers. Polluting water causes many chemical changes in the water. For example, there is not enough air in the water, the fishes and plants don't have much air in the water, so they died. This is a serious problem, not only because human eat fish in the water, but also damage the balance of the environment. My experiment includes: * 2 (150ml) beakers * 50ml water * 50ml polluted water (Chemical changes in the water) * Plants which grow in the water To make my experiment more accurately, I keep the weigh of the plants (I count the numbers of leaves), the same amount of water and the same size of beakers. The details are below: Dates Colour of leaves Beaker1 Colour of leaves Beaker2 Alive leaves in Beaker 1 Alive leaves in Beaker 2 22/11/07(Thursday) All Green All Green 22 22 23/11/07(Friday) All Green All Green 22 22 26/11/07(Monday) All Green 21 Green 1 Yellow 22 21 27/11/07(Tuesday) All Green 21 Green 1 Yellow 22 21 28/11/07(Wednesday) 21 Green 1 Yellow 20 Green 2 Yellow 21 20 29/11/07(Thursday) 21 Green 1 yellow 20 Green 2 Yellow 21 20

• Word count: 465
• Level: International Baccalaureate
• Subject: Chemistry

#### The Activity Series. Aim - To determine the relative activity of several metals

The Activity Series AIM To determine the relative activity of several metals MATERIALS * Calcium metal granules (dry) * Copper metal foil * Zinc metal granules * Lead strips * Tin granules * Iron nails * Magnesium ribbon * Tap water * HCl acid * Forceps * Safety gloves * Safety glasses * Test tubes * Test tube rack * Beakers SAFETY * Do not ingest chemicals * Wear safety glasses and goggles VARIABLES: Control: Molarity and amount of HCl used, amount of water used, amount of metal used, amount of time given to react Dependent: Reactivity of each metal Independent: METHOD . Set up 8 test tubes in a test tube rack 2. Fill 3/4 of each test tube with water 3. Place some of each metal into a test tube, and observe any changes 4. If any metals react, put them on the side as they will no longer be used. 5. Dispose the contents of each test tube and refill them with 3/4 HCl acid 6. Add some of each of the metals that did not previously react into a test tube, and observe and changes 7. If any metals react, put them on the side as they will no longer be used. 8. Finally, add some of each of the materials that did not previously react into a beaker with either copper sulphate, and observe any changes. RESULTS Reaction with water Metal Observations Calcium Ca(s) + 2H2O(g) › Ca(OH)2(aq) + H2(g) Initial reaction is fast, but gradually slows

• Word count: 459
• Level: International Baccalaureate
• Subject: Chemistry

#### Separation of sand (SiO2) and salt NaCl

Separation of sand (SiO2) and salt NaCl Aim I am trying to separate SiO2 and NaCl.lso I will calculate the percentage of SiO2 and NaCl in the mixture. Apparatus/Chemicals * Weighing bottle * Analytical balance * Bunsen Burner * Side arm flask * Thick walled rubber tube * 5 grams of the mixture SiO2 and NaCl * Tripod * 50ml of H2O * Filter paper * Funnel * Spatulas * Evaporating dish * Clamp * Boss * Gauze Method . Collect all apparatus above 2. Measure approximately 5 grams of the SiO2/NaCl mixture on an analytical balance. 3. Measure 50cm3 of H2O 4. Pour SiO2/NaCl mixture into beaker with H2O. 5. Stir the solution so that the salt dissolves. 6. Set up vacuum filtration 7. Pour the mixture in the Buckner funnel. 8. Wait for the H2O to run into the conical flask. 9. Set up Bunsen burner. 0. Heat H2O left in the conical flask in an evaporating dish. 1. The H2O will evaporate leaving behind NaCl. 2. Weigh SiO2 and the NaCl. Results Weight of mixture before experiment 5.5380 After heating mixture there was SiO2 2.5221 NaCl 2.4317 2.5221+2.4317=4.9538 There was some salt and sand lost 5.5380 -4.9538=0.5894 Calculations on the percentage of SiO2 and NaCl in mixture Salt 2.4317 X100 =44% 5.5432 Sand 2.5221 X100 =45% 5.5432 Lost Sand/Salt 0.5894 X100=11% 5.5432 Conclusion My aim was to separate SiO2 and NaCl, which I did

• Word count: 456
• Level: International Baccalaureate
• Subject: Chemistry

#### Chemistry report - The aim of this practical is to observe the difference in the reactivity between Mg and Ca when hot and cold water is added.

Reactivity of Mg and Ca Introduction The aim of this practical is to observe the difference in the reactivity between Mg and Ca when hot and cold water is added. Hypothesis We think they will react in some kind of form. Variables The independent variable was the temperature of the water and also the quantity of metal use in the experiment and the dependent variable was the reaction. Equipment * Mg (s) * Ca (s) * Phenolphthalein * Beaker * Cold and hot water * Sandpaper * Test tubes with rack Procedure . The Magnesium was cleaned with sandpaper. 2. Two pieces of magnesium was put into two test tubes and two pieces of calcium was put into two test tubes. 3. Cold water was added into two test tubes one with calcium one with magnesium. 4. Two droplets of phenolphalein were added in each of the test tubes filled with cold water to observe the change in ph. 5. Water was heated over a Benson burner and poured into the two test tubes without water while observing the reactions. 6. Two droplets of phenolphalein were added in each of the test tubes filled with cold water to observe the change in ph. Observation Relative reactivity of Cold water Hot water Mg There was a minimum reaction. Fig. 1 There is a bigger reaction than with the cold water. Fig. 2 Ca The temperature of the water increases right away. Before we dropped the phenolphthalein the solution

• Word count: 455
• Level: International Baccalaureate
• Subject: Chemistry

#### Lactase Enzyme Lab Design

Lactase Enzyme Lab Introduction: Lactose, the sugar found in milk, is a disaccharide composed of glucose and galactose (both six-sided sugars). Glucose is a six-sided sugar. Lactase is an enzyme that breaks lactose down into galactose and glucose. Lactase can be purchased in pill form by people who are lactose intolerant. These people lack the enzyme, lactase, and cannot break down the sugar lactose into its component parts. Question: How does the temperature of the lactose solution affect the process that Lactose goes through to break down Lactose? Hypothesis: The higher the temperature the less time it takes lactase to break down lactose into Galactose and glucose. Manipulated Variables: Temperature of the lactose solution Responding Variable: The amount of time it takes lactase to break down lactose into galactose and glucose. Controlled Variables: Type of reagent strips; Type of Thermometer; Amount of milk per trial; Type of Lactose drops; Room Temperature; Air Pressure Prediction: The milk with the most temperature (100F) will take the least time to get broken down into galactose and glucose. This is because rise in temperature increases activity of the enzyme, as the higher the temperature the more kinetic energy the enzyme and the substrate molecules have. This means that more enzyme substrate complexes are likely to form. The increase in movement of the

• Word count: 454
• Level: International Baccalaureate
• Subject: Chemistry

#### Synthesis of FeS

SYNTHESIS OF FeS Seren Sapmaz 582 ATC Purpose: To synthesize iron (II) sulfide from iron and sulfur. Materials: * * Iron filings * Sulfur powder * A test tube * Test tube holder * Bunsen burner * A magnet * An equal arm balance * A box of matches * Ethyl alcohol * A set of masses Procedure: . Weigh 2.1g of iron powder and 1.2g of sulfur powder. Put them in different places. 2. First take a magnet and approach to iron fillings. The magnet will attract the iron filings. 3. Then take a little bit sulfur and put in into water. Observe what you see. Sulfur dissolves in water. 4. After that put the iron filings and rest of sulfur in a test tube and mix well. 5. Hold the test tube with the holder in inclined position, and heat gently. Take away the burner as soon as the mixture starts to glow. When reaction stops, heat it strongly for two more minutes. 6. Take the substances out of the test tube onto the watch glass. If necessary, break the test tube to remove the contents from it. 7. Watch the substance formed. Examine the effect of the magnet on the substance. Is it attracted by magnet? Crush the substance. Add a spatula measure of the substance to about 10cm3 of ethyl alcohol. Close the test tube with a stopper and shake it. Does it dissolve? 8. Remember that iron is attracted by magnet and sulfur dissolves in ethyl alcohol. Did the properties of

• Word count: 453
• Level: International Baccalaureate
• Subject: Chemistry