Acid Rain
Acid Rain Acid rain is basically defined as any kind of precipitation that has a pH-level lower than 5.5 (? 5.5), meaning that is more acidic than the usual precipitation. Precipitation is by nature already slightly acidic, because of the natural carbon dioxide in our atmosphere gets dissolved into the rain. The aqueous solution of CO2 is than Carbonic acid, H2CO3: Co2 + H2O - H2CO3. H2CO3 is as an aqueous solution not stable, and thus it ionizes into water, forming the ions H3O+ and HCO3-: CO2 + H2O - HCO3- + H3O+.When the precipitation, which can be for example rain, snow or fog, has a pH-level lower than 5.5 it is considered acidic and is due to human output of nitrogen and sulphur compounds. They both go through many complex steps until they become sulphuric acid (H2SO4) and Nitric acid (HNO3), which one would find in Acid rain. sulphuric acid originates from its compound. The sulphur dioxide reacts with Oxygen to make sulphur trioxide: 2SO2 +O2 -> 2SO3. Than the sulphur trioxide reacts with water to make sulphuric acid: SO3 + H2O -> H2SO4. Nitric acid is made when nitrogen dioxide reacts with water, which also makes nitrous acid. The pH-level of an acidic rain cloud can reach up to 2.6. In areas with a lot of industry the pH-level will be about 4; in Los Angeles the most acidic precipitation had a pH-level of 3. However, in some parts of the world, such as north
Chemistry Revision - ATOMIC STRUCTURE REVIEW TOPIC 2
ATOMIC STRUCTURE REVIEW – TOPIC 2 State the position of the subatomic particles, their relative masses and charges. Protons: 1 atomic mass unit & + charge Neutrons: 1 atomic mass unit & 0 charge Electrons: 1/1836 atomic mass unit & - charge Define the terms of mass number [A] and atomic number [Z] and be able to identify isotopes and ions of elements. AZE Notation: The number at the top left represents the mass number and the bottom left the atomic number Mass Number: The total number of protons and neutrons in a nucleus Atomic Number: The number of protons in the nucleus of an atom, which determines the chemical properties of an element and its place in the periodic table. EXAMPLE: 56 Fe 26 – The mass number = protons + neutrons = 56 The atomic number = number of protons = 26 Therefore the number of neutrons = 56-26 = 30 Isotopes are atoms of the same element with different mass numbers (atoms of the same element must contain the same number of protons, but can contain a different number of neutrons) Be able to identify protons, neutrons and electrons in isotopes and in ions. For a given element the number of protons never changes The number of neutrons depends on the mass number of the isotope in question The number of electrons is the same as the number of protons UNLESS we are dealing with an ion! Compare properties of isotopes of an element Since
Experiment Plan. Chemistry IA: Electrolysis of Metal Sulphate solutions (NiSO4)
Chemistry IA: Electrolysis of Metal Sulphate solutions (NiSO4) Introduction Electrolysis is the chemical decomposition of a compound by applying an electric current through a solution containing ions. Electrolytes are required to conduct electricity. They must be dissolved in water or in molten state for the electrolytes to conduct because then, the ions are free to move allowing the solution to be electrolyzed.[1] In electrolysis, reduction happens at the cathode whilst oxidation happens at the anode. Reduction is the loss of electrons and oxidation is the gain of electrons. Research Question In this experiment, I will be electrolyzing nickel sulphate (NiSO4) solution. To further explain the aim of this experiment, I have formulated a research question: “How does changing the current affect the mass of nickel deposited at the cathode in the electrolysis of nickel sulphate?” Hypothesis I predict that as the electrical charge increases, the mass of nickel deposited at the cathode after electrolysis will also increase. Faraday’s law of electrolysis, which investigates the quantitative relationship on electrochemical, can support this. Faraday’s law states, “The amount of the substance produced by current at an electrode is directly proportional to the quantity of electricity used”.[2] During this electrolysis experiment, the aqueous solution of Nickel
Research into the production of Nitrate Fertillisers.
Part 1: stimulus material: Research and collecting secondary data. Advantages of Ammonium Nitrate Fertilizer Ammonium (NH4+) and nitrate (NO3-) are the two forms in which plants can directly utilize nitrogen from the soil. Upon application of ammonium nitrate fertilizer, it easily dissociates into these two forms. When urea-based fertilizers are used, conversion of urea to ammonium occurs for a period of one day to one week and some amount of nitrogen is lost to the atmosphere in the process. Thus, the immediate availability of nitrate ions makes ammonium nitrate preferable over urea-based ones. Moreover, ammonium nitrate is a stable fertilizer and is not subject to losses due to volatilization. Ammonium nitrate, is widely used in mining and construction industries, in the form of dry blasting agents. Being available in the crystalline dry forms, it is easy to handle and load. Moreover, they are safe and cheap as compared to other blasting agents. Disadvantages of ammonium Nitrate Fertilizer As I said earlier, ammonium nitrate is a strong and explosive agent. In fact it is one of the largest industrial explosive manufactured in the US. Ammonium nitrate is used in the fields of quarrying and mining. Ammonium nitrate is one of the cheapest crop nourishing fertilizer types and hence, it is easily available in the markets and agricultural stores too. Chemical formula for
Chem lab report shampoo experiment
Chemistry Lab Report-6 Shampoo test Teacher’s Instruction: Investigate the Chemical Properties different shampoo brands and determine the best brand of shampoo manufacturers. Apparatus Required: Beaker Measuring cylinder Boiling tube pH meter Reagents Used: Shampoo Ink Distilled water Procedure: Test A (determination of pH): Take your 1% shampoo solution in a beaker and using the pH meter find the pH value of the shampoo. Information: Most shampoos are neutral or slightly acidic. Acidic solutions cause the cuticle of the hair to shrink and lay flatter on the shaft of the hair. Basic solutions cause the cuticle to swell and open up. Acidic solutions make the hair seem smoother. Basic solutions make their hair seem frizzier. Test B (shake test- determination of foam formation): Put approximately 10ml of the 1% shampoo solution into a 250ml graduated cylinder and record the initial volume of the solution. Cover the cylinder with your thumb and shake 10 times. Record the total volume of the contents after shaking. Then calculate the volume of the foam only by subtracting the initial volume of the solution without the foam. Information: The smaller the bubbles the better the shampoo. Test C (foam quality and retention): This test should be done together with the previous test. After preforming the previous test record the amount of foam in each minute
Chemistry Investigation to find the Empirical Formula of Magnesium Oxide
Empirical Formula of Magnesium Oxide 24/14/2012 Empirical Formula of Magnesium Oxide ________________ Data Collection and Processing Diagram Set-up Quantitative observation: During the experiment the crucible used was slightly damaged however did not continue to break or cause any inconvenience to the investigation. For approximately ten minutes, no reaction was visible. The substance contained no odour. After approximately fifteen minutes the piece of magnesium commenced to burn with an overall duration of twenty minutes. When burnt, the magnesium converted to bright orange colour. During the reaction the lid of the crucible was lifted to see any changes or if the magnesium had combusted, and whilst doing this white smoke escaped the crucible which may perhaps be magnesium oxide. By the completion of the reaction, it was noticed that not all of the magnesium was combusted and left some silvery remnants. Raw Data Table 1 – Raw data of the total weight of the crucible including magnesium Trial 1 Group Weight of crucible ± 0.001 (g) Total weight before ± 0.001 (g) Total weight after ± 0.001 (g) 1 37.140 37.250 37.238 2 37.841 37.924 38.038 3 33.405 33.520 33.613 4 33.834 33.979 33.971 5 39.133 39.264 39.322 6 32.784 32.823 32.844 Trial 2 Group* Weight of crucible ± 0.001 (g) Total weight before ± 0.001 (g) Total
Chemistry: Strong Acid and Weak Base Titration Lab
Chemistry: Strong Acid and Weak Base Titration Lab Cherno Okafor Mr. Huang SCH4U7 November 21st, 2012 Data Collection and Processing Concentration of the standard HCl solution: 0.1 M Data Collection: Trial 1 Trial 2 Trial 3 Final HCl Buret Reading ± 0.05 mL 38.3 45 54.5 Initial HCl Buret Reading ± 0.05 mL 29.9 38.3 45 Volume of NaHCO3 used ± 0.1 mL 9.2 9.5 9.8 Qualitative Data: I used the methyl orange indicator which was suitable for my titration because of its clear and distinct colour change from orange to a bright red at the endpoint At the beginning of the titration after I added 3 drops of methyl orange into the base (NaHCO3) and swirled, I began titrating the acid (HCl) slowly, and initially in the methyl orange and base, there was a tiny amount of red colour present, but then it quickly disappeared due to insufficient HCl (H+ ions)then I gradually kept titrating more acid while swirling and there was even more red colour present, until finally I reached the endpoint when the orange-yellow colour had completely transformed into a red colour Changes from an orange-yellow colour (slightly higher pH 4.4) to a bright red colour (at low pH 3.1) at the endpoint and point of equivalence Baking Soda (NaHCO3) absorbed the odour caused by the strong acid of HCl when I mixed the two: bleach-like smell Processing If the concentration of
Measuring the fatty acid percentage of the reused sunflower oil after numerous times of potato frying and determining the effects of it on human health.
ASLAN Özge Cemre D129077 CHEMISTRY EXTENDED ESSAY “Measuring the fatty acid percentage of the reused sunflower oil after numerous times of potato frying and determining the effects of it on human health.” Özge Cemre Aslan D1129077 Session: May 2010 Supervisor: Serenay Tarhan Güler TED Ankara College Foundation High School ASLAN Özge Cemre D129077 TABLE OF CONTENTS Abstract Introduction…………………………………………………………………………………….1 Research Question……………………………………………………………………2 Background Information………………………………………………………………3 Chemistry of Fatty Acids and Oils Information on Effects of Fatty Acids and Oils Experiment Materials…………………………………………………………………………………6 Preparation of KOH and phenolphthalein…………………………………………… 7 Method……………………………………………………………………………………8 Preparing the sunflower oil……………………………………………………………9 Determining the percentage of free fatty acids in the sunflower oil by acid-base titration…………………………………………………………………………………………10 Data
Investigation of Enthalpy Change When Mg is added to HCl
Investigation of Enthalpy Change When Mg is added to HCl Defining the Problem and Selecting Variables Research Question: How is the enthalpy change of the reaction between Mg and HCl affected when different amounts of Mg are added to a constant amount of HCl? Variables: Independent: Mg Dependent: Temperature Control: HCl Controlling Variables I will use three different masses of Mg: 0.02g ± 0.001g, 0.03g ± 0.001g, 0.04g ± 0.001g I will measure the temperature using a colorimeter made from a foam cup, a lid, and a thermometer I will keep HCl at a constant volume of 10mL of 1M HCl Procedure: Step 1: Measure out 10mL of 1M HCl Step 2: Pour HCl into foam cup and place lid on cup Step 3: Measure the initial temperature and record data Step 4: Add in 0.02g of Mg into cup and measure the finial temperature Step 5: Calculate change in temperature by subtracting the initial temperature from the finial temperature Step 6: Repeat steps 1-5 8 more times for a total of 9 trails changing the mass of Mg in step 4 every 3 trails from 0.02g ± 0.001g to 0.03g ± 0.001g to 0.04g ± 0.001g. Data Trail/Mg Mass Trail 1 0.02g ± .001g Mg Trail 2 0.02g ± .001g Mg Trail 3 0.02g ± .001g Mg Trail 4 0.03g ± .001g Mg Trail 5 0.03g ± .001g Mg Trail 6 0.03g ± .001g Mg Trail 7 0.04g ± .001g Mg Trail 8 0.04g ± .001g Mg Trail 9 0.04g ± .001g Mg Initial Temp 22°C
Determining Ka by the half-titration of a weak acid
Determining Ka by the half-titration of a weak acid To get the Ka of acetic acid, HC2H3O2 I will react it with sodium hydroxide. The point when our reaction is half-titrated can be used to determine the pKa. As I have added half as many moles of acetic , as NaOH, Thus, OH- will have reacted with only half of the acetic acid leaving a solution with equal moles of HC2H3O2 and C2H3O2-. Then I will use the Henderson-Hasselblach equation to get pKa. CH3COOH + NaOH H2O + NaCH3COO Results: Below is a table that summarizes our results for the reaction of 1M of acetic acid with 1Molar of NaOH which 50cm3 was used. The table shows the PH record at ½ equivalence and at equivalence. We also recorded the observations we saw during the reaction. PH ±0.1 Qualitative observations At ½ equivalence 5.0 When I recorded this, as we slowly added NaOH to the acid, there was a change of color from colorless to a very slight pink as the Phenolphthalein indicator changed color. At equivalence 8.9 As I added the acetic acid to 250 cm3 of reaction mixture, there was no color change. Also as we measured the PH, the PH changed slowly but then changed very quickly at the solution approached equivalence. At this time, the indicator turned pink, when equivalence was reached Calculating the PKa To calculate PKa, we will use the Henderson-Hasselbalch equation. Hence the