#### experiment Hess Law

Name : Mohd Haziq Al-Hakim Bin Hamirruddin Class : M08F Date : 9 July 2009 Practical : 16 Data Collection and Processing Part A:- Mass of magnesium ribbon = 0.2836 ± 0.0001g Volume of 0.5 M of HCl = 50.0 ± 0.5 cm3 Reactant Initial Temperature, ± 0.5°C Highest Temperature, ± 0.5°C Mg + HCl 27.0 40.5 Table 1 shows the initial and highest temperature for reaction between Mg and HCl To calculate the change in temperature, the following equation will be used:- Reactant Change in Temperature, ± 1.0°C Mg + HCl 3.5 Table 2 shows the change in temperature for reaction between Mg and HCl Equation for part A:- Mg (s) + 2HCl (aq) --> MgCl2 (aq) + H2 (g) Mole of magnesium Mole of hydrochloric acid = mass = MV RMM 1000 = 0.2836 g = 50.0x0.5 24.3 1000 = 0.0118 mol = 0.025 mol Thus, the limiting reagent is magnesium Heat released in the reaction can be calculated by the formula: Q = mc?T Whereby; m = Mass of the solution c = Specific heat of water (4.2 Jmol-1°C-1) ?T = Change in temperature Enthalpy change of the reaction can be determined by the formula:- ?H = Amount of heat released/absorbed, Q Number of moles of reactant, n Calculation: Heat change, Q = mc?T = 50g × 4.2 Jmol-1°C-1 × 13.5 °C = 2835 J ?Hrxn = -2835_ 0.0118 = -240.25 kJmol-1 Uncertainties: Mass = = 1.0 % Temperature = = 7.4

• Word count: 762
• Level: International Baccalaureate
• Subject: Chemistry

#### Rates of Reaction Lab

Rate of Reaction Design Experiment Effect of concentration on the rate of reaction between a metal and an acid Introduction In chemistry, chemical kinetics is the study of the factors affecting the rate of a chemical reaction. By definition, rate is the increase in the concentration of one of the products per unit time or decrease in the concentration of one of the reactants per unit time. Many factors trigger the rate of reaction, such as concentration, surface area and temperature. I will investigate the effect of concentration on the rate of reaction between a metal, zinc (Zn), and an acid, hydrochloric acid (HCl). According to the collision theory and based on my experimental results, I will prove my following hypothesis to either be correct or incorrect in theory. Design (D) Aim To investigate the effect of concentration of HCl on the rate of reaction of zinc (Zn)) by measuring the volume of hydrogen produced. Hypothesis As the concentration of the hydrochloric acid (HCl) is increased, the rate of reaction per unit time will increase up to a certain concentration too, until an increase in the concentration of the acid will no longer effect the reaction rate. According to the collision theory, the more concentrated the reactants the more collisions there will be per second per unit volume. As the reactants get used up, their concentration decreases. This

• Word count: 2387
• Level: International Baccalaureate
• Subject: Chemistry

#### Teflon. This essay will discuss the uses of Teflon, and the myths and truths surrounding it.

STAGE 2 CHEMISTRY SOCIAL & ENVIRONMENTAL RELEVANCE TASK 2 CORNERSTONE COLLEGE Name: NATALIE BAMPTON Topic: TEFLON INTRODUCTION: Teflon is also known as polytetrafluoroethylene and can be found in cookware, electronics, aeronautics and clothing. It was invented by accident by Roy Plunkett, who was originally supposed to invent a CFC gas refrigerant for the company DuPont(tm). He was investigating a substance called TFE as an alternative to toxic refrigerants such as ammonia and sulphur dioxide. Since TFE is expensive, he froze it to keep it safe. While experimenting one day, he tried to empty one of the supposedly full canisters of TFE, when nothing came out. He opened the canister to find the TFE had polymerised into a white powder, because of the high pressure and low temperature. The waxy substance was Teflon; it was slippery, non-corrosive and chemically stable, which was a breakthrough for DuPont(tm) and the rest of the world. However, recent studies have shown that Teflon may be toxic or is even likely to cause cancer. This essay will discuss the uses of Teflon, and the myths and truths surrounding it. CHEMISTRY: Polytetrafluoroethylene is a long chain of tetrafluroethylene monomers bonded to each other. It is a carbon chain backbone with two fluorine atoms attached to it. In a monomer tetrafluroethylene unit, a carbon is double bonded to another

• Word count: 1019
• Level: International Baccalaureate
• Subject: Chemistry

#### Vitamin C in Fruit Juices

INTRODUCTION Vitamin C resembles a monosaccharide, but its structure has several unusual features. The compound has a five-membraned unsaturated lactone ring with two hydroxyl groups attached to the doubly bonded carbons1. This enediol structure is relatively uncommon. As a consequence of this structure feature, ascorbic acid is easily oxidized to dehydroascorbic acid. Both forms are biologically effective as a vitamin. There is no carboxyl group in ascorbic acid, but it is nevertheless an acid with a pKa of 4.17. The proton of the hydroxyl group is acidic, because the anion that results from its loss is resonance stabilized and similar to a carboxylate anion2. Humans and a few other vertebrates lack an enzyme that is essential for the biosynthesis of ascorbic acid from D-glucose. Hence ascorbic acid must be included in the daily diet of humans and these other species. Ascorbic acid is abundant in citrus fruits and tomatoes. Its lack in the diet causes scurvy, a disease that results in weak blood vessels, hemorrhaging, loosening of teeth, lack of ability to heal wounds, and eventually death. Ascorbic acid is needed for collagen synthesis. Vitamin C -2D structure3 AIM The aim of this experiment was to see how different juices of the same brand have varying vitamin C content even though on the container it states the same RDI%(Recommended Daily Intake Percnetage).

• Word count: 2006
• Level: International Baccalaureate
• Subject: Chemistry

#### IB IA: Determination of Heat of Neutralization

NAME: Nur Amira Rozali SUBJECT: Chemistry HL ASSESSMENT CRITERIA: DC, DP, EV INTERNAL ASSESSMENT CHEMISTRY HIGHER LEVEL PRACTICAL 16: DETERMINATION OF HEAT OF NEUTRALISATION Data collection: EXPERIMENT REACTANTS INITIAL TEMPERATURE AVERAGE INITIAL TEMPERATURE,T (x±0.5)°C MAXIMUM TEMPERATURE,T (x±0.5)°C ACID BASE ACID BASE HNO3 NaOH 32.0 31.0 31.50 37.0 2 HNO3 KOH 32.0 30.5 31.25 37.0 3 HCl NaOH 32.0 30.0 31.00 37.0 4 HCl KOH 32.0 30.0 31.00 37.0 5 H2SO4 NaOH 32.0 30.0 31.00 43.0 6 H2SO4 KOH 32.0 30.0 31.00 43.5 TABLE 1 Volume of acid used for each experiment = 50cm3 Volume of base used for each experiment = 50cm3 Specific heat capacity of water = 4.18 kg-1K-1 Concentration of each reactant = 1 M Data processing: EXPERIMENT AVERAGE INITIAL TEMPERATURE, T1, (x±0.5) °C MAXIMUM TEMPERATURE,T2, (x±0.5)°C TEMPERATURE DIFFERENCE, T3, (T1 - T2) °C 31.50 37.0 5.50 2 31.25 37.0 5.75 3 31.00 37.0 6.00 4 31.00 37.0 6.00 5 31.00 43.0 2.00 6 31.00 43.5 2.50 TABLE 2 Assumption: The acid and base solutions, each has the same density and specific heat capacity as water, so the mass of "water", m in this case = the total amount of solutions used = 50cm3 + 50cm3 = 100cm3 Specific heat capacity of water, c = 4.18 kJ kg-1 K-1 Temperature change, ? = Temperature difference, T3 Since the

• Word count: 1387
• Level: International Baccalaureate
• Subject: Chemistry

#### IB Chemistry - Boyle's Law Lab Report

Boyle's law The relationship between pressure and volume Raw data . Average book weight = (1441 + 1439)/2 = 1440 ±0.5g 2. The total pressure on the piston equals the pressure from the books plus the atmospheric pressure = 1034g/cm² (it does not have uncertainty because it was given) ? of book Volume (±0.5ml) Mass book(±0.5g) Diameter of syringe (±0.05cm) Total pressure on piston equals the pressure from the books plus the atmospheric pressure (g/cm²) 0 36 0 2.7 034 36 440 2 30 2880 3 26 4320 4 23 5760 5 20 7200 6 8 8640 Data processing . Radius= diameter/2 = 2.7/2 = 1.35 ±0.05cm 2. Area of the contact = ?r² = ?*(1. 35 ±0.05)² = 5.73 cm² 3. % of uncertainty on Area › r = 0.05/1.35*100 = 3.7% › ? doesn't have uncertainty but r² has 2 r so I need to add uncertainty of 2 r. › 3.7 + 3.7 = ±7.4% so the absolute uncertainty is 5.73*7.4% = ±0.42 cm² 4. % of uncertainty on Mass of book = 0.5/1440*100 = 0.04% = ±0% 5. Pressure = mass(of book)/area = 1440/5.73 = 251g/cm² 6. % of uncertainty on pressure = % of uncertainty on Mass + % of uncertainty on Area + % of uncertainty on the total pressure on the piston equals the pressure from the books plus the atmospheric pressure = 0 + 7 + 0 = ±7% 7. Pressure with 7% of uncertainty = The total pressure on the piston equals the pressure from the books plus the atmospheric pressure

• Word count: 1402
• Level: International Baccalaureate
• Subject: Chemistry

#### liquids polarity

LIQUIDS DEFLECTION /cm STRUCTURE Explanation Heptane 0.5-1 -Non polar -tetrahedral -It has an elctronegativity of 0.3, so it becomes covalent non-polar. Deflection is too small, and no attraction towards charged ruler is seen. Ethanol 2-3 -Polar -Tetrahedral -Oxygen with its two lone pairs becomes the negative charge as electronegativity is the biggest; while hydrogen and carbon atoms become positive, having two different charges giving polarity and dipole moments. This molecule as others, as Propanol, Ethane -1,2-diol and Ethanoic acid have Oh bonding and will have a high boiling points. Propanol -2.5 -Polar -Tetrahedral -Because of the two lone pairs of oxygen, and its high electronegativity between hydrogen and carbon atoms, attraction will be lead to him and become negatively charged, while hydrogen with the lower electronegativity will lose electrons easier and become positively charged; having a polarity in charges. However charges are balanced with the big positive charge, having a polar stream but no to strong to deflect it enough as shown. Propanone 2-3 -Polar -Tetrahedral -The electrons in both bonds are strongly attracted to the more electronegative oxygen atom than the less electronegative carbon. This results oxygen to be negatively charged and a carbon to be positively charged. Ethanoic acid 4-5 -Polar -Tetrahedral -In this

• Word count: 1133
• Level: International Baccalaureate
• Subject: Chemistry

#### Chemistry Lab

Physical & Chemical Porperties of two Metals ( Mg & Cu) Physical and Chemical Properties of two Metals (Mg & Cu) " Design Design Purpose The purpose of this lab is to examine the physical and chemical properties of Magnesium and Copper. Definitions: Chemical change: A chemical change is a kind of change in which a specific type of matter is converted to a new one. Physical change: Physical changes are those in which a change in the form, but not the composition of matter occurs. Physical property: Physical properties can be observed or measured without changing the composition of matter, they are generally used to observe and describe matter. Chemical property: Chemical properties can not be determined unless a substance undergoes a change in its composition. If you answer the question "Can I get the original substance back?" with no, then you are observing chemical properties. Physical and Chemical Properties of two Metals (Mg & Cu) " Data Collection Data Collection Observations for Magnesium: Process Observations a) Examine a piece of magnesium ribbon. Silver in color, shiny and smooth surface b) Attempt to bend a piece of magnesium ribbon. Breaks easily c) Obtain two 3 cm pieces of magnesium. Clean the surface of each with a piece of steel wool. Still silver in color, still shiny but became more smooth d) Place one of the pieces of

• Word count: 1130
• Level: International Baccalaureate
• Subject: Chemistry

#### Green House Effect

Greenhouse Effect & Global Warming: Evidence for Global Warming: The average global temperature has risen about 0.50C in the last century, and scientists expect another increase of 1 to 40C in the next 100 years due to increasing pollution in our atmosphere. Greenhouse Effect: * Greenhouse gases allow the passage of incoming solar short-wavelength radiation but absorb some of the reflected infrared radiation and reradiate it back to the Earth's surface. * Energy is then reradiated in the lower atmosphere causing an increase in temperature and climatic change. Hence, contributing to global warming. http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch09/studyplan.asp Greenhouse Gases: . Water (H2O) 2. Carbon dioxide (CO2) 3. Methane (CH4) 4. Nitrous oxide (N2O) 5. CFCs 6. Ozone (O3) Greenhouse Gas Sources: Greenhouse Gas Human Sources Natural Sources Water .Combustion of hydrocarbons .Evaporation of oceans and lakes Carbon dioxide Burning of- .Fossil fuels 2. Forest fires 3. Wood 4. Waste. .Respiration. 2.Decay of organic matter. 3.Natural forest fires Methane .Cattle farming 2.Rice paddies 3.Petroleum 4.Natural gas production Decay of organic matter- .Swamps 2.Marshes Nitrous oxide .Use of nitrogen based fertilizers. 2.Combustion of biomass. .Bacterial Action CFCs .Refrigerants 2.Propellants Ozone .Secondary pollutant

• Word count: 571
• Level: International Baccalaureate
• Subject: Chemistry

#### Determination of the relative molecular mass of sodium carbonate by titration.

Title : Determination of the relative molecular mass of sodium carbonate Data collection: (A) Qualitative o Colour of methyl orange changes from reddish pink to orange (B) Quantitative o Tare weight of sodium carbonate: 2.5061g ± 0.0001g o Volume of sodium carbonate in volumetric flask: 250.00cm³ ±0.15cm³ o Volume of sodium carbonate in pipette: 25.00cm³ ± 0.25cm³ o Volume of 0.09677M hydrochloric acid used: Burette reading (cm³) Titration 1 Titration 2 Titration 3 Initial reading ± 0.05cm³ 0.00 0.00 0.00 Final reading ± 0.05cm³ 27.50 27.30 27.70 Volume of HCL used ± 0.10cm³ 27.50 27.30 27.70 Average reading for volume of 0.09677M hydrochloric acid used 27.50 + 27.30 + 27.70 = 27.50cm³ ± 0.10cm³ 3 Data processing: Subject Calculation % uncertainties Tare weight of sodium carbonate 0.0001 g x 100% 2.5061 g 0.004% Volume of sodium carbonate in volumetric flask 0.15 cm³ x 100% 250.00 cm³ 0.060% Volume of sodium carbonate in pipette 0.25 cm³ x 100% 25.00 cm³ .00% Volume of hydrochloric acid used 0.10 cm³ x 100% 27.50 cm³ 0.3636% Standardization equation H2NSO3H + NaOH H2NSO3Na + H2O From the equation, we know that one mole of H2NSO3H reacted with one mole of HCL to form one mole of H2NSO3Na and one mole of H2O Mole of HCL used = MV 1000 = 27.50cm³ ± 0.3636% x 0.09677 M 1000 =

• Word count: 618
• Level: International Baccalaureate
• Subject: Chemistry