Deborah Bell April 17, 2012 Chemistry 1212K Lab Synthesis Report Introduction In this Chemistry Lab the main objective is to perform accurate chemical analysis for the quantity of elements and compounds in a sample. There will be a compound made then synthesized. The methods used were acid-base titrations, redox
CHM130 Lab 4 Calorimetry Name: Data Table: (12 points) ALUMINUM METAL Pre-weighed Aluminum metal sample mass (mmetal) 20.09 g Temperature of boiling water and metal sample in the pot (Ti(metal)) dsdfa(Ti 99°C Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water)) 24°C Maximum Temperature of water/metal in calorimeter after mixing (Tf) 28°C LEAD METAL Pre-weighed Lead metal sample mass (mmetal) 20.03g Temperature of boiling water and metal sample in the pot (Ti(metal)) 103°C Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water)) 25°C Maximum Temperature of water/metal in calorimeter after mixing (Tf) 26°C IRON METAL Pre-weighed Iron metal sample mass You have the mass of water from calculation #9, the specific heat of water is 4.184 J/g(oC), and the temperature change of water
Data/ Results: See attached report sheets Sample Calculations: Value of heat capacity 〖-(q〗_cu+q_water)=q_calorimeter q=C_s*m*ΔT q_water=58.7150g*4.18J/(g*˚C)*7.9˚C q_calorimeter=-(-1950+1938)=12.0 Molar enthalpy of neutral ΔH_1=q_rx1/moles=-65800J/mole -q_rx1=q_HCl+q_NaOH+q_calorimeterl=1870J+59.5J+90J=2000J Heat of rxn of Mg2+ and acid -q_rx1=q_Soln+q_calorimeter=2270J+122J=2390J ΔH3 , Mg(s) + 2H+(aq) →Mg2+(aq) + H2(g) 65.8kj/mole 〖〖2H〗_2 O(l)→2H〗^+ (aq)+〖2OH〗^- (aq) -362kj/mole Mg(s) + 2 H2¬O(l) → Mg2+(aq) + 2OH¬-(aq)+ H2(g) -296kj/mole ΔH4 Mg2+(aq) + 2OH¬-(aq)+ H2(g) →Mg(s) + 2 H2¬O(l) 296kj/mole Mg(s) + O2(g) +H2(g) →MgOH2(s) -924.7kj/mole 2H2O(l) → 2H2(g) + O2(g) 571.6kj/mole Mg2+(aq) + 2OH¬-(aq) →MgOH2(s) -57kj/mole Discussion/ Conclusions: The lab used methods of calorimetry in order to measure the temperature change of reactions and calculate the changes in
3. Calculate the total heat released in each reaction, assuming that the specific heat of the solution is the same as for pure water (4.18J/gK). Use q=mcΔT. Show work here and record your answer in Data Table 2.
Analysis: The average deviation describes the precision of the results. It was determined the results our group obtained for were very precise. This is because our average deviation for Keq was only 6.8 which comes out to be a 6. percent error. Due to our deviation being so low it indicates
The first part of this lab’s objective was to find the calorimeter constant using DI water. We accomplished this by first checking the temperature and then adding 20 mL of cold DI water into our calorimeter. Next we collected 20 more mL of DI water in a 50 mL beaker and placed it onto a burner in order to heat it. We removed the beaker once the temperature of the water reached 60º celsius. After we removed the water, we poured it into the calorimeter with the cold water and took the temperature. The temperature ended up being 37º celsius. From this information we were able to calculate for qhot, qcold, qcal and Ccal. To be as accurate as possible we conducted this same test three more times and used the averages from all Ccal calculations as the final Ccal.
0.100 mol Na2O2 2 mol H2O/2 mol Na2O2, 0.100 ________________ mol H2O 6. How many moles of oxygen are produced if the reaction produces 0.600 mol sodium hydroxide?
4) Heating favours endothermic reactions. As the reactant “Co(H₂O)₆²⁺” was heated, the solution changed from light red to dark purple. This suggests that the equilibrium shifted right. Therefore, the forward reaction is endothermic.
formula Kw[Fex(C2O4)y]·zH2O. The variables x, y, and z were determined through the duration of the
Lab 5Cellular Respiration Introduction: Cellular respiration is an ATP-producing catabolic process in which the ultimate electron acceptor is an inorganic molecule, such as oxygen. It is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria within each cell. Carbohydrates, proteins, and fats can all be metabolized as fuel,
3. Record the mass of the Also include any observations you made over the course of Part II. Metal: | Metal A | Metal B | Metal C | Mass of metal: | 15.262 g | 25.605 g | 20.484 g | Volume of water in the calorimeter: | 24.0 mL | 24.0 mL | 24.0 mL | Initial temperature of water in calorimeter: | 25.2 °C | 25.5 °C | 25.2 °C | Temperature of hot water and metal in hot water bath: | 100.3 °C | 100.5 °C | 100.3 °C | Final temperature reached in the calorimeter: | 27.5 °C | 32.2 °C | 28.0 °C | Calculations: Show your work and write a short explanation with each calculation. Part I: 1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation qwater = m × c × ΔT. We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.
Introduction: The theory behind this experiment is the heat of a reaction (∆E) plus the work (W) done by a reaction is equal to
Fuels Investigation Aim: To find out which fuel gives out the most energy. Planning We will be using 6 different fuels to heat up 100ml of water, and find out the changes of the temperature. We will measure the temperatures of the water before and after the experiment. We will burn heat the water for exactly 2 minutes, and check the changes in temperature. The change in temperature will allow us to work out the energy given off the fuel by using this formula:
The aim of this experiment was to test the heat of combustion over a period of time, and the energy required to combust alcohols with different carbon chain levels. It was hypothesised that the higher the carbon chain of the alcohol present, the faster the heat of combustion will occur.
To find the value of the pre-exponential factor, ������������ , and activation energy, ������, would require linearizing the Arrhenius equation given as: k = k o ������ −������������ , where R is the gas constant and T is the temperature the given k is at in degrees Kelvin. Equation 8 is linearized by taking the natural log of both sides: ln(k) = ln(k o ) − E . RT