Oxygen balance is another critical property to consider for energetic materials.
Not only does it affect a compound’s explosive or propulsive power but also determines, in part, what chemical products might be produced upon detonation or deflagration. The oxygen balance is a weight percent of how much oxygen is either in excess or lacking for an ideal combustion reaction of an energetic material. In a perfect combustion reaction, there is enough oxygen to completely react with all the hydrogen, carbon, sulfur, and metal atoms present to form H2O, CO2, SO2 and metal oxides. Based on the products from an ideal combustion reaction of an organic energetic material CaHbNcOdXeSf:
C_a H_b N_c O_d X_e S_f aCO_2+〖1/2(b-e)H〗_2 O+〖c/2N〗_2+eHX+fSO_2-(a+f+1/4(b-e-d/2)O_2 (1.3)
The oxygen balance can be calculated using the following equation: [16]
(32(a+f+1⁄4 (b-e)-d⁄2))/MW (1.4)
Where a, b, d, e, and f represent the number of carbon, hydrogen, oxygen, halogen and
Sulfur atoms present respectively.
For explosives containing metals, this equation can be modified to reflect the amount of oxygen required to fully oxidize the metal atom present, an amount that varies by metal. If the oxygen balance is close to zero, then the heat of
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Among important thermal properties to consider are the melting point, if any, and the decomposition temperature of the energetic materials. Another critical property to consider is the volatility of energetic materials. If energetic materials are excessively volatile this adds undesirable instability into munitions. And finally, chemical stability in hot conditions and sensitivity to thermal shocks are important to understand as unstable materials cannot be safely
The purpose of this experiment is to verify the formula of magnesium oxide based on the masses of magnesium and the product (MgO). We verify the formula firstly by calculating the empirical formula of magnesium oxide and then calculating creating the magnesium oxide itself- a magnesium ribbon is combined with oxygen in the presence of air through combustion and this forms MgO. The empirical formula of a compound is the simplest method of expressing a chemical formula in whole-number ratios of the constituent atoms that are consistent with masses measured in the experiment; whereas the molecular formula expresses the chemical formula using the actual number of atoms. For example, the molecular formula of anthracene is C14H10 while the empirical formula is C7H5.
Introduction Oximetry calculates the oxygen level (saturation) of the hemoglobin in your red blood cells. Red blood cells carry oxygen. This measurement is taken with a sensor. The sensor is connected by a cable to either a device called an oximeter or a monitor with multiple displays. The measurement helps to assess:
The purpose of this experiment was to observe combustion, oxidation-reaction, and catalyzed decomposition reactions as they work to create rockets that run on a fuel of the reaction H2+1/2O2→H2O2. Instead of being in partners, the partners were put into groups of 4 to create an experiment that tested rocket performance by manipulating a variable such as the order of the collection of gases or amounts of fuel. It was evident during the lab that the reaction of hydrochloric acid and mossy zinc [Zn (s) + 2 HCL (aq) → ZnCl2 (aq) + H2 (g)] reacted much faster than the reaction with the yeast and hydrogen peroxide [2H2O2 (aq) →catalyze 2H2O (l) + O2 (g)]. Both reaction continuously produced gas just at different rates. It can be inferred that the reactivity of the stoichiometric reaction was higher relative to the catalytic reaction.
In this week’s lab experiment, the rate of decomposition of hydrogen peroxide forming oxygen gas will be observed and studied. Since the rate of a chemical reaction is dependent on two things; the concentrations of the reactants and the temperature at which the process is performed, the rate can be measured at which a reactant disappears or at which a product appears. When measuring the rate, the rate law will be applied. The objective of this lab is to demonstrate how the rate changes with varying initial concentrations of hydrogen peroxide by measuring the rate at which oxygen is evolved.
Fuel is usually consistent of octane, ethyl benzene, Trimethylpentane, toluene, and brutane. The reactions of 3.51kg of oxygen to one kilogram of fuel causes the production of 1.42kg pf water and 3.09kg of carbon dioxide. Simply written out, it would form an equation similar to C8H18+O2+CO2+H2O —> CO2+H2O. Dismally, this doesn’t work as a proper equation, as one would find it impossible to balance out. The balanced equation would be closer to C8H18+O2 —> CO2+H2O. Now that one is able to balance the equation, we’ll multiply the carbon dioxide and water by eight and nine respectively. Now there are too many oxygen on the right side of the equation (product). We move to the left side of the equation (reactant) and change the O2 to 25/2O2. Balanced, the equation is C8H18+25/2O2 —> 8CO2+9H2O. In terms of heat through energy, this equation can be converted to [(8 x -393.5) + 9 x -241.8)] - 250.1. What this means is for each time the reactant produces the CO2+H2O mixture, -5270kj of energy are produced.
The balanced reaction of Thiazole when combusted perfectly is; 4C3H3NS + 13O2 12CO2 + 2H2O + 4H2S + 2N2 The requirements for a perfect combustion to take place in the air is the enough supply of oxygen. In this case, only oxygen reacts with the thiazole and other elements does not play any role thus yielding products such as carbon dioxide, water, hydrogen sulphide and nitrogen. The RMM of thiazole; C3H3NS (12*3)
The objective of this lab was to find the empirical formula of magnesium oxide. In order to produce magnesium oxide. In this experiment, magnesium reacted in an oxygen-rich environment while inside of a crucible. The masses before and after the oxidation were both measured. The masses that were resulted were then used to calculate the empirical formula of magnesium oxide. The simplest whole-number in the form of a ratio in which atoms join and form a compound is known as the empirical formula. By finding the percent composition of each element in the compound you are able to determine the empirical formula. When finding the empirical formula it is important to know how to find things such as the limiting reagent, percent yield along with the theoretical yield. The limiting reagent is a substance that will not let the reaction proceed because of its presence. The percent yield and the theoretical yield are both found by comparing the reaction and the experimental yield and recording those results.
The purpose of this lab was to investigate the process of cellular respiration and use our understanding to determine how the type of seed affects the rate of cellular respiration and why. We used six different kinds of seeds and categorized them into three general groups. Tomato and squash were vegetable seeds, Alaskan peas and Red Bush kidney beans were legumes, and wheat and barley were cereals. We determined whether the vegetable seeds, legumes, or cereals had the greatest rate of cellular respiration, and how their seed type affected the rate of cellular respiration.
This report discusses how varying the mole ratio of the reactants affects the amount of the product that is produced and the amount of the reactants that remain at the end of the chemical reaction. We know that you can’t simply add another atom of nitrogen and take an atom of hydrogen away from ammonia because it would change to another substance. We also know that in a chemical reaction atoms cannot be created or destroyed but can make a new substance. A guiding question was made so they can figure out how to solve/find out what is going on. The atomic theory is connected to this lab because that the atoms in the lab cannot be created of destroyed during a chemical reaction. Meaning that each side the of the equation must include the same number of each type of atom. The relationship between the amounts of any two compounds, in moles, that are involved in a chemical reaction. In this lab we are showing how different ratios of moles of sodium bicarbonate and acetic acid and how the different moles leave and percentage out of the product but still remains in the lab. How this lab works we will combine sodium bicarbonate and acetic acid and we are see the amount of CO2 leftover after the reaction happens. Knowing this.
The Cycle of Cellular Respiration and Fermentation Introduction: In this Cellular Respiration and Fermentation lab, we discuss about the cellular respiration which is classified as an aerobic process and fermentation which is anaerobic process. An aerobic process is where it contains oxygen unlike the anaerobic process does not. In the aerobic process there are 4 steps that should be recognized: glycolysis, conversion, citric acid cycle, and the electron transport chain. This experiment contains or uses the yeast concentration to see how much CO2 developed over 20 minutes, counting per 2 minutes and measuring how much has progressed.
Abstract In this experiment, microscale quantities of hydrogen and oxygen were generated to test their explosive nature both separately and in mixtures of various proportions in order to identify the mixture with the most explosive energy, via a pop test that measured relative loudness, thus determining the optimal ratio for their combustion reaction to give water, and determine a mole ratio for the reaction in a balanced equation. Each gas mixture was then tested for power as fuel through the launching of a micro mole rocket. The gases were produced via two separate generators which isolated a chemical reaction to either produce oxygen or hydrogen gas, the reactions included the single- replacement reaction reaction of zinc and hydrochloric acid to generate hydrogen gas, and the yeast catalyzed decomposition of hydrogen peroxide to generate oxygen gas
During this lab the temperature and Oxygen levels were measured at a multitude of depths. The temperature at 0 meters was 27.26 °C. The temperature reading was then done at 1 meter below was 27.2 °C. The Temperature than began to drop constantly until about 10 meters down where the temperature was at 10.09°C. From this point the temperature dropped less dramatically until settling on 7 degree range. The lowest Temperature was at the bottom of the lake being 22 meters deep with a temperature at 7.35°C (figure 1).
When the fuels are burnt, energy is given off. I will be calculating the energy given off using the formula above. The specific heat capacity is the energy
1-Hexanol has a chemical formula of CH3(CH2)5OH and a melting point of: -53 to -41 °C and a boiling point of: 155 to 159 °C The heat of combustion of 1-hexanol is 4638 kJ mol-1 and it has a combustion equation of 2C6H13OH(l) + 18O2(g) → 12CO2(g) +
Also in the production of weapons such as guns and bombs, it is required for chemicals to react extremely vigorously to make the explosion go off and have the desired effect.6