Arrhenius equation is a mathematical expression which illustrates the effect of temperature on the rate of a chemical reaction and is used to calculate reaction-rate constants\cite{logan}. In the equation, we have $k$ as the reaction-rate constant, R as the thermodynamic gas constant, A as the pre-exponential factor, E$_a$ as the activation energy and finally $T$ as the absolute temperature. Generally, the equation is represented in exponential form: \begin{equation} k = A \exp[-E_a/RT]
1. For the demo experiment, the balanced chemical equation is as follows: (NH4)2Cr2O7(s)=Cr2O3(s)+N2(g)+4H2O(g). After the lightning of Ammonium dichromate, Chromium (III) oxide was formed while the Nitrogen and Water escaped into the atmosphere in a gaseous phase. Ammonium dichromate((NH4)2Cr2O7) gave rise to Chromium (III) oxide (Cr2O3), Nitrogen Gas(N2) and water (H2O) In terms of microscopic level, the ratio between reactants and products is as follows. One mole of Ammonium dichromate will give
MTH08B_GAMath 8_Semester 2 Systems of Equations Project (15 points) DUE Friday, March 2, 2018 at 4:00 Student Name: Jordan J. Pinckney Directions: Use what you know about solving systems of equations to help Ms. Chen solve her dilemma. If you have completed the OMS MTH Project Equations Check in then you’ve already got a head start. You may use those equations in helping you solve this problem. You must complete all parts (A-G) of the project to receive full credit. This project is worth 15 points
In the not so short story “the cold equations” the author presents a hard decision to kill one or many the killing of one to save many is decided. Tom godwin wrote a story about an EDS pilot who find a stowaway-- an 18 year old girl named Marilyn. Marilyn snuck on the little ship halling fever medicine for 6 men that were sent to the planet Woden. The pilot gets to know Marilyn who snuck on the ship to go see her brother who was on the other side of Woden. The rule was any sort of stowaway was to
Using equation (1) we obtain the total time to be Tt=∑(0.30+0.08+1.42+0.39+0.34+0.50+0.09+0.16+0.32+0.50+0.07+0.30+0.09+1.57+0.30+0.09+1.57+0.21+0.43+0.41+1.03+0.28+0.13+.06+28.5+1.29+0.22+0.30+0.56+0.39+0.40) = 42.3 (min). For: obtaining the cards from the front of the class, taking out six cards, sorting, searching for the missing cards, shuffling four times, placing back into the box, opening the box of cards, obtaining six cards, organizing, sorting, searching for the missing six cards, shuffling
linear equations and linear system of equations (with two unknow variables). Students will also solve linear equations with one unknow variable. The central focus leads students to make connections between algebra, geometry, reading, and application of mathematics in real world context. In lesson1, students will learn how to solve linear equation in one unknown variable. I will give examples of linear equations with one unknown variable with one solution. students will also learn linear equations with
the linear attenuation coefficient, the attenuation equation can be rewritten: I=I_0 e^(-μ_m ρx) 4.2.3.3. Half Value Layer The half value layer (HVL) is the thickness of a shielding material required to reduce the intensity of radiation at a point to one half of its original intensity. It can be calculated by setting I = ½ I_0 and solving the attenuation equation for x:
For the output layer, the input values oi, and the output values oo (also denoted by y) are given by equation 5.4 and 5.5. ξk in equation 5.4 is a threshold value which is also adapted during the training procedure. Equation 5.5 sigmoid transfer function is applied. oi_k=∑_(j=1)^(n hidden)▒ who_(j,k).ho_j+ξ_k (5.4) y_k=〖oo〗_k=1/(1+exp(-〖oi〗_k))
solutions are going to be complex, which methods can be used to solve quadratic equations? Explain your reasoning. When the answers to a quadratic problem are complicated, the quadratic formula or completing the square can be used to solve it. This is because these approaches are designed to produce complicated solutions when the exponent is negative. The quadratic formula is a generic solution for any quadratic equation that provides the two solutions in terms of the problem's coefficients. Completing
(VCAA), 2016a, VCMNA336) Relevant prior VCM codes - year 7: Solve simple linear equations (VCAA, 2016b, VCMNA256) - year 8: Solve linear equations using algebraic and graphical techniques. Verify solutions by substitution (VCAA, 2016c, VCMNA284) - year 9: Sketch linear graphs using the coordinates of two points and solve linear equations (VCAA, 2016d, VCMNA310) - year 10: Solve problems involving linear equations, including those derived from formulas (VCAA, 2016e, VCMNA335) Students’ prior knowledge
to find out the respective enthalpies of the reactions. These two values completed the Table of Thermochemical Equations given and with respect to Hess’s Law, the heat of formation of solid NaCl was computed by adding the enthalpies in the table. Two Styrofoam cups and a thermometer through its lid served as the calorimeter where the reactions took place. Using the heat transfer equation, the enthalpy of the first reaction was computed to be -1.080 kJ/mol. On the other hand, the enthalpy of the second
The unit A.2 test is my best assignment because I find question involving linear functions the easiest to solve. When I learned the four ways to solve linear functions, I learned that if I’m unsure of an answer I can just use a different method to check if I have the same answer. One way to solve the y=mx+b formula when given only two points is to start by plotting the points; then draw a straight line from one of the points to the other. Then count up or down on the Y axis and count right or left
clock reaction are shown below. Equation 1: H2O2 + 3 I- + 2 H+ → I3- + 2 H2O • H2O2 = Hydrogen peroxide • I- = Iodide ion (from potassium iodide) • H+ = A proton, from hydrochloric acid (HCL) • I3- = Triiodide • H2O = Water Equation 1 shows that hydrogen peroxide reacts with iodide ions in acid solution
to the concentrations of reactants (and catalysts) raised to various powers. The rate law is shown in Equation 11. Rate = k[A]x[B]y[C]z (Eq.1) The rate constant (k) is a proportionality constant in the relationship between rate and concentrations, but changes when temperature changes. The effect of temperature on a reaction rate is given by the Arrhenius equation2 (Equation 2): k=Ae-Ea/RT (Eq.2) where A is the collision frequency factor, Ea the activation energy
affects the time the mixed chemicals stay clear before turning blue. The reactions that form the basis for the iodine clock reaction are shown below. Equation 1: H2O2 + 3 I- + 2 H+ → I3- + 2 H2O • H2O2 = Hydrogen peroxide • I- = Iodide ion (from potassium iodide) • H+ = A proton, from hydrochloric acid (HCL) • I3- = Triiodide • H2O = Water Equation
target and first focal point (fs) of the standard lens were measured to give χ. The focimeter equation〖[F〗_t=F_(s^2 ) x] was used to work out the correct power of the lenses (Ft). A graph was plotted with Ft being the Y value (in dioptres) and χ being the X variable (in metres). Fs2 remained constant. A line of best fit was drawn from the results which gave the power of the unknown lenses. The equation of the line of best fit was[Y = 100x]
differentiation formulas since integration is for the reverse of finding the function. A separable differential equation is any differential equation that we can write in the following form, N(y)dy/dx = M(x) Note that in order for a differential equation to be separable all the y's in the differential equation must be multiplied by the derivative and all the x's in the differential equation must be on
Will Encounter Alien Life” by David Haugen and Zack Lewis, astronomer Frank Drake is interviewed on the search for extraterrestrial life. The Drake equation and the organization (SETI) are also discussed. Finding alien life seems almost a certainty. At least, that 's what astronomer Frank Drake, creator of the Drake equation, thinks. The Drake equation, created by its namesake, is a formula to estimate the number of extraterrestrial civilizations in the Milky Way Galaxy. Frank Drake is also the
the second order reaction rates of Ace and IBP with HO•. By dividing Equations 6 and 7, the following relation can be obtained: (Eq. 8) Solving Equation 8 for k, the following expression is obtained: (Eq. 9) However, it can be approximated that the second order rate constants of Ace and IBP each multiplied by [HO] are their respective first order reaction rate constants: and . By substitution to Equation 9, the following relation is obtained: (Eq. 10) As
We started off by watching a video on artistic choice that talked about color choices, lines, forms, shapes, textures, value, and space. After that, we were all given the same equations and were told to make points out of them. We chose 0, 2, 4, and 6 for the x-axis and we kept them the same for all the eleven equations. Before we plotted the points we had to figure out where our origin and scale factor would be, then we had to draw our x and y-axis. After that, we had to figure out the spacing