Results With Respect to Potassium Bromate(V) The table below shows my results when varying the concentration of potassium bromate(V):    Concentration of Potassium Bromate(V) (mol/dm-3) Time Take For the Mixture to Turn Colourless (seconds)    Trial 1 Trial 2 Trial 3 Trial 3 Trial 5 Average Reaction Rate (seconds-I)   0.01 31.9 ±0.01 32.2±0.01 31.7±0.01 31.8±0.01 32.0±0.01 31.92±0.05 0.0313   0.008 45.6±0.01 45.1±0.01 45±0.01.3 45.4±0.01 45.2±0.01 45.32±0.05 0.0221   0.006 62.8±0.01 63.4±0.01 ±0.0162.7 62.9±0.01 6±0.013.2 63.00±0.05 0.0159   0.005 88.8±0.01 88.3±0.01 89.0±0.01 88.5±0.01 88.7±0.01 88.66±0.05 0.0113   0.004 …show more content…
This is called the rate constant and its value changes with temperature. Using results of certain temperature I should be able to calculate a value for 'k'. Using the following values: [Br03] = 0.008 [Br -] = 0.01 [H+]2 = (0.1)2 = 0.01 At room temperature, calculated reaction rate for the values are to be 0.0221 seconds-1. Substituting the above values into the rate equation in order to calculate the constant 'k', Firstly we need to rearrange the equation to make 'k' the subject: Reaction Rate = k[Br03] [Br-] [H+]2 k = k = k = !"#$%&'( !"#$ [!"!"] [!"!][!!]! !.!""# (!.!!") (!.!") (!.!") !.!""# (! ! !"!!) k = 27625 The units for 'k' vary depending on the reaction investigated. In this case they are worked out as
The reaction rate will be measured by the rate of production of oxygen gas as hydrogen peroxide is
\begin{eqnarray*} d_2^{fBm} & = & \frac{\ln{\frac{S}{K}} + \frac{1}{2}(r ( T - t) - \frac{\sigma^2{( T^{2H}
75 90 Doc 1 1.31556 1.80357 1.80357 Doc 2 1.75182 1.87012 1.87012 Doc 3 2.13338 2.27178 2.27178 Doc 4 1.56941 1.0743 1.0867
The molar heat of combustion of a compound is 1250 kJ/mole. If 0.115 moles of this compound in a bomb calorimeter with 2.50 L of water, what would the temperature increase be? If change in heat is positive it is an _______ reaction If change in heat is negative it is an_______ reaction
The goal of the experiment is to synthesize a bromohexane compound from 1-hexene and HBr(aq) under reflux conditions and use the silver nitrate and sodium iodide tests to determine if the product is a primary or secondary hydrocarbon. The heterogeneous reaction mixture contains 1-hexene, 48% HBr(aq), and tetrabutylammonium bromide and was heated to under reflux conditions. Heating under reflux means that the reaction mixture is heated at its boiling point so that the reaction can proceed at a faster rate. The attached reflux condenser allows volatile substances to return to the reaction flask so that no material is lost. Since alkenes are immiscible with concentrated HBr, tetrabutylammonium bromide is used as a phase-transfer catalyst.
Table 1 Results DDA Concentration Initial Mass(g) Time Interval Recovered Mass Cumulative Mass (g) Cumulative Recovery (%) Ln[(Rinf -R)/ Rinf] R=Rinf(1-e-kt) (M) (g) 10^(-5) 160 0 0
To determine the rate of reaction there are many method to be used for example, measuring the mass after the product has been added and measuring the difference in mass on the duration of a digital scale. Another method, which will be used in this experiment is using a gas syringe to measure the volume of the gas which has been produced. The cylinder inside, will be pushed out to show a quantitative presentation of the volume produced by the reaction. Hypothesis
In this experiment rate of reaction with different reactants concentration: KI (0.010 M), KBrO3 (0.040 M), and HCl (0.10 M) will be observed. So, this is reaction between iodide and bromate ion under acidic conditions: 6 I- (aq) +BrO3-(aq)+ H+(aq)→ 3I2(aq) + Br-(aq)+ 3H2O The end of the reaction, will be determined by observing color change of solution. Thus, solution should shange color to blue.
Dependent The time taken for the bluish -black color to fade away (color of Iodine solution mix with starch solution ). The rate of enzyme reaction Minutes (min) Table 1.1 – Table shows the controlled variables in the experiment variables Units Measures of controlled variables.
The percent yield was calculated to be 88.1%. Some amount of the product was lost when transferring the product from the Buchner funnel to the balance to measure its mass. To ensure the formation of the desired product, melting point of the product was measured to be 119.8-121.90c, which is in the range of the normal melting point of 2,4,6-tribromoanilne, 120-1220c. Thus, the product was indeed
Predict/ roughly determine the Vmax and ½ Vmax values from the peak of the graph, where the slope of the graph levels off (the asymptotical line). Predict/ roughly determine the Km by reading off of the graph the corresponding substrate concentration on the x-axis for the ½ Vmax value. Plot a Lineweaver-Burke graph (the inverse of the velocity of the reaction vs. the inverse of the substrate concentration). Calculate accurate Vmax and Km values using the following equation for the Lineweaver-Burk
Chemistry IA Background information: Introduction: Electrolysis it’s a chemical process that when you pass an electric current into a solution or a liquid that contains ions to separate substances back to their original form. The main components that are required for electrolysis to take a place are: Electrolyte: it’s a substance that when dissolved in water it ionize and then it will contain free moving ions and without these moving ions the process of electrolysis won’t take place. Direct current (DC): This current provides the energy needed to discharge the ions in the electrolyte Electrodes: it’s an object that conducts electricity and it’s used in electrolysis as a bridge between the solution and power supply. A great example
n=cV n=1.00 mol dm^(-3)×(25 dm^3±0.16%)/1000=0.025 mol±0.16 % The enthalpy of neutralization is then calculated. ∆H=(-1356.5 J±3.104% )/(0.025 mol±0.16%)=-54260 J 〖mol〗^(-1)±3.3 % ∆H=-54260 J m〖ol〗^(-1)±3.3%÷1000=-54 kJ 〖mol〗^(-1)±3.3 %
The gradient gave the value of K, the rate constant for the reaction. Figure 2 shows the plotted graph Figure 2. From the
The 250 mL beaker was rinsed well with the distilled water. The titration procedure above was repeated 2 more times with fresh potassium acid