From equation , the Absorbance, Abs = ebC. Thus, C = Abs/e*b e*b = 1 dm ×100cm/dm × 1.38 × 〖10〗^5 cm^(-1) M^(-1) L. =1.38 × 〖10〗^7 C=Abs/(1.38 × 〖10〗^7 ) C=(2.507)/(1.38 × 〖10〗^(-3) )=1.82×〖10〗^(-7) M The value of K: From the calculations above, the rate law equation  reduces to: Rate = K_observed 〖[Blue dye]〗^x Where Kobserved = K [NaoCl]1 and Kobserved is the experimentally obtained value of K = 0.1963 M-1 min-1. Thus, Rate =0.1963/(M.min) 〖[Blue dye]〗^x From Figure 3, it was established that the reaction was in first order. Thus, the overall rate law equation reduces to: Rate = 0.1963〖M^(-1) min^(-1)× [Blue dye]〗^1 Rate = 0.1963M^(-1) min^(-1)× 〖[1.82×〖10〗^(-7) M]〗^1 Discussion From Part B, the optimal reactants combination selected for Part C was 1.00 ml of Clorox + 40 ml of Blue dye because the mixture turned colorless in ~15 minutes (15:32.58 s). In addition, the graph of In[A] versus time was linear.
Step 1: Calculate the mean, median, and standard deviation for ounces in the bottles. Answer: Mean 14.87 Median 14.8 Standard Deveiation 0.55033 For the full calculation, refer to Appendix #1 at the end of the essay. Step 2: Create a 95% Confidence Interval for the ounces in the bottles. Answer: x ̅=14.87 ,s=0.5503 , n=30 , α=0.05 The level of confidence is at 95%. Use the following formula to determine the confidence interval: (x ̅-t_(α/2) (s/√n),x ̅+t_(α/2) (s/√n)) t_(α/2)=t_0.025=2.045 Substitute the values into the formula: (14.87-2.045(0.5503/√30),14.872.045(0.5503/√30)) = (14.665,15.075) The calculation above clearly states that the confidence interval at 95% confidence is approximately 14.665 - 15.075 ounces.
The theoretical yield for Zinc Sulfide is 0.49 grams but the actual yield is 0.38 grams. So if 0.38 is divided by 0.49 and multiplied by 100 then the percent yield for Zinc Sulfide would be 77.6%. When it comes to Sodium Chloride, the theoretical yield is 0.58 grams and the actual yield is 0.45 grams. So when 0.45 grams is divided by 0.58 grams and multiplied by 100, the percent yield would be 77.5% of Sodium chloride. The actual yield is directly taken from the mass of the products in the experiment while the theoretical yield is determined by using stoichiometric calculations.
The following is a copy of the calculations done using decimals: 5.The equivalence point of the titration curve measured in step 1 was 25.25 mL of NaOH. Half of this value is 12.63 mL. By interpolating the graph, the pH at this volume was 4.80, which is equivalent to the pKa of acetic acid. According to the tabulated data, the pKa was 4.90 at 15 mL of NaOH. At this point, the change in pH with respect to volume was minimal since these values were far from the equivalence point, which occurred experimentally at 27.41 mL.
After identifying reagents and finding the theoretic yield, it’s possible to find the excess reagent mass and number of moles for each test. To do this, the smallest mole number of Ca(OH)2 was subtracted from the highest one. The result is used to find the amount of moles excess, by multiplying it to the corresponding number of moles of excess reagent and dividing then by 1 mole of Ca(OH)2. After finding the answer in moles, it’s possible to find the number of grams by following the rules of conversion factors from moles to grams. 5.
1.15585 1.29561 8.84900 11.47416 C.V.% 1.89 1.18 1.71 1.53 % Nominal 101.74 99.06 102.41 100.25 N 6 6 6 6 Recovery The mean overall recovery of Baclofen was 68.98% with a precision range of 1.31% to 9.70%. The mean recovery of internal standard Baclofen d4 was 69.31% with a precision ranging from 3.17% to 7.79%. Recovery of Baclofen from Human Plasma LQC Response MQC2 Response HQC Response Extracted QC Non Extracted QC Extracted QC Non Extracted QC Extracted QC Non Extracted QC LQC (19-24) LQC(01-06) MQC2 (19-24) MQC2(01-06) HQC(19-24) HQC(01-06) 15854 18425 108932 173808 169921 237625 13754 17998 124540 167729 163087 241625 13705 17375 102897 168326 167543 238566 12588 18846 111066 169105 164131 246917 14014 19071 93004 169214 174716 240865 13120 19279 104781 168300 158808 248580 Mean 13839.2 18499.0 107536.7 169413.7 166367.7 242363.0 S.D. 1113.37 717.75 10429.14 2222.91 5596.72 4450.72 C.V.(%) 8.05 3.88 9.70 1.31 3.36 1.84 N 6 6 6 6 6 6 % Recovery 74.81 63.48 68.64 Overall Recovery of Baclofen from Human Plasma Mean 68.98 S.D. 5.675 % Difference 11.33 N 3 Recovery of Baclofen d4 from Human
Statistical Analysis The Chi-square test was implemented in order to calculate the differences in the allotment of the treatment and control groups. The percent changes of the epileptiform discharges were calculated using the paired t-test and ANOVA of the data collected from the EEG examinations after 1, 2, and 6 months of musical treatment. A value of p < 0.05 was considered significant . Results First
The values of K for solutions 1-5 and U were 4.39E4, 4.53E4, 4.23E4, 4.70E4, 6.35E4, and 4.03E4 respectively. The average K for the lab was found to be 4.71E4 and the standard deviation was 8.302E3. The range then that all experimental values of K must fall under is 3.05E4 to 6.37E4. All experimental values of K for this trial fell within the range. Therefore, K can be determined a true constant.
Initial results from 30 patients (verum acupuncture n=15; sham acupuncture n=15) are presented. All the following items are scaled from 0 to 100. A high score in a symptom scale represents an aggravation of symptoms. A high score in QoL shows an improvement. Concerning nausea/vomiting and QOL QoL no there can not be seen asignificant difference can be observed between the sham-acupuncture group and the verum group.
To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) / ((320.5 + 315.8) / 2)) * 100]) and solved to get (1.58%). For the second station we had to determine the distance required to balance the system and the percent difference. To find the unknown distance we set up the equation Fleft*dleft = Fright*dright. We then plugged in the values (11.35 N * x cm = 48cm *