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.
In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). 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%).
Measurement of lipid peroxidation TBARS, a measure of lipid per oxidation, was measured as described by Ohkawa . Briefly, 1 ml of suspension medium was taken from the 10% tissue homogenate. 0.5 ml of 30% Trichloroacetic acid (TCA) was added to it, followed by 0.5 ml of 0.8% thiobarbituric acid (TBA) reagent. The tubes were covered with aluminium foil and kept in shaking water bath for 30 minutes at 80°C. After 30 minutes, tubes were taken out and kept in ice-cold water for 30 minutes.
Repeat stet 4 to 9. Repeat step 21 to 24 for 5 times. Diagram Results In different concentration of HCL, the time it took to react Concentration of HCl (Mol) Time it takes to react (pop) (s) Average (Time) (s) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 0.5 48.05 97.72 156.07 126.94 52.74 96.30 1.0 15.54 8.64 9.92 44.17 2.85 16.22 1.5 6.92 5.67 3.95 31.10 30.06 15.54 2.0 6.17 11.04 6.30 5.16 18.95 9.524 To calculate the average (Time) (s) (Sum of all trial in one mole)/5 = (48.05+97.72+156.07+126.94+52.74)/5 = 481.97/5 = 96.304 (4sigfig) = 96.30s I have calculated the average of the time it took to react with my hands
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.
To then find the percent water divide the water mass by the hydrate mass and multiply by 100 since the number is a percent. The water percent is determined to be 42.06%. To find a percent error, a theoretical percent water must be used. To find the theoretical percent error divide the mass of water by the mass of magnesium sulfate heptahydrate and multiply by 100 to get a percent. The theoretical percent water is
Grade) in 1 l water, standardize this solution with 0.0192(N) silver nitrate solution. The solution losses strength gradually and must be rechecked every week. 1 ml of solution = 1 mg CN (ix) Standard cyanide solution: Dilute 10 ml stock cyanide solution to 1 litre with distilled water, mix and make a second dilution of 10 ml to 100 ml. Prepare fresh solution daily. 1 ml = 1 µg CN (x) Chloramine –T: Dissolve 1 gm chloramine – T in 100 ml distilled water.
Figure 1 shows the diluted and actual concentrations of salicylic acid, the concentration and log value of aspirin at various times. Time / min C1 / mol L-1 C2 (actual) / mol L-1 [Aspirin] / mol L-1 ln([Aspirin]) 0 1.996*10-5 9.98*10-4 0.0225 -3.79 10th 6.925*10-5 3.46*10-3 0.0200 -3.91 20th 1.135*10-4 5.68*10-3 0.0178 -4.03 30th 1.372*10-4 6.86*10-3 0.0166 -4.10 40th 1.653*10-4 8.27*10-3 0.0152 -4.18 50th 1.828*10-4 9.14*10-3 0.0144 -4.24 60th 1.953*10-4 9.77*10-3 0.0137 -4.29 Figure 1. A graph of ln([aspirin]t) against time (min) was plotted. The gradient gave the value of K, the rate constant for the reaction. Figure 2 shows the plotted graph Figure 2.
Three trials has been done for this experiment. For Trial 1, Trial 2 and Trial 3, the apparent mass for each trial are 99.330g, 99.261g and 98.741g respectively. After calculation, the true mass has been obtained and for each trial, the mass recorded are 99.449g, 99.380g and 99.859g respectively. As the temperature for all trial is 24.0˚C, the density of water being recorded is 0.997296g/ml. Finally, the actual volume of water being transferred after calculations are 99.717ml, 99.649ml and 99.127ml respectively.
• It which may not sound like much but this is actually a 30% increase in the concentration of hydrogen ions. The ocean is predicted to have decreased by another 0.3-0.5 units by 2100 from it current ph of 8.1 • Firstly, carbonate ions are used by marine animals such as crabs, lobsters and starfish to form shells and skeletons. • However, as the concentration of hydrogen ions increases and bond with the available carbonate ions to form bicarbonate • This means There are less available carbonate ions for animals to create shells and skeletons. • Scott Doney and his colleagues at the Woods Hole Oceanographic Institution studied 11 species of which they found that only 2 increased their carbonate building in more acidic