Trial I Trial II Final reading HCI 26. 40 ml 27.10 ml Initial reading HCI 1.2 ml 1.8 ml Volume of HCl used ? ? Final reading NaOH 33.7 ml 32.8 ml Initial reading NaOH 0.8 ml 0.5 ml Volume of NaOH used ? ?
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- Trial one Trial two KIO3 solution taken (mL) 1.70 16.5 Moles of KIO3 1.06 x 10^-3 1.06 x 10^-3 Final buret reading (mL) 17.0 33.5 Initial buret reading 0 17.0 Volume of Na2S2O3 solution (mL) 17.0 16.5 Molarity of Na2S2O3 solution (MOL/L) 0.374 0.389 MEAN molarity (mol/L) 0.382 Filtrate Trail one Trail two Final buret reading (mL) 24.0 47.9 Initial buret reading (mL) 0 24.0 Volume of Na2S2O3 solution (mL) 24.0 23.9 Moles of Na2S2O3 used 0.00917 0.00913 Moles if IO3^- present initially 7.05 x 10^-4 7.02 x 10^-4 Molarity of IO3^- (mol/L) 3.53 x 10^-2 3.51 x 10^-2 Average 1.7 x 10^-4 1a. Calculate the solubility of Ca(IO3)2 from your results with the first and second filtrates 1c. Calculate Ksp for Ca(IO3)2 using the mean solubilitity4. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value.Density of solution:Trial 1: 1.2 g/mLTrial 2: 1.2 g/mLTrial 3: 1.2 g/mL Average density = 1.2 g/mL What is the relative average deviaion, %?
- EX 2.Given-> Volume of Na+ = 500 ml Molarity of Na+= 0.0100M Molar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volume Number of millimole = 0.0100 × 500 = 5 millimole Na2CO3 ---> 2Na+ + CO32- Millimole of Na2CO3 = millimole of Na+/2 Millimole of Na2CO3 =5/2 = 2.5 millimole Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole) Weight of Na2CO3 required = mole × molar mass = 2.5 × 10-3 × 105.99 =0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.A 50.00 (±0.03) mL portion of an HCl solution required 29.71(±0.03) mL of 0.01963(±0.0030) M Ba(OH)2 to reach an end point with bromocresol green indicator. The molar concentration of the HCl is calculated using the equation below (attached image): a.) Calculate the uncertainty of the result (absolute error). M=0.02333(±?????) M b.) Calculate the coefficient of variation for the result. CV= (Sy/y) x 100%Table of caffeine standards concentration . Sample Conc, ppm Std1 16 Std2 32 Std3 48 Std4 64 Std5 80 If the volume used to make 100 mL of std 1 is 2 uL what is the concentration in M used to make std calibration curve ? The standards are going to be used to build calibration curve to analyze caffeine in an energy drink. if 500 mL of the energy drink has target of 400 mg , how will you prepare the sample if you need 10 mL for the analysis ? Caffeine MM=194.19 g/mol.
- The gravimetric factor used to express CoCBr6·H20 in a sample that is finally weighed as PbClBr is choose below: FW PbClBr / 6 x FW CoCBr6·H20 FW CoCBr6·H20 / FW PbClBr FW CoCBr6·H20 / 6 x FW PbClBr 6 x FW PbClBr / FW CoCBr6·H20A pipet delivers 10.9 mLmL , 10.1 mLmL , and 10.9 mLmL in consecutive trials. Find the mean volume of the samples. = Answer = 10.6 mL Part B What is the average deviation from the mean for the above samples?Please answer fast it’s very important and urgent I say very urgent so please answer super super fast please For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466
- You are absolutely the BEST. Finally someone that dosen't calculate this exercise as 836f66dgsffjgvkik!!!! I understand everything you did, but I have one question. Some experts says that km can be calculated by saying Vmax from date divided by 2 and then they find Km by going down on x-axis on the graph. But, you used the MM equation, to solve for Km. Is both method the same? Or is one of them more precise than the other? *And if theres no need to reply back after you answering my question, then I will thank you again for your help. I have more these types of exam questions that I will probably upload later, and I hope someone like you will look at it.Sample AnalysisSource of Water sample:Tap Water Molarity of Na2S2O3 1 2 3 0.0243 0.0245 0.0228 Average Molarity of Na2S2O3 0.0239 Trials 1 2 3 Volume of sample used 100.00 100.00 100.00 Final Volume Reading Na2S2O3 (ml) 6.70 10.00 13.10 Initial Volume Reading Na2S2O3 (ml) 3.40 6.70 10.00 Net VolumeNa2S2O3 used (ml) 3.30 3.30 3.10 mg O2 in sample ppm O2 in sample average ppm O2 in sampleCalculations for experimentally determining R Trial 1 Trial 2 Mass of Mg ribbon (g) 0.031 0.039 Temperature of H2(g) (°C) 24.0 24.0 Volume of H2 collected (mL) 31.80 40.10 Atmospheric pressure (torr) 754 754 Vapor pressure of water (torr) Volume of H2 gas in liters (L) Temperature of H2 gas in Kelvin (K) Moles of H2 gas (mol) Pressure of H2 gas in atmospheres (atm) Experimental value of R (L·atm/(mol·K)) Average value of R (L·atm/(mol·K)): Percent error (%)Use 0.08206 L·atm/(mol·K) for the theoretical value of R: The hydrogen generated in this lab was a product of the reaction between magnesium and hydrochloric acid. Which of these reactants is the limiting reactant?: Magnesium or Hydrochloric Acid? Explain reasonsing.