Compare Statement A with Statement B The molar absorptivity of X is 2.50. Statement A: 0.50 M X in a 0.50-cm cuvette Statement B: 0.05 M X in a 1.5-cm cuvette Choose A if % T of Statement A is greater than B B if % T of Statement B is greater than A C if % T of both is equal D Cannot Be Determined
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- (3) In a titration, the % analyte is computed from the following equation: Vol (titrant) x M (titrant) x MW (analyte) % analyte = x 100 Weight (sample) Calculate the absolute error in the % analyte from the following data. Vol (titrant)= 38.04 t 0.02 ml M (titrant) MW (analyte) = 74.116 t 0.005 mg/mmol Wt (sample) = 800.0 ± 0.2 mg = 0.1137 ± 0.0003 mmol/ml %3!An analysis of city drinking water for total hardness was done by two students in the laboratory and produced the following results (in ppm CaCO3): Student A: 228.3, 226.4, 226.9, 227.1, and 228.6. Student B: 229.5, 226.1, 230.7, 223.8, and 227.5 what is the coefficient variation?4. 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.
- The percentage of an additive in gasoline was measured six times with the following results: 0.13; 0.12; 0.16; 0.17; 0.20 and 0.11%. What is the 99% confidence interval for the additive percentage?Chemistry The amount of V was found to be 0.19% (+/-) 0.02% V (N=6). Calculate t-calc, assuming 95% confidence level.Chemical Reactions: Molecular Equation: HCl(aq) + NaOH(aq) NaCl(aq) + H₂O( Net lonic Equation: H(aq) + OH(a (aq) Trial 1 Details Molarity of HCI Standard Solution Volume (mL) HCI Standard Solution No. of mmoles |HC| = No. of mmoles H+ No. of mmoles OH reacted with HCI No. of mmoles NaOH reacted with HCI Final Reading (mL) NaOH Buret Initial Reading (mL) NaOH Buret 25.00 22.17 0.05 → H₂O() Trial 2 0.1024 25.00 22.15 0.05
- Based on the given data, what is the λmax of cobalt? 0.1M Cobalt 300 0.451 325 0.467 350 0.584 375 0.629 400 0.694 425 0.673 450 0.541 475 0.433 500 0.423 525 0.327 550 0.264 575 0.245 600 0.233 625 0.211 650 0.189 675 0.187 700 0.165Chemistry help solve part d to h. the standard deviation is 0.0043m and the mean is 0.1052m.College of Teacher Bachelor of Secondary Education *promoting pedagogical excellence elor of Sec omoting pedage RIN ht: (a) (b) Br. C. CH3 b1. HNO3, H2SO4 2. Fe, H30*r H2/Pd s / oD Br NO2 (c) KMN04 7 (d) Cl CH3CH2CH2CI AICI3 H20 7. OCH3 4. Predict the major product(s) of the following reactions: OM (a) CI (b) CH CH CH3CH2C! AICI3 CH3CH2COCI AICI3 (c) CO2H HNO3 (d) N(CH2CH3)2 H2SO4 SO, H2SO4 ASITY
- An analysis of city drinking water for total hardness was done by two students in the laboratory and produced the following results (in ppm CaCO3): Student A: 228.3, 226.4, 226.9, 227.1, and 228.6. Student B: 229.5, 226.1, 230.7, 223.8, and 227.5What is the 95% confidence interval for the mean?Sum of coefficients C7H8 + O2 --> CO2 + H2O after balancingTable 1: Standardization Data Trial 1 Trial 2 O145504 Mass of KHP 04536 Initial burette reading 0.05ML 12.3 ML Final burette reading 12.8 a6.omL Volume of base used ר.12 13.2 mL Data Analysis: KH Cg Hg Da (MOIAR MASS 2041ag) 1. Calculate the molarity of base for each trial. You must use dimensional analysis – show all unit conversions (hint: start with grams of KHP used). Review the balanced equation for KHP and NaOH you wrote in Expt 11 part 1. (204.29 TRIAI 1: 0.4550g. Average: Trial 2: Trial 1: with 1875 mL of your base (use the average