/(0.124M)+6.90 x 104 4Y3.45x104): 1 2.84 y0 is what I Keep Suting bt The Correct answer As- 0.124 M K=5.50 x 10-44's-' (0.124)+( -4-1 t- 3.454104 5.90x1043.45x10) is 3.52 X102 How?
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I keep getting wring answers even when I plug in the equation exactly. Can you help me figure out what I'm doing wrong?
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- 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?Which procedure(s) decrease(s) the random error of a measure-ment: (1) taking the average of more measurements; (2) calibrat-ing the instrument; (3) taking fewer measurements? ExplainChemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
- Sample IdentificationCodeConcentration of M%TA=2-log(%T)Q50004.00 x 10-417.90.75Q50013.20 x 10-425.00.6Q50022.40 x 10-435.70.46Q50031.60 x 10-450.20.3Q50048.000 x 10-570.80.15SampleIdentificationCode%TA=2-log(%T)AMQ0210150143.70.359518560.360.000192Q0210150244.10.355561410.360.00018Q0210150343.80.358525890.360.00017Q0210150444.10.355561410.360.00018Q0210150543.80.358525890.360.00017What was their percent error?43%Does Batch 021015 meet legal requirements?No, because it is not between 2.85 * 10(4) and 3.15 * 10(4)Well #DropsBluedye1234567891012345678910Drops 9Distilled water876543210Concne 0.26tration0.52Test Tube #0.781.041.3Solutions3Concentration (M)2.082.32.6Concentration (ppm)1:1 dilution11.82Starting Dilution21.562:1 dilution0A.Zero standard0Was your calibration curve as linear as you expected?B.Did you experience any âdriftâ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…Chemistry The levels of an organic pollutant (P) in the groundwater at the perimeter of a plant were a cause for concern. A 10 mL sample of the water was taken and the pollutant was extracted with 95% efficiency using 25 mL of diethyl ether. GC was used to analyse the concentration of P in diethyl ether. A calibration curve was plotted for a series of standards of P which yielded the following results: Peak Area Toluene Conc. (µg/ml) 12,000 2.6 23,700 5.0 35,500 7.7 46,800 9.9 31,250 Sample Determine the concentration of P in ppb in the initial groundwater sample.To test the quality of the work of a commercial laboratory, duplicate analyses of apurified benzoic acid (68.8% C, 4.953% H) sample were requested. It is assumed that therelative standard deviation of the method is sr → = 4 ppt for carbon and 6 ppt forhydrogen. The means of the reported results are 68.5% C and 4.882% H. At the 95%confidence level, is there any indication of systematic error in either analysis?
- 1) Report the results of this calculation along with both the percent relative and absolute uncertainties. 417.3 ± 0.4 × 71.8 ± 0.7(647.21 ± 0.2 − 182.3 ± 0.4) Absolute uncertainty:Percent uncertainty:1. 500.0 mL 0.2 M KI2. 500.0 mL 0.2 M KCl3. 500.0 mL 0.1 M K2S2O84. 500.0 mL 0.1 M K2SO45. 500.0 mL 4.0 mM Na2S2O3 (from Na2S2O3∙5H2O)pls compelete the 2nd table given the data