3-16. Find the absolute and percent relative uncertainty, and express each answer with a reasonable number of significant figures. a. 9.23 (+0.03) mL+4.21 (+0.02) mL-3.26 (+0.06) ml=?9.23 (±0.03) mL + 4.21 (±0.02) mL - 3.26 (±0.06) mL = ? b. 91.3 (±1.0) mM×[40.3 (+0.2) mL]+[21.1 (±0.2) mL]=?91.3 (±1.0) mM x [40.3 (±0.2) mL] ÷ [21.1 (±0.2) mL] = ? c. [4.97 (±0.05) mmol-1.86 (±0.01) mmol]+[21.1 (±0.2) mL]=? [4.97 (±0.05) mmol - 1.86 (±0.01) mmol] [21.1 (±0.2) mL] =? 120011214 61 +0.01 ) —?

[8.47 (+0.06)]13 = absolute uncertainty: + percent relative uncertainty: + % log [6.56 (+0.05)] = absolute uncertainty: percent relative uncertainty + %

Estimate the absolute deviation and the coefficient of variation for the results of the calculation below. Round off the result so that it contains only the significant digits. The numbers in parentheses are absolute standard deviations.

K 2

Estimate the absolute deviation(or uncertainty) for the results of the following calculations. Round each result so that it contains only significant digits. The numbers in parentheses are absolute standard deviations. Finally, write the answer and its uncertainty.a.) Y= 6.75 (± 0.03) + 0.843(±0.001) - 5.021 (±0.001) = 2.572b.) Y= 19.97(± 0.04) + 0.0030(±0.0001) + 4.29(±0.08) = 24.263c.) Y= 143(± 6) - 64(±3) = 5.9578 x10-21249 (±1) +77 (±8)

3-3 Types of Error; 3-4 Propagation of Uncertainty from Random Error (30 min) If A = 3.475 (+0.002), B = 87.336 (±0.001), C = 10.004 5 (±0.000 5), D = 11.8 (+0.2), and E = 5.10 (±0.03), report the answers of the following calculations with both the absolute uncertainty and the percent relative uncertainty. a) (A - B) XE c) b) (C+D)/(AXE) d) [(A+B+C) x (B-C-E)] / [DXE] (10-D)/(E/1000) Answer w/ absolute uncertainty: -428 (13) or -427.7 (±2.5) Answer w/% relative uncertainty: -428 (±0.6%) or -427.7 (±0.5⁹%) b) Answer w/ absolute uncertainty: 1.23 (±0.01) or 1.230 (+0.013) Answer w/ % relative uncertainty: 1.23 (±1%) or 1.230 (+1.1%) Answer w/ absolute uncertainty: 3 (±1) x 10-10 or 3.1 (±1.4) × 10-10 Answer w/ % relative uncertainty: 3 (±50%) x 10-10 or 3.1 (±46%) × 10-10 Answer w/ absolute uncertainty: 121 (±2) or 121.0 (+1.4) Answer w/ % relative uncertainty: 121 (±2%) or 121.0 (±1.8%)

Analysis was run on samples from a known CRM with 8.85% Fe. The following results for %Fe were obtained by an analyst: 8.75, 8.51, 8.25, 8.61. Which of the following statements is true assuming none of the data points are outliers?   Group of answer choices   Tcalc = 63.15 The analysis is not accurate at the 95% confidence interval.   Tcalc = 3.03 The analysis is accurate at the 95% confidence interval.   Not enough information to determine   Tcalc = 63.15 The analysis is accurate at the 95% confidence interval.   Tcalc = 3.03 The analysis is not accurate at the 95% confidence interval.

3-3 Types of Error; 3-4 Propagation of Uncertainty from Random Error ¹. If A = 3.475 (±0.002), B = 87.336 (±0.001), C = 10.004 5 (±0.000 5), D = 11.8 (±0.2), and E = 5.10 (±0.03), report the answers of the following calculations with both the absolute uncertainty and the percent relative uncertainty. a) (A - B) X E c) (10-) / (E/1000) b) (C+D)/(A X E) d) [ (A + B + C) x (B - C - E)] / [DXE]

1.  How many milliliters of 53.40(±0.06) (w/w) % NaOH with a density of 1.52(±0.01)g/mL will you need to prepare 1.000 L of 0.200 M NaOH?  Keep 1 extra sig fig in your answer (4). 2. If the uncertainty in delivering NaOH is ±0.01 mL, calculate the absolute uncertainty in molarity (0.200 M). Assume there is negligible uncertainty in the formula mass of NaOH and in the final volume (1.000 L) Hint:  You will need to calculate the molarity of the 53.40% NaOH.  One way to solve this is to assume that you have 100.00 grams of the initial NaOH solution.  Determine how many grams are NaOH.  Convert that value to moles NaOH.  Using the density, determine the volume of your 100.00 grams of solution.

Find the missing data with the given information for trial 1 Number (1-10)