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Is this correct or does the true value need the same sig figs as the estimated value?
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- Estimate the absolute standard 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. Show your solution and express the final answer and its corresponding uncertainty. a) y= 2.998(±0.002) - 3.98 (±0.15) + 9.035 (±0.002) = 8.0531 b) y= 39.2(±0.3) x 3.054 (±0.022) x 10^ -2 = 1.197 c) y= [198(±6) - 89 (±3)] / 1335 (±2) + 64 (±7)] = 7.791 x 10^-2Estimate the absolute deviation and the coefficient of variation 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. (1.203 (+0.004) × 103 + 1.625 (±0.005) × 102 – 34.012(±0.001) y = - %3D 3.9201 (±0.0006) × 10-3confused in this part why is the answer 0.0005666? Therefore, average error/ deviation = d1 + d2 + d33= (0.0006+0.0008+0.0003)3= 0.0005666 ≈ 0.00057 error/ deviation in parts per thousand(ppt) = average error or deviationaverage concentration×1000 = 0.000570.9548×1000 = 0.596 So, the final answer is = 0.9548 M ± 0.596 ppt
- A quia.com/quiz/7898892.html?AP_rand%3D1784472953 UIA FAQ About Log in Subscribe now 30-day free trial Home 201208 Assessment for Scoutlier Significant Figures Calculations Go to 1 - Not answered v Go stion 1 of 20 1. When expressed as 7.5x10^4, only the significant figures of are to.be considered. (1 point) O 7.5 7.05 O 7.005 O None of the, above Submit answer Skip for now Save progress or end quiz1. Calculate the experimental density of a salt solution and the percent error (same as relative error percent) using some or all the data given below. solubility of NaCl salt in water: 0.357 g/mLmass of empty graduated cylinder: 25.19g mass of graduated cylinder + salt solution: 30.47g total volume of salt solution: 4.98 mLtrue density of salt solution: 1.07 g/mLIf you were pipetting 17.0 mL, what is the percent error in the pipet? I know that Volumetric pipet with 10.0ml has precision of +-.02 and a Volumetric pipet of 20.0 ml has a precision of +- 0.06, but I'm not sure how to use this information to get the answer.
- Based on the given data, what is the λmax of copper? 0.1 M Copper 300 0.157 325 0.233 350 0.465 375 0.575 400 0.622 425 0.708 450 0.864 475 0.897 500 0.876 525 0.769 550 0.687 575 0.603 600 0.499 625 0.363 650 0.327 675 0.228 700 0.184Estimate 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)4. Apply the real rule of significant figures to express the final result for each of the values given: (a) 0.250869 + 0.0018443 M KOH => (b) 5.7837 wt% = 0.0676 => (c) 15.04611 ± 0.035 mL =>
- ← Aktiv Chemistry STARTING AMOUNT + CLE SOD X X Aktiv Chemistry O8 https://app.101edu.co MAR 30 o Mail THOMPSON, X Convert 5.88 x 10-6 ADD FACTOR 10 x( ) 5.88 1 10-6 10⁹ m 10-⁹ 5.88 x 10-4 5.88 x 10-6 (IC) My.UAFS - Students: X Question 5 of 58 to nanometers. Use only the metric system. 5.88 x 104 5.88 x 10³ 0.0588 nm Zall = m alt ANSWER 5.88 x 10-15 5.88 x 10-4 0.01 M Inbox (11) - hthomp0 X 1000 588 0.1 RESET 2 58.8 0.001 106 tv A W Home - LioNetIn a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assume that we want 96% confidence that the error is no more than 4 percentage points.how do i know if my measurement is significantly different from the exact value, at least 2 sentences. exact value was 0.25 the average was 0.223 and standard deviation was 0.025. the RSD was 11.2%