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Chapter 3: Q17P (page 62)

Verify the following calculations:

a)3.1415(±0.0011)=1.77243(±0.00033);b)log[3.1415±0.0011]=0.49714(±0.00015);c)antilog[3.1415±0.0011]=1.14470(±0.00035);d)In[3.1415±0.0011]=1.14470(±0.00035);e)log(0.104±0.0060.0511±0.0009)=0.800(±0.015)

Short Answer

Expert verified

a)y=x1/2%e=1/2100×0.0011/3.1415=0.0175%1.75×10-43.1415=3.1×10-4

Step by step solution

01

Verification of solutions.

a)y=x1/2%e=1/2100×0.0011/3.1415=0.0175%1.75×10-43.1415=3.1×10-4

02

Step 2:

b)y=logx%ey=0.43429×0.0011/3.1415=1.52×10-4

03

Step 3:

c)y=antilogx=10xey=y×2.3026ex=103.14152.30260.0011=3.51

04

Step 4:

d)y=Inxey=0.0011/3.1415=3.5×10-4

05

Step 5:

e)Thenumeratoroflogterm:y=x1/2ey=1/20.006/0.104×100=2.88%0.3225±2.88%0.0511±0.0009=0.3225±2.88%0.0511±1.76%=6.311±3.3755=6.311±0.213y=logxey=o.42429ex/x=0.015

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Most popular questions from this chapter

You can calculate the time it will take for an object to fall from the top of a building to the ground if you know the height of the building. If the height has an uncertainty of 1.0% ,what is the uncertainty in time?

3-1. How many significant figures are there in the following numbers?

(a) 1.903 0

(b)0.0391 0

(c) 1.40×104

Round each number as indicated:

  1. 1.236 7 to 4 significant figures
  2. 1.238 4 to 4 significant figures
  3. 0.315 2 to 3 significant figures
  4. 2.051 to 2 significant figures
  5. 2.005to 3 significant figures

You have a stock solution certified by a manufacturer to contain 150.0±0.3μgSO42-/mL. You would like to dilute it by a factor of 100 to obtain 1.500μg/mL. Two possible methods of dilution are stated below. For each method, calculate the resulting uncertainty in concentration. Use manufacturer's tolerances in Tables 2-3 and 2-4 for uncertainties. Explain why one method is more precise than the other.

(a) Dilute 10.00mL up to 100mL with a transfer pipet and volumetric flask. Then take 10.00mL of the dilute solution and dilute it again to 100mL.

(b) Dilute 1.000mL up to 100mL with a transfer pipet and volumetric flask.

Vernier scale. The figure below shows a scale found on instruments such as a micrometer caliper used for accurately measuring dimensions of objects. The lower scale slides along the upper scale and is used to interpolate between the markings on the upper scale. In (a), the reading (at the left-hand 0 of the lower scale) is between 1.4and 1.5on the upper scale. To find the exact reading, observe which mark on the lower scale is aligned with a mark on the upper scale. Because theon the lower scale is aligned with the upper scale, the correct reading is 1.46. Write the correct readings in (b) and (c) and indicate how many significant figures are in each reading.
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