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Chapter 3: Q3 P (page 61)

Round each number to three significant figures:

(a) 0.21674

(b) 0.2165

(c) 0.2165003

Short Answer

Expert verified

a) 0.217

b) 0.217

c) 0.217

Step by step solution

01

Definition of significant figure.

In positional notation, significant figures are digits in a number that are dependable and essential to express the quantity of something.

02

Find the rounded number.

(a)0.21674≈0.2167≈0.217

(b)0.2165≈0.217

(C)0.2165003≈0.2165≈0.217

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Most popular questions from this chapter

What is the true mass of water in vacuum if the apparent mass weighed in air at 24°Cis 1.0346±0.0002g? The density of air is 0.0012±0.0001g/mL and the density of balance weights is 8.0±0.5g/mL. The uncertainty in the density of water in Table 2-7 is negligible in comparison to the uncertainty in the density of air.

We can measure the concentration of HCI solution by reaction with pure sodium carbonate: 2H++Na2CO3→2Na++H20+105.9884±0.0007required 27.35±0.04mLof HCI .

(a) Find the formula mass (and its uncertainty) for Na2CO3.

(b) Find the molarity of the HCI and its absolute uncertainty.

(c) The purity of primary standard Na2CO3is stated to be 99.95 to 100.5wt % , which means that it can react with (100.00±0.05)%of the theoretical amount of H+. Recalculate your answer to (b) with this additional uncertainty.

Rewrite the number 3.123 56 (±0.167 89%) in the forms (a) number (± absolute uncertainty) and (b) number (± percent relative uncertainty) with an appropriate number of digits.

Write each answer with a reasonable number of figures. Find the absolute and percent relative uncertainty for each answer.

(a)[12.41(±0.09)÷4.16(±0.01)]×7.0682(±0.0004)=?

(b)[3.26(±0.10)×8.47(±0.05)]-0.18(±0.06)=?(c)6.843(±0.008)×104÷[2.09(±0.04)-1.63(±0.01)]=?(d)3.24±0.08=?(e)(3.24±0.08)4=?(f)log(3.24±0.08)=?(g)103.24±0.08=?

Find the absolute and percent relative uncertainty and express each answer with a reasonable number of significant figures.

a)9.23±0.03+4.21±0.02-3.26±0.06=?b)91.3±1.0×40.3±0.02/21.1±0.02=?c)4.97±0.05-1.86±0.01/21.1±0.02=?d)2.0164±0.0008+1.233±0.002+4.61±0.01=?e)2.0164±0.0008×103+1.233±0.002×102+4.61±0.01×101=?f)3.14±0.051/3=?g)log3.14±0.051/3=?
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