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Chapter 3: Q14P (page 62)

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

Short Answer

Expert verified

$Absolute uncertainty = 3.12 (卤 0.005) and % relative uncertainty = 0.168 %$

Step by step solution

01

To solve for:

猞� Number $卤$ absolute uncertainty =?

猞� Number $卤$ percentage relative uncertainty =?

02

Calculation of absolute uncertainty

Given information:

3.12356 $(卤 0.16789 %)$

猞� To solve for the absolute uncertainty as shown below:

$absolute uncertainty = 3.12356 脳 0.16789 % 鈥� = 3.12356 脳 \frac{0.16789}{100} = 0.00524415 Absolute uncertainty = 3.12 (卤 0.005)$

03

Calculation of percentage relative uncertainty

b) To solve the percentage relative uncertainty as shown:

% relative uncertainty = 0.16789%
% relative uncertainty = 0.168%

Therefore, the $Absolute uncertainty = 3.12 (卤 0.005)$ and $% relative uncertainty = 0.168 %$

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Most popular questions from this chapter

Explain the difference between systematic and random error.

We can measure the concentration of HCI solution by reaction with pure sodium carbonate: $2 H^{+} + {Na}_{2} {CO}_{3} 鈫� 2 {Na}^{+} + H_{2} 0 + 105.9884 卤 0.0007$ required $27.35 卤 0.04 mL$ of HCI .

(a) Find the formula mass (and its uncertainty) for ${Na}_{2} {CO}_{3}$ .

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

(c) The purity of primary standard ${Na}_{2} {CO}_{3}$ is stated to be 99.95 to 100.5wt % , which means that it can react with $(100.00 卤 0.05) %$ of the theoretical amount of $H^{+}$ . Recalculate your answer to (b) with this additional uncertainty.

(a) How many millilitres of $53.4 (卤 0.4) wt % NaOH$ with a density of $1.52 (卤 0.01) g / mL$ will you need to prepare 2.000L of 0.169M NaOH ?

(b) If the uncertainty in delivering NaOH is $卤 0.01 mL$ , calculate the absolute uncertainty in the molarity (0.169M). Assume there is negligible uncertainty in the formula mass of NaOH and in the final volume (2.000L).

Write each answer with the correct number of significant figures.

(a) $1.0 + 2.1 + 3.4 + 5.8 = 12.3000$

(b) 106.9 - 31.4 = 75.5000

(c) $107.868 (2.113 脳 10^{2}) + (5.623 脳 10^{3}) = 5519.568$

(d) $(26.14 / 37.62) 脳 4.38 = 3.043413$

(e $(26.14 / (37.62 脳 10^{8})) 脳 (4.38 脳 10^{- 2}) = 3.043413 脳 10^{- 10}$

(f) $(26.14 / 3.38) + 4.2 = 4.5999$

(g) $l o g (3.98 脳 10^{4}) = 4.5999$

(h) $10^{- 6.31} = 4.89779 脳 10^{- 7}$

You have a stock solution certified by a manufacturer to contain $150.0 卤 0.3 {渭驳厂翱}_{4}^{2 -} / mL$ . You would like to dilute it by a factor of 100 to obtain $1.500 渭 g / m L$ . 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.

See all solutions

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