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Art [367]
1 year ago
12

Practice

Chemistry
1 answer:
krek1111 [17]1 year ago
4 0

The number of significant figures in each of the following numbers are as follows:

  • 640 cm³ - number of significant figures: 2
  • 200.0 mL - number of significant figures: 4
  • 0.5200 g - number of significant figures: 4
  • 1.005 kg - number of significant Figures: 4
  • 10,000 L - number of significant figures: 1
  • 20.900 cm - number of significant figures: 5

<h3>What is significant figures?</h3>

Significant figures are those figures that are meaningful with respect to the precision of a measurement.

In other words, a digit that is nonzero, followed by a nonzero digit,or (for trailing zeroes) justified by the precision of the derivation or measurement is a significant figure.

Based on the question, the number of significant figures in each of the following numbers are as follows:

  • 640 cm³ - number of significant figures: 2
  • 200.0 mL - number of significant figures: 4
  • 0.5200 g - number of significant figures: 4
  • 1.005 kg - number of significant figures: 4
  • 10,000 L - number of significant figures: 1
  • 20.900 cm - number of significant figures: 5

Learn more about significant figures at: brainly.com/question/14359464

#SPJ1

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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
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Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

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