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Module 4: Liquid Characteristics - Review Exercises

Features
Instructions:
Work these problems on a sheet of paper and check your answers against those provided below.

Helpful Calculators:
The following calculator may be useful in solving these problems. You can access it either from the "Calculators" link in the Features box or from the link below.

Gas Flow Rate Converter (Actual Standard Conditions)

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Q icon #1
A liquid with 22% suspended solids by weight is used in an acid gas scrubber. What is the general term used to describe this type of liquid?

  1. City water
  2. Aqueous liquid
  3. Slurry
  4. Ionic solution
A icon
Answer: c. Slurry

Slurries have high levels of suspended solids ranging from 2% to 30% of the liquid weight.

To review material, see Module 4 lesson on Types of Liquids.

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Q icon #2
Two-liter liquid samples are taken from various points along an industrial process as a QA/QC (quality assurance/quality control) check. The liquid in the process line contains suspended solids. To determine the percent of suspended solids at the different sampling points, the 2-liter samples were placed in a beaker and boiled to remove the water. Determine the classification of two of the samples based on the results provided below.

  1. How would you classify a sample if the volume of the suspended solids equals 10 mL and the density of the suspended solids equals 2.5 gm/cm3?

  1. City water
  2. Aqueous liquid
  3. Slurry
  4. Ionic solution

  1. How would you classify a sample if the volume of the suspended solids equals 390 mL and the density of the suspended solids equals 1.75 gm/cm3?

  1. City water
  2. Aqueous liquid
  3. Slurry
  4. Ionic solution
A icon
Answer: i. b. Aqueous liquid

Solution:
  1. Determine the volume of water in the sample.

    Equation 2i(1)

  2. Determine the weight of the water in the sample.

    Equation 2i(2)

  3. Determine the weight of the suspended solids based on the density data provided.

    Equation 2i(3)

  4. Determine the percent of suspended solids in the sample by weight.

    Equation 2i(4)

Since aqueous liquids contain 0.2 to 2% suspended solids by weight, the sample is classified as an aqueous liquid.

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Answer: ii. c. Slurry

Solution:
  1. Determine the volume of water in the sample.

    Equation 2ii(1)

  2. Determine the weight of the water in the sample.

    Equation 2ii(2)

  3. Determine the weight of the suspended solids based on the density data provided.

    Equation 2ii(3)

  4. Determine the percent of suspended solids in the sample by weight.

    Equation 2ii(4)

This sample is classified as a slurry because slurries contain suspended solids that range from 2% to over 30% by weight.

To review material, see Module 4 lesson on Types of Liquids.

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Q icon #3
Following the methodology of EPA Method 12 for "Determination of Inorganic Lead Emissions from Stationary Sources," an air emission sampling train sampled the flue gas exiting a cement kiln for four hours. The first two impingers each contained approximately 100 mL of 0.1 N HNO3 (nitric acid). The solution in the two impingers captured a total of 0.0001 lbm of inorganic lead (Pb). The specific gravity of nitric acid is 1.48(68°F/4°C). Link to periodic table of elements. See Figure 1 for an example of a sampling train with impingers.
Figure 1

  1. Determine the mass of HNO3 to be added to 200-mL of water to prepare the proper concentration of impinger solution required for the test. (The change in the oxidation number during the reaction is 1.)

  1. 1.26 gm HNO3
  2. 6.3 gm HNO3
  3. 12.6 gm HNO3
  4. 63.0 gm HNO3

  1. What is the specific gravity of the impinger solution at 68°F relative to pure water at 4°C?

  1. 1.002(68°F/4°C Ref. T)
  2. 1.0063(68°F/4°C Ref. T)
  3. 1.000(68°F/4°C Ref. T)
  4. 0.9937(68°F/4°C Ref. T)

  1. What is the concentration of Pb in the solution (in ppm)?

  1. 0.000442 ppm
  2. 442 ppm
  3. 225 ppm
  4. 0.000225 ppm
A icon
Answer: i. a. 1.26 gm HNO3

Solution:

Note that the volume of HNO3 added to the solution can be neglected in the denominator.

Equation 3i

  1. Determine the approximate molecular weight (MW) of HNO3.

    Equation 3i(1)

  2. The change in oxidation number is given as 1.

  3. Calculate the mass of HNO3 that should be added to 200 mL of water to create a 0.1 N HNO3 solution.

    Equation 3i(3)

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Answer: ii. a. 1.002(68°F/4°C Ref. T)

Solution:

Equation 3ii

Step 1. Express the density of the impinger solution as a mass per volume.

Equation 3ii(1)

Step 2. Determine the volume of the HNO3 added to the impinger solution (in mL).
  1. Determine the density of the HNO3 added to the impinger solution.

    Equation 3ii(2)(1)

  2. Determine the volume of HNO3 added (in mL).

    Equation 3ii(2)(2)

Step 3. Determine the specific gravity of the impinger solution.
  1. Determine the density of the impinger solution.

    Equation 3ii(3)(1)

  2. Determine the specific gravity of the impinger solution.

    Equation 3ii(3)(2)

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Answer: iii. c. 225 ppm

Solution:

Equation 3iii
  1. Determine the mass of the impinger solution in lbm.

    Equation 3iii(1)

  2. Calculate the concentration of Pb in ppm.

    Equation 3iii(2)

To review material, see Module 4 lessons on Density/Specific Gravity and Concentration.

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Q icon #4
A facility that manufactures fiber optic cable emits hydrogen chloride (HCl) from its production line. To remove these emissions a packed bed wet scrubber was installed. Liquid from an alkaline storage tank is added to the recirculating scrubbing liquid in order to maintain the desired pH of the scrubber liquid. The cylindrical storage tank has a diameter of 8 ft. The storage tank contains only water and is presently filled 12 ft high. The desired solution of the alkaline storage tank is 0.5 N NaOH. Link to periodic table of elements.

Figure 2

Note that the following reaction occurs as sodium hydroxide is added to water.

Equation 4

  1. How many kilograms of NaOH should be added to the storage tank to produce a 0.5 N NaOH solution?

  1. 341.5 kg of NaOH
  2. 468.4 kg of NaOH
  3. 427 kg of NaOH
  4. 4.68 kg of NaOH

  1. Calculate the molarity.

  1. 20 M NaOH
  2. 1.5 M NaOH
  3. 0.781 M NaOH
  4. 0.5 M NaOH

A icon
Answer: i. a. 341.5 kg of NaOH

Solution:

Step 1. Calculate the volume of water in the storage tank in liters.
Equation 4i(1)

Step 2. Calculate the amount of NaOH (in kg) that should be added to the water in the storage tank.

Equation 4i(2)
  1. Calculate the equivalent weight of NaOH.

    Equation 4i(2)(1a)

    The approximate molecular weight (MW) of NaOH is calculated as follows:

    Equation 4i(2)(1b)

    Use the following equation to determine the change in oxidation number that occurs when NaOH and H2O react.

    Equation 4i(2)(1c)

    The neutral NaOH molecule added to H2O results in a single-charged state as shown above. Therefore the change in the oxidation number is 1.

    Equation 4i(2)(1d)

  2. Calculate the mass of NaOH that should be added to the water in the storage tank.

    Equation 4i(2)(2)

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Answer: ii. d. 0.5 M NaOH

Solution:

Equation 4ii
  1. Calculate the number of gm moles of NaOH added to the water. (MW represents molecular weight.)

    Equation 4ii(1)

  2. Calculate the molarity of the NaOH solution.

    Equation 4ii(2)

To review material, see Module 1 lesson on Geometry and Module 4 lessons on Concentration and pH.

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Q icon #5
When equal parts of Compound A and Compound B are combined, what is the pH of the resulting solution?

Equation 5a

  1. 8.26
  2. 6.3
  3. 5.26
  4. 4.96

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Answer: c. 5.26

Solution:

Equation 5b

To review material, see Module 4 lesson on pH.

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Q icon #6
How does raising the temperature affect the absolute viscosity of a liquid?

  1. Raising the temperature lowers the viscosity of the liquid.
  2. Raising the temperature increases the viscosity of the liquid.
  3. Temperature changes have no measurable effect on liquid viscosity.

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Answer: a. Raising the temperature lowers the viscosity of the liquid.

To review material, see Module 4 lesson on Viscosity.

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Q icon #7
In which one of the following situations would you most likely use a liquid of relatively high viscosity?

  1. The liquid will be used to absorb gaseous pollutants in a venturi scrubber.
  2. The liquid will be used in condensers and evaporative cooling towers to cool a hot gas stream.
  3. The liquid will be used in a sedimentation tank to settle out the suspended solids.
  4. The liquid will be used to form droplets with high surface tension.

A icon
Answer: d. The liquid will be used to form droplets with high surface tension.

The surface tension of a liquid increases as its viscosity increases.

Options (a) and (b) are incorrect for the following reason. The performance of the liquids described in situations (a) and (b) improves as the total surface area of the droplets increases. Liquids of lower viscosity tend to form smaller droplets than liquids of higher viscosity. Given the same mass of liquid, smaller droplets have greater surface area than larger droplets. Therefore, these situations would benefit from using a liquid of lower viscosity.

Option (c ) is incorrect because suspended solids would settle more slowly in a highly viscous liquid.

To review material, see Module 4 lesson on Viscosity.

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Q icon #8
A pulp and paper facility uses an evaporative cooling tower to cool a 10,000 ACFM gas stream (see Figure 3). The instrument readings for the evaporative cooling tower provide temperature and pressure data to the control room. The control room operators need to cool the outlet gas stream to 500°F.

Equation 8

Answer the following questions assuming that the gas stream is similar to air and all the water evaporates during cooling. Use Tables 1 and 2 as needed.

Table 1 Enthalpies of Gases, Btu/ib

Table 2 Values for the Enthalpy of Vaporization for Water (Approx.) at 14.7 psia

  1. What is the inlet gas flow rate in lbm/min? The average molecular weight of air is 29 lbm/lb mole.

  1. 378 lbm/min
  2. 3220 lbm/min
  3. 176 lbm/min
  4. 2340 lbm/min

  1. What is the change in enthalpy of the gas stream as it cools from 1800°F to 500°F? See Table 1 (Enthalpies of Gases, Btu/lbm). See Table 2 [Values for the Enthalpy of Vaporization of Water (Approximate) at 14.7 psia].

  1. 61,000 Btu/min
  2. 18,800 Btu/min
  3. 45,000 Btu/min
  4. 79,700 Btu/min

  1. What is the change in enthalpy for one pound of water as it is heated from 90°F to 500°F?

    See Table 1 (Enthalpies of Gases, Btu/lbm). See Table 2 [Values for the Enthalpy of Vaporization of Water (Approximate) at 14.7 psia].

  1. 1,140 Btu/lbm
  2. 1,050 Btu/lbm
  3. 1,400 Btu/lbm
  4. 1,180 Btu/lbm

  1. What is the total quantity of the water (in lbm/min) required for cooling the inlet gas stream from 1800°F to 500°F?

  1. 151 lbm/min
  2. 59,610 lbm/min
  3. 0.0207 lbm/min
  4. 51.7 lbm/min
A icon
Answer: i. c. 176 lbm/min
Solution:
  1. Convert the gas flow rate from ACFM to SCFM.

    Equation 8i(1)

  2. Calculate the inlet gas flow rate (lbm/min) using the characteristics of air.

    Equation 8i(2)

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Answer: ii. a. 61,000 Btu/min

Solution:

Equation 8ii
  1. Interpolate data from Table 1 (Enthalpies of Gases, Btu/lbm) to calculate the value for the enthalpy of air at 1800°F.

    Equation 8ii(1)

  2. From Table 1, H of Air at 500°F equals 106.7 Btu/lbm.

  3. Calculate the change in enthalpy of air on a Btu/min basis.

    Equation 8ii(3)

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Answer: iii. d. 1,180 Btu/lbm

Solution:

Equation 8iii

The boiling point of water is 212°F.
  1. Calculate the sensible energy as the temperature of the water rises from 90°F to 212°F.

    Equation 8iii(1)

  2. From Table 2, the enthalpy of vaporization of water is 970.3 Btu/lbm.

  3. Calculate the sensible energy as the temperature of the water rises from 212°F to 500°F. For enthalpy values see Table 1 (Enthalpies of Gases, Btu/lbm). The enthalpy of water at 212°F is provided in the footnote.

    Equation 8iii(3)

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Answer: iv. d. 51.7 lbm /min

Solution:

Equation 8iv

To review material, see Module 2 lessons on Flow Rate and Heat Capacity/Enthalpy and Module 4 lesson on Enthalpy.

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Q icon #9
The solute concentrations of Gas A and Gas B are dilute enough so that Henry's law can determine their solubility. The molecular weights of Gas A and Gas B are 30 and 36, respectively. The following solubility data was collected for these two gases.

The partial pressure of Gas A, measured while the water was at 25°C and 720 mm Hg, is given below:

Equation 9a

The partial pressure of Gas B, measured while the water was at 25°C and 690 mm Hg, is given below:

Equation 9b

  1. What is Henry's law constant for Gas A in water?
Hint: Solve on a 100 gm of water basis.

  1. 0.135
  2. 3.56
  3. 20.1
  4. 14.3

  1. What is Henry's law constant for Gas B in water?

  1. 0.130
  2. 4.03
  3. 4.22
  4. 22.5

  1. Which gas is more soluble in water?

  1. Gas A is more soluble than Gas B.
  2. Gas B is more soluble than Gas A.
  3. Gas A and Gas B have the same solubility.
  4. Not enough information is provided.
A icon
Answer: i. c. 20.1

Solution:

Step 1. Convert partial pressure measurements of Gas A to 100 gm of water basis. Note that 100 mL of water is equivalent to 100 gm of water.

Equation 9i(1)

This information is presented in tabular form below.

Table 3 Calculate Mole Fractions (Gas A)

The two points defined by (x1, y1) and (x2, y2) can be plotted on a graph to determine the slope of the line, which will determine Henry's law constant.

Step 2. Calculate values for y1 and y2, the mole fractions of Gas A in air.

Equation 9i(2)

Step 3. Calculate values for x1 and x2, the mole fractions of Gas A in water.

  1. Calculate the value for x1 (Gas A).

    Equation 9i(3)(1)

  2. Calculate the value for x2 (Gas A).

    Equation 9i(3)(2)

Step 4. Calculate the Henry's law constant for Gas A. [The Henry's law constant is the slope of the solubility graph created from points (x1, y1) and (x2, y2).]

Equation 9i(4)

Figure 4 Henry's Law Solubility Curve for Gas A

Answer: ii. d. 22.5

Solution:

Step 1. Convert partial pressure measurements of Gas B to 100 grams of water basis. Note that 100 mL of water is equivalent to 100 gm of water.

Equation 9ii(1)

This information is presented in tabular form below.

Table 4 Calculate Mole Fractions (Gas B)

The two points defined by (x1, y1) and (x2, y2) can be plotted on a graph to determine the slope of the line, which will determine Henry's law constant.

Step 2. Calculate values for y1 and y2, the mole fractions of Gas B in air.

Equation 9ii(2)

Step 3. Calculate values for x1 and x2, the mole fractions of Gas B in water.

  1. Calculate the value for x1 (Gas B).

    Equation 9ii(3)(1)

  2. Calculate the value for x2 (Gas B).

    Equation 9ii(3)(2)

Step 4. Calculate the Henry's law constant for Gas B. [The Henry's law constant is the slope of the solubility graph created from points (x1, y1) and (x2, y2).]

Equation 9ii(4)

Figure 5 Henry's Law Solubility Curve for Gas B

Answer: iii. a. Gas A is more soluble than Gas B.

Henry's law constant for Gas A is 20.1 and Henry's law constant for Gas B is 22.5 at 25°C. Since Compound A has a lower Henry's law constant than Compound B, Compound A is more soluble in water.

To review material, see Module 1 lesson on Moles, Module 2 lesson on Gas Concentration, and Module 4 lessons on Liquid Concentration, Vapor Pressure, and Gas Absorption.

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Q icon #10
A wet scrubber uses a scrubbing liquid to remove gaseous pollutants through the process of absorption. Suppose the removal efficiency of a wet scrubber is lower than desired. Which of the following options would you suggest to increase the removal efficiency?

  1. Increase Henry's law constant.
  2. Lower the pH of the scrubbing liquid to 4.5.
  3. Increase the pH of scrubbing liquid to 10.
  4. Lower the temperature of the scrubbing liquid.

A icon
Answer: d. Lower the temperature of the scrubbing liquid.

Lowering the temperature of a liquid increases the solubility of the gas in the scrubbing liquid.

Option (a) is incorrect because increasing Henry's law constant will lower the solubility of the gaseous pollutant.

Option (b) is incorrect because a liquid pH of less than 5 may cause corrosion damage to components in contact with the scrubber liquid.

Option (c) is incorrect because a liquid pH greater than 9 may cause solids buildup in the spray nozzles, packing material, and piping.

To review material, see Module 4 lessons on Vapor Pressure, pH, and Gas Absorption.

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