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3 years ago

A scrubber on a coal-fired power plant is useful to remove

Engineering

2 answers:

solong [7]3 years ago

5 0

Answer:

sulfur dioxide.

Explanation:

Eddi Din [679]3 years ago

4 0

Answer:

sulfur dioxide

Explanation:

The scrubber is an apparatus installed in a coal-fired power plant to clean the passing gas through the smokestack. Due to the norm enacted through the clean air Act, almost all the scrubber used in the U.S is used to remove sulfur concentration from coal. it can remove approx 90-95% SO_2 from the smokestack.

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What is Not considered as metric system ? a. Length b. Mass c. Time d. Volume e. Temperature

vampirchik [111]

Answer:

D.) Volume

Explanation:

7 0

3 years ago

Read 2 more answers

A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m

natima [27]

Answer:

a. 0.4544 N

b. $5.112 \times 10^{-5 M}$

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

$H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O$

$NaOH\ Mass = Normality \times equivalent\ weight \times\ volume$

$= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL$

= 0.27264 g

$NaOH\ mass = \frac{mass}{molecular\ weight}$

$= \frac{0.27264\ g}{40g/mol}$

= 0.006816 mol

Now

Moles of $H_2SO_4$ needed is

$= \frac{0.006816}{2}$

= 0.003408 mol

$Mass\ of\ H_2SO_4 = moles \times molecular\ weight$

$= 0.003408\ mol \times 98g/mol$

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

$= \frac{acid\ mass}{equivalent\ weight \times volume}$

$= \frac{0.333984 g}{49 \times 0.015}$

= 0.4544 N

b. And, the acid solution molarity is

$= \frac{moles}{Volume}$

$= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}$

= 0.00005112

= $5.112 \times 10^{-5 M}$

We simply applied the above formulas

4 0

3 years ago

The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u

zimovet [89]

Answer:

F = 0.0022N

Explanation:

Given:

Surface area (A) = 4,000mm² = 0.004m²

Viscosity = µ = 0.55 N.s/m²

u = (5y-0.5y²) mm/s

Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

8 0

3 years ago

provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki

Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

percentage of solar power absorbed by refrigerant = 60%

Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0

3 years ago

Create a flowchart that describes the following algorithm and then write Python code to implement the algorithm. Write the Pytho

Naya [18.7K]

Answer:

<em>Python code is as follows: </em>

********************************************************************************

#function to get number up to any number of decimal places

def toFixed(value, digits):

return "%.*f" % (digits, value)

print("Enter the price: ", end='', flush=True) #prompt for the input of price

price = float(input()) #taken input

totalCost = price + 0.05 * price #calculating cost

print("Total Cost after the sales tax of 5% is applied: " + toFixed(totalCost,2)) #print and output totalCost

************************************************************************************

3 0

3 years ago