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3 years ago

Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.

Engineering

1 answer:

marin [14]3 years ago

8 0

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61

= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l

[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

Read more about Technology

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5 0

2 years ago

you are given the following data set: 15, 12, 11, 13, 10, 23, 16, 18, 11, 8. find the values of the lower quartile, median, uppe

Sav [38]

First quartile: Q1=11
second: Q2=12.5
third: Q3=16
interquartile: 5
median: Q2=12.5
minimum: 8
maximum: 23
range: 15

8 0

3 years ago

Suppose we are managing a consulting team of expert computer hackers, and each week we have to choose a job for them to undertak

lina2011 [118]

Answer:

if number == 1

then

tempSolution= max(l[number],h[number])

else if number == 2 then

tempSolution= max(optimalPlan(1, l, h)+ l[2], h[2])

else

tempSolution= max(optimalPlan(number − 1, l, h) + l[number], optimalPlan(number − 2, l, h) + h[number])

end if

return Value

FindOptimalValue(number, l, h)

for itterator = 1 ! number do

tempSolution[itterator] = 0

end for

for itterator = 1 ! number do

if itterator == 1 then

tempSolution[itterator] max(l[itterator], h[itterator])

else if itterator == 2 then

tempSolution[itterator] max(tempSolution[1] + l[2], h[2])

else

tempSolution[itterator] max(tempSolution[itterator − 1] + l[itterator], tempSolution[itterator − 2] + h[itterator])

end if

end for

return Value[number]

OPtimalPlan(number, l, h, Value)

for itterator = 1 ! number do

WeekVal[itterator]

end for

if tempSolution[number] − l[number] = tempSolution[number − 1] then

WeekVal[number] ”Low stress”

OPtimalPlan(number-1, l, h, Value)

else

WeekVal[number] ”High stress”

OPtimalPlan(number-2, l, h, Value)

end if

return WeekVal

7 0

3 years ago

Risks are Not Perceived Differently from What is Happening<br> False<br> True

Gnoma [55]

Answer:

False......................

8 0

3 years ago

2. A thin vertical panel L = 3 m high and w = 1.5 m wide is thermally insulated on one side and exposed to a solar radiation flu

n200080 [17]

Answer: 383.22K

Explanation:

L = 3m, w = 1.5m

Area A = 3 x 1.5 = 4.5m2

Q' = 750W/m2 (heat from sun) ,

& = 0.87

Q = &Q' = 0. 87x750 = 652.5W/m2

E = QA = 652.5 x 4.5 = 2936.25W

T(sur) = 300K, T(panel) = ?

Using E = §€A(T^4(panel) - T^4(sur))

§ = Stefan constant = 5.7x10^-8

€ = emmisivity = 0.85

2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)

T(panel) = 383.22K

See image for further details.

5 0

3 years ago