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A brine solution of salt flows at a constant rate of 9 L/min into a large tank that initially held 100 L of brine solution in w

Valentin [98]

Answer:

a) C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) The concentration of salt in the tank attains the value of 0.01 kg/L at time, t = 0.0713 min = 4.28s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the Concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F

Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dC/dt = FᵢCᵢ - FC

Fᵢ = 9 L/min, Cᵢ = 0.02 kg/L, F = 9 L/min

dC/dt = 0.18 - 9C

dC/(0.18 - 9C) = dt

∫ dC/(0.18 - 9C) = ∫ dt

(-1/9) In (0.18 - 9C) = t + k

In (0.18 - 9C) = -9t - 9k

-9k = K

In (0.18 - 9C) = K - 9t

At t = 0, C = 0.1/100 = 0.001 kg/L

In (0.18 - 9(0.001)) = K

In 0.171 = K

K = - 1.766

So, the equation describing concentration of salt at anytime in the tank is

In (0.18 - 9C) = -1.766 - 9t

In (0.18 - 9C) = - (9t + 1.766)

0.18 - 9C = e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾

9C = 0.18 - (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) when C = 0.01 kg/L

0.01 = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

0.09 = (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

- (9t + 1.766) = In 0.09

- (9t + 1.766) = -2.408

(9t + 1.766) = 2.408

9t = 2.408 - 1.766 = 0.642

t = 0.642/9 = 0.0713 min = 4.28s

4 0

3 years ago

ELEMENTS ARE CLASSIFIED AS METALS, NONMETALS, AND METALLOIDS. USE THE PERIODIC TABLE AND IDENTIFY TWO OF EACH CLASSIFICATION.

alekssr [168]

Answer:

Elements are classified into- metals – Nonmetals- Metalloids – Noble gases. State which of A, B, C, D is a:

1) Metallic element

2) Non-metallic element

3) Metalloid

4) Noble gas.

A) Is non-malleable, non-ductile and a poor conductor of electricity

B) Has lustre, is malleable and ductile and a good conductor of electricity

C) Is unreactive and inert and present in traces in air

D) Shows properties of both metals and nonmetals

Explanation:

3 0

2 years ago

Draw the skeletal structure for the monohalogenated alkene formed when trans-1-chloro-4-iodocyclohexane is reacted with 1 equiva

Ksivusya [100]

Because I (iodide) is better leaving group than Cl, so it will leave when this molecule is reacted with strong base (sodium tert-butyl oxide) giving the elimination product provided in picture<span />

5 0

3 years ago

The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

5 0

2 years ago

Determine the mass of 2.75 moles of CaSO4. Record your work and your answer.

maw [93]

Explanation:

$RFM \: = 160 \: g \\ 1 \: mole \: weighs \: 160 \: g \\ 2.75 \: moles \: weighs \: \frac{2.75 \times 160}{1} g \\ = 440 \: g$

6 0

3 years ago