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4 years ago

15

1 gallon of benzene having a density of 0.88 g/mL is spilled in 200 feet by 150 foot lake having an average depth of 6 feet. Wha

t is the concentration of benzene in this contaminated lake?

1 answer:

Irina18 [472]4 years ago

4 0

Answer:

The concentration of benzene in this contaminated lake would be 653549ng/L.

Explanation:

The concentration C of a contaminant over a body of water of volume V can be obtained as:

C=Mass contaminant / V

In this case, the volume of the body of water:

$V=200ft\cdot150ft\cdot6ft=5097.033m^3$

The mass of the contaminant (Benzene):

$m=\delta V_{ben}=0.88\frac{g}{ml} 1gal(us)=3.3311kg$

Therefore:

$C=\frac{m_{ben}}{V_{lake}}=6.53549\cdot10^{-4}kg/m^3=653549ng/l$

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A(n) ___ is an electric device that uses electromagnetism to change (step up or step down) AC voltage from one level to another.

Pie

Answer:

A transformer is an electric device that uses electromagnetism to change voltage from one level to another or to isolate one voltage from another.

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3 years ago

Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow

Black_prince [1.1K]

Answer:

$\omega_y,\omega_x,\omega_Z$ all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0$

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function $\phi$ given as follows

$u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}$

Rotationality of fluid is given by $\omega$

$\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z$

$\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x$

$\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y$

So now putting value in the above equations ,we will find

$\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},$

$\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}$

So $\omega_y$ =0

Like this all $\omega_y,\omega_x,\omega_Z$ all are zero.

That is why velocity potential flow is irroational flow.

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4 years ago

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

so

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

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3 years ago

How many kg / day of NaOH must be added to neutralize a waste stream generated by an industry producing 90,800 kg / day of sulfu

hram777 [196]

Answer:

74.12kg/day

Explanation:

Equation of reaction for the neutralization reaction of sodium hydroxide and sulfuric acid: 2NaOH + H2SO4 yields Na2SO4 + 2H2O

Mass of sulfuric acid produced per day = 90,800kg

Percentage of sulfuric acid in wastewater = 0.1%

Mass of sulfuric acid that ends up in wastewater per day = 0.1/100 × 90,800 = 90.8kg

From the equation of reaction, 2 moles of NaOH (80kg of NaOH) is required to neutralize 1 mole of H2SO4 (98kg of H2SO4)

80kg of NaOH is required to neutralize 98kg of sulfuric acid

90.8kg of sulfuric acid would be neutralized by (90.8×80)/98kg of NaOH = 74.14kg/day of NaOH

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3 years ago

Read 2 more answers

How much work is performed if a 400 lb weight is lifted 10 ft ?

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W = M*G*H

Work done = 400*10

w = 4000 ft.lb

hope it helps!

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4 years ago