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stealth61 [152]

3 years ago

11

A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50

mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?

1 answer:

Leno4ka [110]3 years ago

8 0

Answer:Decay rate constant,k = 0.00376/hr

Explanation:

IsT Order Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time t and [A]o is the inital concentration at time 0, and k is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k = 0.00376/hr

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Use the image below to measure the blocks. Enter your measurement in centimeters accurate to the 0.1 cm.

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Answer:

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4 years ago

What are the two types of furnaces used in steel production?

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Explanation:

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3 years ago

Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state

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Answer:

W=-940.36 KJ

Explanation:

Given that

$P_1=1\ bar,V_1=4.25 {m^3}$

$P_2=6.8\ bar$

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

Now by putting the values (1.4 bar =140 KPa)

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W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

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4 years ago

Neon is compressed from 100 kPa and 20◦C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

PIT_PIT [208]

Answer:

The specific volume is reduced in 80 per cent due to isothermal compression.

Specific enthalpy remains constant.

Explanation:

Let suppose that neon behaves ideally, the equation of state for ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in cubic meters.

$n$ - Molar quantity, measured in kilomoles,

$T$ - Temperature, measured in kelvins.

$R_{u}$ - Ideal gas constant, measured in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$ .

On the other hand, the molar quantity ( $n$ ) and specific volume ( $\nu$ ), measured in cubic meter per kilogram, are defined as:

$n = \frac{m}{M}$ and $\nu = \frac{V}{m}$

Where:

$m$ - Mass of neon, measured in kilograms.

$M$ - Molar mass of neon, measured in kilograms per kilomoles.

After replacing in the equation of state, the resulting expression is therefore simplified in term of specific volume:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$P\cdot \nu = \frac{R_{u}\cdot T}{M}$

Since the neon is compressed isothermally, the following relation is constructed herein:

$P_{1}\cdot \nu_{1} = P_{2}\cdot \nu_{2}$

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$\nu_{1}$ , $\nu_{2}$ - Initial and final specific volume, measured in cubic meters per kilogram.

The change in specific volume is given by the following expression:

$\frac{\nu_{2}}{\nu_{1}} = \frac{P_{1}}{P_{2}}$

Given that $P_{1} = 100\,kPa$ and $P_{2} = 500\,kPa$ , the change in specific volume is:

$\frac{\nu_{2}}{\nu_{1}} = \frac{100\,kPa}{500\,kPa}$

$\frac{\nu_{2}}{\nu_{1}} = \frac{1}{5}$

The specific volume is reduced in 80 per cent due to isothermal compression.

Under the ideal gas supposition, specific enthalpy is only function of temperature, as neon experiments an isothermal process, temperature remains constant and, hence, there is no change in specific enthalpy.

Specific enthalpy remains constant.

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3 years ago