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slava [35]
4 years ago
14

In the first-order process,a blue dye reacts to form a purple dye. The amount of blue dye at the end of 1 hr is 480 g and the en

d of 3 hr is 120 g . Estimate the amount of the blue dye present initially
Engineering
1 answer:
Tasya [4]4 years ago
6 0

Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 =  ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is  960 kg

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dimulka [17.4K]

The purpose of the pin-indexing system for compressed gas cylinders is to ensure that the correct regulator is used for the cylinder. Thus, Option A is the correct statement.

<h3>What is the purpose of PIN indexing system for cylinders?</h3>

The pin index system refers to a safety system designed to make sure the suitable gas is filled into the suitable cylinder, and that the cylinder will most effectively connect with the correct equipment. The positions of the holes at the cylinder valve correspond with the pins suited for the yoke connected to the equipment.

Therefore, The purpose of the pin-indexing system for compressed gas cylinders is to ensure that the correct regulator is used for the cylinder. Thus, Option A is the correct statement.

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4 0
2 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
Implement a program that requests four numbers (integer or floating-point) from the user. Your program should compute the averag
maks197457 [2]

Answer:

#include<iostream>

int main() {

  float num_1, num_2, num_3, num_4, average;

  //Taking input for four numbers

  std::cout << "Enter first number(integer or floating-point)";

  std::cin >> num_1;

  std::cout << "Enter second number(integer or floating-point)";

  std::cin >> num_2;

  std::cout << "Enter third number(integer or floating-point)";

  std::cin >> num_3;

 

  std::cout << "Enter fourth number(integer or floating-point)";

  std::cin >> num_4;

  average= (num_1+num_2+num_3)/3;

 // Comparing average with fourth number

 if (average==num_4)

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  std::cout << "Equal";

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3 0
3 years ago
A mass of 1.9 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com
emmainna [20.7K]

Answer:

W=-260.66 kJ (negative answer means, that the work was done on the gas)

Explanation:

1) Convert temperature from C to K- T=24+273=297K- all temperature in the gas problems should be used in Kelvins;

2) We need to analyse type of the process- it is given, that the temperature is constant, so it is an Isothermal process, which means, that the equation of the process is: pV=const (constant);

3) Work, done on the system, should be calculated using the following equation: W=\int\limits^{Vb}_{Va} {p} \, dV

4) To calculate initical and final volumes (Va and Vb), we can use the following equation: pV=mRT, so V=mRT/p. Note, that the pressure is changing, thus we can calculate volumes for the both cases- initial and final, using initial (120kPa) and final (600kPa) pressures, in addition, we can find equation for the pressure, as function of the volume, which we need to use for the integration in step 3: p=mRT/V;

5) Now we can calculate the integral, given in the step 3: W=mRT ln(\frac{Vb}{Va}). As we have pressure as a known values, we can re-write the equation, using pressures: W=mRT ln(\frac{pa}{pb})=1.9*0.287*279*ln(\frac{120}{160})=-260.66 kJ

Note, that natural logarithm (ln) yields negative answer, which supports the question, that the work was done on the gas, not by the gas.

6 0
3 years ago
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