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iVinArrow [24]
3 years ago
6

The organic acid, ACOOH, reacts reversibly with the alcohol BOH, to form the ester ACOOB according to the stoichiometric equatio

n: ACOOH BOH ACOOB H20 The reaction will be carried out in an ideal batch reactor and the water will be removed rapidly by stripping with an inert gas as the reaction proceeds. Therefore, the reverse reaction can be neglected. The rate equation for the forward reaction is: The value of the rate constant s k = 0.16 M-2h-1 at 373 K. The reaction is carried out in solution in a 2000 L reactor. The initial concentration of ACOOH is ACOOHJo = 2 M and the initial concentration of BOH is [BOHo 3 M. The reactor is operated isothermally at 373 K. How long must the liquid phase reaction run in an ideal, isothermal batch reactor to obtain a fractional conversion of A of 0.9?

Engineering
1 answer:
kompoz [17]3 years ago
4 0

Answer:

The time taken to achieve the 0.9 fractional conversion, t = 28.125 hours

Explanation:

The first image attached is the picture showing the question clearly.

The 2nd attached image is page 1 of the solution and the 3rd attached image is the page 3 of the solution.

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Explain what a margin of safety is in driving as well as how it can help minimize risk.
Yakvenalex [24]

Answer:

A safety margin is the space left between your vehicle and the next to provide room, time and visibility at every instant

Explanation:

A safety margin is defined as an allowance given between your vehicle and the next vehicle in front to provide enough room, visibility and time to move in a safe manner to prevent the occurrence of an accident at anytime the frontal vehicle suddenly stops or slows down

Safety margins help minimize risks in the following way

1) A common knowledge of safety margins, improves predictability among road users, thereby minimizing the risk traffic accidents caused due to late communication

2) The use of safety margins helps minimize the risk due to a change in driving conditions such as when the road becomes more slippery from being covered with fluid that is being wetted

3) Safety margin can help prevent the occurrence of an accident between vehicles due to failure of a car system, such as a punctured tire or failed breaking system

4) Safety margin helps to protect road users from the introduction of obstacles on the main roads such as ongoing road construction, broken down vehicles, road blockage by vehicles involved in an accident etc

5) Safety margin help protect road users from being involved in an accident due to the loss of driving focus of the driver of the frontal vehicle

6 0
3 years ago
A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

5 0
4 years ago
What are the three elementary parts of a vibrating system?
zhenek [66]

Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring -  the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and  zero elasticity.

7 0
3 years ago
A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp
iragen [17]

Answer:

Elastic modulus of steel  = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm  = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × 10^{-4} m

solution

we know that Elastic modulus is express as

Elastic modulus = \frac{stress}{strain}    ................1

here stress is

Stress = \frac{Force}{area}       .................2

Area = (0.022)²

and  

Strain = \frac{extension}{length}       .............3

so here put value in equation 1 we get

Elastic modulus = \frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }  

Elastic modulus of steel = 202.27 × 10^9 Pa

Elastic modulus of steel  = 202.27 GPa

3 0
4 years ago
we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw
gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0
3 years ago
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