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2 years ago

15

Software that is released to have users test out the "bugs" is known as Ransomeware O Break-in software 2 O Flim flam software O

Beta software

1 answer:

Sophie [7]2 years ago

5 0

Answer:

Beta software

Explanation:

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Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

<u>HCl: 0.4 kmol/h</u>

<u>Cl₂: 1.6 kmol/h</u>

<u>H₂O: 1.6 kmol/h</u>

<u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

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brainly.com/question/82989

7 0

2 years ago

A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con

Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0

3 years ago

A(n) ____ is an object setting used to control the visible display of objects.

KatRina [158]

Remote?? maybe I’m not really sure

3 0

3 years ago

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal

7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )

m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04

now lets find the value of h/mk

h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb = (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]

so we substitute

0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

4 0

3 years ago

1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b

Rainbow [258]

Answer:

1. b. False

2. b. False

3. b. False

4. b. False

5. a. True

6. a. True

7. b. False

8. b. False

9. a. True

Explanation:

1. The fatigue properties of a material are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6. The bending fatigue could be handled with specific load requirements for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the maximum stress, as in torsional fatigue, which can be performed on axial-type specially designed machines also, using the proper fixtures if the maximum twist required is small, in which linear motion is changed to rotational motion.

7. A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0

3 years ago