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tia_tia [17]
3 years ago
5

If the products of a certain reaction are favored (i.e., the Δ G value of the reaction is negative), one would expect a signific

ant amount of product to form when the reactants are combined. Which of the choices could explain why such a reaction might produce very little product?
Engineering
1 answer:
EastWind [94]3 years ago
4 0

Answer:

B. high Ea

The available choices are as below:

A. low Ea

B. high Ea

C. positive AH

D. negative AS

Explanation:

ΔG is a measure of how favorable a reaction is, it also relates to the equilibrium constant.  It is also the measure of spontaneity of a chemical reaction.

A reaction with a negative  ΔG, is very favorable.

A reaction with a positive  ΔG is not favorable.

A reaction with  ΔG = 0 is at equilibrium.

Now a molecule must acquire sufficient activation energy before they can react. However, the higher the activation energy, the slower the chemical reaction will be which leads to production of very little product.

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Consider a magnetic disk consisting of 16 heads and 400 cylinders. This disk has four 100-cylinder zones with the cylinders in d
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Answer:

Disk Capacity  = 45056000 bytes

Optimal skew = 26.455 ≈ 27

Maximum Data transfer rate =  262.5 GB per second

Explanation:

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disk capacity, optimal track skew, and maximum data transfer rate

solution

first we get total number of sectors that is  

total number of sector = number of zones × (number of sectors in different zones

total number of sector = 100 × (160+200+240+280)

total number of sector = 88000

so

Disk Capacity = total number of sectors  ×  size of each sector

Disk Capacity =  88000 × 512

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and

Rotation time = \frac{60}{7200}

Rotation time = 8.33 milli seconds

so Optimal number of sectors in a track = average of ( 160,200,240,280 )

Optimal number of sectors in a track  = 220

now New sector is read every  \frac{8.33}{220} i.e = 0.0378 ms

here Optimal skew = seek time ÷ new sector read time

Optimal skew = \frac{1}{0.0378}

Optimal skew = 26.455 ≈ 27

and

here we know that for maximum transfer rate we will select cylinder with maximum number of sectors i.e here  280 sectors

so

capacity of one track with maximum = 280 × 512

capacity of one track with maximum =  = 143360 bytes

and Number of rotations in 1 second is = \frac{7200}{60}

Number of rotations in 1 second is = 120

so Data transfer rate = Number of heads × Capacity of one track × Number of rotations in one second

Data transfer rate =  16 × 143360 × 120

Data transfer rate = 275251200 bytes per second

Data transfer rate =  262.5 GB per second

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