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e-lub [12.9K]
3 years ago
7

Individual pyrrole molecules readily join together into larger molecules called polypyrroles. If polypyrroles form from pyrrole

in the presence of zeolites, they do so by attaching to the zeolite either in lumps on the outer surface of the zeolite or in delicate chains within the zeolite's inner channels. When zeolite changes color from yellow to black, it means that on or in that zeolite polypyrroles have formed from pyrrole. Yellow zeolite free of any pyrrole was submerged in dissolved pyrrole. The zeolite, turned black even though no polypyrroles formed on its outer surface.
If the statements above are true, which one of the following must on the basis of them be true?
(A) Polypyrroles had already formed on or in the zeolite before it was submerged.
(B) Lumps of polypyrrole attached to the zeolite were responsible for its color change.
(C) At least some of the pyrrole in which the zeolite was submerged formed polypyrrole chains.
(D) None of the pyrrole in which the zeolite was submerged attached itself to the zeolite.
(E) Little, if any, of the pyrrole in which the zeolite was submerged reached the zeolite's inner channels.
Physics
1 answer:
konstantin123 [22]3 years ago
4 0
<h2>Hey there!</h2>

<h2>I guess the correct option is:</h2>

<h3>"(A) Polypyrroles had already formed on or in the zeolite before it was submerged."</h3>

<h2>Hope it help you</h2>
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For an atom to be neutral the amount of ? and? must be equal
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Read 2 more answers
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

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x = √(0.7^2 + 2.4^2)
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