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mestny [16]
4 years ago
15

A chemical reaction takes place in which energy is absorbed. Arrange the characteristics of the reaction in order from start to

finish.
lower energy of reactants
higher energy of products
higher energy of reactants
transition state
lower energy of products

Physics
2 answers:
myrzilka [38]4 years ago
8 0

Let’s create a diagram that shows the chemical reaction absorb energy. Take a look at the diagram I created that is attached to this answer.


Now, the start of the graph line is pretty flat and low leveled. This is the start, which is the reactants. The beginning of a chemical reaction that absorbs energy will almost always start with reactants. I labeled this this “Start: Reactants” and colored the text light green so you can easily detect the part I’m talking about.


Moving on, we see there’s a light blue arrow pointing from the the near bottom to the top curve of the graph line. This is the where the activation energy is rising. Activation energy is the difference between the reactants and products. When the graph line begins to cure, this is when the transition state is beginning. I labeled the transition state, “Transition State” and colored the text a light orange. I colored the line a light blue and also labeled it “Activation Energy” and colored the text light blue as well. I hope it’s easy enough for you to follow along. :)


Next we see that the graph line dips and then goes flat, but it doesn’t go low leveled. It says fairly higher than where it started. This is the end product. I labeled this “End: Product” and colored the text light purple so you can easily detect it.


Okay, now that that’s clear, let’s go ahead and fill in the first blank. According to the diagram we see that the start should be a reactant, and since the graph line starts out pretty low. So, this means that the first tile should be “Low energy of reactants."


Now the second tile should definitely be “Transition State.” This is because the second part in the diagram is transition state. Again, this is mainly where the activation energy is starting to rise.


As for the last and final tile, we should chose “High energy of products.” This is because the diagram is left off on a high leveled part at the end of the graph line.


So, when you place the tiles they should look something like this.


Low energy of reactants

↓

Transition state

↓

High energy of products



- Marlon Nunez

lozanna [386]4 years ago
6 0

Answer: I'm not sure what but I just used the answer from this verified answer and the top answer "low energy of reactants" and the bottom answer "high energy of products" were wrong in Plato. Transition was the correct answer for the middle spot.

Explanation: I just took the test and got all but the middle spot "transition" wrong. Good luck finding the correct answer.

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Annette [7]

The magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Impulse:

The effect of applied force acting for a very small interval of time is known as Impulse.

Given data:

The initial magnitude of the force on Ball A is, F = 50 N.

The time interval for the applied force on Ball A is, t = 1.25 s.

And the impulse at initial on ball A is, I = 62.5 N-s.

Now, if force on ball A doubles, which means new magnitude of force is,

F' = 2F

F' = 2 × 50

F' = 100 N

For same time interval, the new impulse on Ball A is given as,

I' = F' × t

Solving as,

I' = 100 × 1.25

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Taking the ratio of both the impulses as,

I' / I = 125/62.5

I' = 2 I

Thus, we can conclude that the magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

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3 years ago
In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain
Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m.  In all the races the  rest starts, so the initial speed is zero (Vo = 0)

   Vf1= Vo + a1 t1    

   Vf1 = x/t                

   XT  = X1 + X2

   X1 = Vo t1 + ½ a1 t1²  

   X1 = ½ a1 t1²  

   X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

    XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

     XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

   a1= XT / [ ½  t1² + t1 (t-t1)]

   a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

   a1=  5.79m/s2  

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2    

Vf2 = x/t                

XT  = X3 + X4

X3 = Vo t2 + ½ a2 t2²  

X3 = ½ a2 t2²  

X4 = Vf2 (t-t2)

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XT = ½ a2 t2² + a2 t2 (t-t2)

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a2 = XT / [½ t2² + t2 (t-t2)]

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We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

          Vf1 = 0 + 5.79 1.82

          Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

            Vf2 = 0 + 3.67 3.07

            Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)    

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Healan  

XH= ½ a2 t2² + Vf2 (t3-t2)

             XH= ½  3.67 3.07² + 11.27 (6.15-3.07)

             XH= 52.01 m

             (XL -XH)= 55.23- 52.01

             (XH -XL)=  3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values ​​of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

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      XL= ½ 5.79 1.822

      XL= 9.59 m

      XH = ½ a2 t12

      XH= ½ 3.67 1.822

      XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

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