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Drupady [299]
3 years ago
13

In a chemical reaction, activation energy increases the __________ of the reactants. This outcome causes the particles to collid

e, which results in the __________ of new products.
first blank-
a-amount of chemical energy
b- number of particles
c- temperature

second blank-
a- breaking up
b-formation
c-reactions

Physics
2 answers:
atroni [7]3 years ago
4 0

The answer is;

first blank-

A

Second blank-

B


Activation energy raised the potential energy of reactants to begin a chemical reaction. This requires an initial input of energy into the reaction before the reaction can then proceed spontaneously. That delta G of the activation energy is usually positive. This activation energy destabilized the stable  bonds of the reactants so they can form new bonds. This results to a transition state that is a high-energy state.


Stolb23 [73]3 years ago
4 0

Answer:

The first answer is c- temperature

the second is b-formation

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Answer:

a) The total displacement of the trip was 5.32 × 10³ m

b) The average speeds were:

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Explanation:

The position and velocity equations for an object moving along a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the velocity is constant, then a = 0 and x = x0 + v · t where "v" is the velocity.

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Phase 1:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 0 m/s · t + 1/2 · 2.45 m/s² · (20.0 s)²

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The velocity reached in that phase is:

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v = 0 m/s +  2.45 m/s² · 20.0 s

v = 49.0 m/s

Phase 2:

x = x0 + v · t

x = 490 m + 49.0 m/s · 96.0 s

x = 5.19 × 10³ m

Phase 3:

x = x0 + v0 · t + 1/2 · a · t²

x =  5.19 × 10³ m + 49 m/s · 5.44 s - 1/2 · 9.00 m/s² · (5.44 s)²

x = 5.32 × 10³ m

The total displacement of the trip was 5.32 × 10³ m

b) The average speed is calculated as the traveled distance divided by the elapsed time:

average speed v = final position - initial position / (final time- initial time)

Phase 1:

v = 490 m - 0 m / 20.0 s = 24.5 m/s

Phase 2:

v = 5.19 × 10³ m - 490 / 96.0 s

v = 48.9 m/s   (without rounding the final position the result is 49.0 m/s)

Phase 3:

v =  5.32 × 10³ m -  5.19 × 10³ m / 5.44 s = 23.9 m/s

For the complete trip:

v =  5.32 × 10³ m  - 0 m / (20.0 s + 96.0 s + 5.44 s)

v = 43.8 m/s

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Momentum after the collision

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Since the momentum is conserved we have that

153.36= 60u_1

u_1 = \frac{153.36}{60}

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The velocity of mass m_2 after the collision is given by

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