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77julia77 [94]
2 years ago
6

For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou

ld NOT be the change in enthalpy (ΔH) and the activation energy (Ea), respectively, for the catalyzed reaction? Select all that apply.
Physics
1 answer:
Ronch [10]2 years ago
5 0

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

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she is pulling with 40 N force

Explanation:

The ring does not move means that the forces are equal.

Let's call Jo's force x

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140 = x + 100

x = 40

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What is the risk when a pwc passes too closely behind another boat?
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6) A map in a ship’s log gives directions to the location of a buried treasure. The starting location is an old oak tree. Accord
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A flying mosquito hits the windshield of a moving car and gets smashed, but the car is intact. Which of the following statements
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Action and reaction are the same

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A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
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