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3 years ago

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The activation energy for the isomerization reaction of CH3NC in Problem 14 is 161 kJ mol 1, and the reaction rate constant at 6

00 K is 0.41 s 1. (a) Calculate the Arrhenius factor A for this reaction. (b) Calculate the rate constant for this reaction at 1000 K.

1 answer:

astraxan [27]3 years ago

5 0

Answer:

Explanation:

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Does any one know the answer

valina [46]

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3 years ago

A vertical spring has a mass hanging from it, which is displaced from the equilibrium position and begins to oscillate. At what

Orlov [11]

Answer:

the object has least potential energy at mean position of the SHM

Explanation:

If a block is connected with a spring and there is no resistive force on the system

In this case the total energy of the system is always conserved and it will change from one form to another form

So here we will say that

Kinetic energy + Potential energy = Total Mechanical energy

As we can say that total energy is conserved so here we have least potential energy when the system has maximum kinetic energy

So here we also know that at mean position of the SHM the system has maximum speed and hence maximum kinetic energy.

So the object has least potential energy at mean position of the SHM

5 0

3 years ago

One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =

blagie [28]

Answer:

vf₁ = 6.86 m/s , to the right

vf₂ = 2.96 m/s, to the right

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁= 0.220 kg : mass of object₁

m₂= 0.345 kg : mass of object₂

v₀₁ = 2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂= 6 m/s, to the right i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

$e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }$

1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)

(2.1 - 6 ) = (vf₂ -vf₁)

-3.9 = (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁ = (3.8775) / (0.565)

vf₁ = 6.86 m/s, to the right

We replace vf₁ = 6.86 m/s in the Equation (2)

vf₂ = 6.86 - 3.9

vf₂ = 2.96 m/s, to the right

8 0

3 years ago

If 100ml of cold water is added to 100ml of hot water what happens

denis23 [38]

The water will be cool and steam will be created by the hot and cold water reacting together

4 0

3 years ago

Find the unit vector in the direction of vector B=4i+2j-5k

solmaris [256]

Answer:

$\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k$

Explanation:

for the unit vector, we need to divide the given vector by its norm, because it should be in the SAME direction as the original vector, but of magnitude "1".

We notice that the norm of the given vector is:

$\sqrt{4^2+2^2+(-5)^2} =\sqrt{45}$

Then, the unit vector becomes:

$\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k$

7 0

3 years ago