Do not see the statements
By definition we have to:
LOG (k2 / k1)=(-Ea/R)*(1/T1-1/T2)
Where,
k1 = 0.0117 s-1
K2 = 0.689 s-1
T1 = 400.0 k
T2 = 450.0 k
R is the ideal gas constant
R = 8.314 KJ / (Kmol * K)
Substituting
ln (0.0117/0.689)=-Ea/(8.314)*((1/400)-(1/450))
Clearing Ea:
Ea = 122 kJ
answer
<span> the activation energy in kilojoules for this reaction is
</span> Ea = 122 kJ
<span>
</span>
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
Answer: False, I think
Explanation:
Well it depends. Its hard to say without a function of the graph. Given a velocity graph, the first derivative is the acceleration.
Negative slope on a velocity graph means you are going in the negative direction. And over time, your velocity per unit of time is also decreasing. That means we are accelerating in the negative direction