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.
Answer is 6.84 approx
reason:-
(2.78^2+6.25^2)^1/2=6.84 approx
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>
Answer:
46.4 s
Explanation:
5 minutes = 60 * 5 = 300 seconds
Let g = 9.8 m/s2. And
be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration
Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]
Jim equation of motion is
As both of them cover the same distance
So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time
Answer:

Explanation:
Given that:
mass of object A, 
mass of object B, 
speed of object A, 
So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.



