Answer: equation for the reaction is given below
PCL2+PCL3=PCL5
Where pcl2=0.40atm,pcl3=0.27atm
Pcl5=0.0029atm
Using ∆G=-RTin(PCL5/PCl2*PCL3)
Where R=8.314J/K/mol and T=298K
∆G=-8.314*298in(0.0029/0.40*.27)
∆G=8962.6J/mol
Explanation:
The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude
while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude
where <em>x</em> is measured in m.
The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have
Solve for <em>x</em> :
The solution with the negative square root is negative, so we throw it out. The other is the one we want,
2* v1 * v2 / (v2+v1) = V avg2*56* v2 / (v2+56) = 33.5
v2 = 23.94 m/s
Answer:
2917.4 m/s
Explanation:
From the question given above, the following data were:
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
Radius (r) of the Moon = 1737 Km
Escape velocity (v) =?
Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:
Gravitational acceleration of the earth = 9.8 m/s²
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
= 0.25 × 9.8 = 2.45 m/s²
Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:
1 Km = 1000 m
Therefore,
1737 Km = 1737 Km × 1000 m / 1 Km
1737 Km = 1737000 m
Thus, 1737 Km is equivalent to 1737000 m
Finally, we shall determine the escape velocity of the rocket as shown below:
Gravitational acceleration of the Moon (g) = 2.45 m/s²
Radius (r) of the moon = 1737000 m
Escape velocity (v) =?
v = √2gr
v = √(2 × 2.45 × 1737000)
v = √8511300
v = 2917.4 m/s
Thus, the escape velocity is 2917.4 m/s
