Explanation:
First, we will calculate fuel consumption is as follows.

= 4526 g/s
Now, we will calculate the power as follows.
Power = Fuel consumption rate × -enthalpy of combustion
= 
=
kW
Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is
kW.
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
Pluto is a dwarf planet, but one of the largest known members, in the Kuiper belt.
The Kuiper Belt extends between 30 AU and 55 AU from the Sun
(1 AU = 1.5 × 10^8 km = distance from Earth to Sun).
Pluto's orbit is highly elliptical. It ranges from 30 AU to 50 AU. When Pluto is closest to the Sun, it is inside the orbit of Neptune (30 AU).
Astronomers class Pluto as a <em>resonant Kuiper belt object</em> (KBO). Because it gets so close to Neptune, its orbit is in <em>resonance</em> with that of Neptune. Pluto makes two orbits for every three of Neptune.
In an atom there are the same number of protons as electrons to start with. The answer would be 65 aswell.