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algol [13]
2 years ago
11

Imagine that a new planet is discovered with two moons of equal mass: Moon A and Moon B. The mass of the new planet is greater t

han the combined mass of its moons. Moon A is farther away from the new planet than Moon B. What is the planet's gravitational pull on Moon A compared to the planet's gravitational pull on Moon B?
The planet's gravity repels Moon A with a greater force than it repels Moon B, which is why Moon A is farther away.

The gravitational pull on Moon B is greater than on Moon A because Moon B is closer to the new planet than Moon A.

The gravitational pull on Moon B is greater than on Moon A because Moon B is farther away from the new planet than Moon A.

The gravitational pull on Moon A is the same as the gravitational pull on Moon B because distance does not affect the planet's gravity.
Chemistry
1 answer:
kobusy [5.1K]2 years ago
3 0

Answer: The gravitational pull on Moon B is greater than on Moon A because Moon B is closer to the new planet than Moon A.

Explanation:

The gravitational force exerted by the two objects on each other is inversely proportional to the square of the distance between the objects.

F=G\frac{mM}{r^2}

F = gravitational force or pull

G = gravitational constant on that planet

M =  mass of the object-1

m = mass of object-2(Mass of Moon-A or Moon-B)

r = distance between two objects

Force\propto \frac{1}{r^2}

With decease in distance 'r' the force between the object increases or vice versa.So, from this we can say that the gravitational pull on Moon-B is more than the gravitational pull on Moon-A because the Moon B is closer than the Moon-A from the new planet.

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What is the value for AG at 100 K if AH = 27 kJ/mol and AS = 0.09 kJ/mol.K)?
Aleonysh [2.5K]

The value for ΔG = 18 kJ/mol

<h3>Further explanation</h3>

Given

ΔH = 27 kJ/mol and ΔS = 0.09 kJ/mol.K

T = 100 K

Required

the value for ΔG

Solution

The spontaneous process of a reaction is based on 2 factors :

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ΔG=ΔH-T.ΔS

\tt \Delta G=27~kJ/mol-100\times 0.09~kJ/mol.K\\\\\Delta G=18~kJ/mol

5 0
2 years ago
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3 years ago
A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the
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Answer:

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Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}

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Given, \Delta{G^0}=15\ kJ/mol

So,  Applying in the equation as:-

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

Thus,

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}

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5 0
2 years ago
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