<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.
<u>Explanation:</u>
The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

Where,
= change in Gibb's free energy
= change in enthalpy
T = temperature
= change in entropy
It is given that freezing of methane is taking place, which means that entropy is decreasing and
is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the
is also negative.
For a reaction to be spontaneous,
must be negative.
![-ve=-ve-[T(-ve)]\\\\-ve=-ve+T](https://tex.z-dn.net/?f=-ve%3D-ve-%5BT%28-ve%29%5D%5C%5C%5C%5C-ve%3D-ve%2BT)
From above equations, it is visible that
will be negative only when the temperature will be low.
Hence, the correct statement is low temperature only, because entropy decreases during freezing.
Answer:
Head loss in turbulent flow is varying as square of velocity.
Explanation:
As we know that head loss in turbulent flow given as

Where
F is the friction factor.
L is the length of pipe
V is the flow velocity
D is the diameter of pipe.
So from above equation we can say that

It means that head loss in turbulent flow is varying as square of velocity.
We know that loss in flow are of two types
1.Major loss :Due to surface property of pipe
2.Minor loss :Due to change in momentum of fluid.
<h2>The required balanced equation:</h2><h2>

</h2>
Explanation:
• Propane when reacts with oxygen it gives carbon dioxide and water.
- In order to make the equation balanced at both ends of the equation,
Add 4 as a coefficient before on L.H.S (Left Hand Side) of Equation.
• Therefore, the required balanced equation is:
Propane, when added with 5 molecules of Water, yields 3 molecules of Carbon Dioxide and 4 molecules of water, as shown in the below equation -
