<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:
Sodium hydroxide
Explanation:
Sodium hydroxide ( NaOH ) -
Sodium hydroxide is an inorganic compound , and is also called caustic soda and lye .
It is an ionic compound , which is white in color and is in solid state .
The cation and anion of this salt are the sodium cation Na⁺ and the hydroxide anions OH⁻ .
<u>It is highly basic in nature and is soluble in water , and when left open in air it can readily absorb moisture from the air , to form a hydrated sodium hydroxide .</u>
Answer:
The 18th-century German physicist Daniel Gabriel Fahrenheit originally took as the zero of his scale the temperature of an equal ice-salt mixture
Explanation:
<em />If 100 brownies require 6 eggs, 5 cups of flour, and 2 sticks of butter. Then, 50 brownies should require half of those required to make 100 brownies. Use ratio and proportion to determine the number of eggs needed:
100/50 = 6/x
x = 3<span />
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
