An acid-base reaction or a neutralization reaction is a <u>chemical reaction that occurs between an acid and a base producing a salt and water</u>. The acids and bases can be strong or weak depending on their degree of ionization in water.
Butyric acid is a weak acid and in water it is ionized in the following way, loosing a proton (H+):
HC4H7O2 (aq) ⇆ H+ (aq) + C4H7O2- (aq)
On the other hand, potassium hydroxide is a strong base, so it will be completely ionized in water:
KOH(aq) → K+(aq) + OH-(aq)
Then the <u>net acid-base reaction</u> between butyric acid and KOH is:
HC4H7O2 (aq) + OH- (aq) ⇆ H2O + C4H7O2- (aq)
It is valid to consider only the OH- produced from the ionization of KOH in water since, as mentioned, this molecule is completely ionized. Also, we do not include the K + in the net equation since it is a spectator ion, it does not undergo chemical changes.
Answer:
2726.85 °C
Explanation:
Given data:
Initial pressure = 565 torr
Initial temperature = 27°C
Final temperature = ?
Final pressure = 5650 torr
Solution:
Initial temperature = 27°C (27+273 = 300 K)
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
T₂ = P₂T₁ /P₁
T₂ = 5650 torr × 300 K / 565 torr
T₂ = 1695000 torr. K /565 torr
T₂
= 3000 K
Kelvin to degree Celsius:
3000 K - 273.15 = 2726.85 °C
Answer:
The new equilibrium total pressure will be increased to one-half to initial total pressure.
Explanation:
From the information given :
The equation of the reaction can be represented as;

From above equation:
2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide
So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.
So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.
Let the total pressure at the initial equilibrium be 
and the total pressure at the final equilibrium be 
According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Thus;
P ∝ 1/V
P = K/V
PV = K
where K = constant
So;
PV = constant
Hence;

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

also;
Thus ;



Dividing both sides by 


From ;




Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.
i think it would be Earth's albedo hope this helps
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