Answer:
It would Newton's third law.
Explanation:
Answer:
A) increasing tension in the vibrating object.
Answer:
P₂ = 299.11 KPa
Explanation:
Given data:
Initial volume = 600 mL
Initial pressure = 70.00 KPa
Initial temperature = 20 °C (20 +273 = 293 K)
Final temperature = 40°C (40+273 = 313 K)
Final volume = 150.0 mL
Final pressure = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 70 KPa × 600 mL × 313 K / 293K ×150 mL
P₂ = 13146000 KPa .mL. K /43950 K.mL
P₂ = 299.11 KPa
Explanation:
It is given that the total volume is (10 mL + 60 mL) = 70 mL.
Also, it is known that
= 
Where,
= total volume
= initial volume
Therefore, new concentration of
= 
= 
= 0.43 M
New concentration of NaOH = 
= 
= 0.14 M
So, the given reaction will be as follows.

Initial: 0.43 0.14 0
Change: -0.14 -0.14 0.14
Equilibrium: 0.29 0 0.14
As it is known that value of
= 4.74
Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.
pH = ![pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 
= 4.74 + (-0.316)
= 4.42
Therefore, we can conclude that the pH of given reaction is 4.42.