Answer:
Yard . I hope this helped:))
<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
<u>Explanation:</u>
The chemical equation for the dissociation of butanoic acid follows:

The expression of
for above equation follows:
![K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BCH_3CH_2CH_2COO%5E-%5D%5BH%5E%2B%5D%7D%7B%5BCH_3CH_2CH_2COOH%5D%7D)
We are given:
![[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COOH%5D%3D0.888M%5C%5CK_a%3D1.54%5Ctimes%2010%5E%7B-5%7D)
![[CH_3CH_2CH_2COO^-]=[H^+]](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COO%5E-%5D%3D%5BH%5E%2B%5D)
Putting values in above expression, we get:
![1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}](https://tex.z-dn.net/?f=1.54%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH%5E%2B%5D%5E2%7D%7B0.888%7D)
![[H^+]=-0.0037,0.0037](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D-0.0037%2C0.0037)
Neglecting the negative value because concentration cannot be negative
To calculate the volume of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is butanoic acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
66 mph
Explanation:
Average speed is the rate of change of distance with time:
Average speed = 
Given parameters:
Starting point = 88mile marker
End of journey = 121mile marker
time = 30minutes = 0.5hr(60min = 1hr)
Solution:
The distance covered in this journey:
Distance = End of journey - starting point = 121 mile - 88 mile = 33mile
Average speed = 
= 66mph
The average speed of Sara's mum is 66mph
Learn more:
Average speed brainly.com/question/8893949
#learnwithBrainly
Answer:
Methane -1,Ethene 2 , Ethane 2,Pentane 5,Propene 3,Hexane 6,Ethyne 2,Propane 3,Octane 8,Heptane 7 Decane 10,Butyne 4,Butane 4,Propyne 3,Butene 4
Answer:
An important example of chemical potential energy is the energy stored in the food that you eat. Inside your body, this food is distributed around your body and is reacted in your cells to release the energy stored in the chemical bonds.
Explanation: