<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
Answer:
1.52V
Explanation:
Oxidation half equation:
2Al(s)−→2Al^3+(aq) + 6e
Reduction half equation
3Sn2^+(aq) + 6e−→3Sn(s)
E°cell= E°cathode - E°anode
E°cathode= −0.140 V
E°anode= −1.66 V
E°cell=-0.140-(-1.66)
E°cell= 1.52V
W = F * s
W = 100 * 10
W = 1000 J
In short, Your Answer would be 1000 Joules
Hope this helps!
Answer:
44.95408163 kg
44.95 kg (2 decimal places)
Explanation:
Weight=Mass*Gravity
440.55=m*9.8
Make m the subject
M=440.55/9.8
M= 44.95 kg
Answer:
See explanation
Explanation:
The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.
On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.
