<h2>Heptene formed is -</h2><h2>
</h2>
Explanation:
The two possibilities when the peroxide is not present
+ HBr →
In presence peroxide,
≡
+ HBr →
- When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
- This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
- One mole of HBr adds to one mole of 1-heptane.
- The structure of heptene formed is -
Answer:
where is the drawing? I can't help if I can't see it sorry
(the heat bomb calorimeter gained)=(the heat material A gave)
r.h.s=25.57 kJ/⁰C * (26.80-24.33)= 63.16 kJ
2.741g of material A made 63.16kJ
the heat of combustion per gram = 63.16/2.741 kJ/g =24.04kJ/g
The answer will be because of the unequal sharing of electrons between the hydrogen and the oxygen atoms of a water molecule.
Hope this helps.
I'd appreciate brainliest answer from u please.
Answer:
4.77 is the pH of the given buffer .
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BK_a%5D%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BK_a%5D%2B%5Clog%28%5Cfrac%7B%5BCH_3CH_2COONa%5D%7D%7B%5BCH_3CH_2COOH%5D%7D%29)
We are given:
= Dissociation constant of propanoic acid = 
![[CH_3CH_2COONa]=0.254 M](https://tex.z-dn.net/?f=%5BCH_3CH_2COONa%5D%3D0.254%20M)
![[CH_3CH_2COOH]=0.329 M](https://tex.z-dn.net/?f=%5BCH_3CH_2COOH%5D%3D0.329%20M)
pH = ?
Putting values in above equation, we get:
![pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B1.3%5Ctimes%2010%5E%7B-5%7D%5D%2B%5Clog%28%5Cfrac%7B%5B0.254%20M%5D%7D%7B%5B0.329%5D%7D%29)
pH = 4.77
4.77 is the pH of the given buffer .
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