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dalvyx [7]
3 years ago
13

Calculate the pH at of a 0.10 Msolution of anilinium chloride . Note that aniline is a weak base with a of . Round your answer t

o decimal place. Clears your work. Undoes your last action. Provides information about entering answers.
Chemistry
1 answer:
notsponge [240]3 years ago
8 0

The question is incomplete, here is the complete question:

Calculate the pH at of a 0.10 M solution of anilinium chloride (C_6H_5NH_3Cl) . Note that aniline (C6H5NH2) is a weak base with a pK_b of 4.87. Round your answer to 1 decimal place.

<u>Answer:</u> The pH of the solution is 5.1

<u>Explanation:</u>

Anilinium chloride is the salt formed by the combination of a weak base (aniline) and a strong acid (HCl).

To calculate the pH of the solution, we use the equation:

pH=7-\frac{1}{2}[pK_b+\log C]

where,

pK_b = negative logarithm of weak base which is aniline = 4.87

C = concentration of the salt = 0.10 M

Putting values in above equation, we get:

pH=7-\frac{1}{2}[4.87+\log (0.10)]\\\\pH=5.06=5.1

Hence, the pH of the solution is 5.1

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Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. include hydrogen atoms,
gogolik [260]
<h2>Answer</h2>

Bromination:

Any reaction or process in which bromine (and no other elements) are introduced into a molecule.

Bromonium Ion:

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)

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In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".

Step 2:

When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.

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6 0
3 years ago
A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike and the rider is 85 kg. Find the kinetic ener
krek1111 [17]

KE=3070.625 J

Height = 3.686 m

<h3>Further explanation</h3>

mass of bike+rider=85 kg

velocity = 8.5 m/s

\tt KE=\dfrac{1}{2}mv^2\\\\KE=\dfrac{1}{2}\times 85\times 8.5^2\\\\KE=3070.625~J

Conservation of energy :

(KE+PE)₁ (downhill) = (KE+PE)₂ (up the hill)

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KE₁=PE₂

\tt KE_1=3070.625\\\\KE_1=mgh_2\\\\3070.625=85\times 9.8\times h\Rightarrow h=3.686~m

5 0
3 years ago
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