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Vitek1552 [10]
3 years ago
5

What is the IUPAC name for the following compound?

Chemistry
1 answer:
djyliett [7]3 years ago
5 0

Answer:

Option a. Benzenecarbonitrile

Explanation:

You are not providing the picture of the compound. However, I manage to find the picture of it, so I'm gonna answer this with the picture. Its attached here at the end.

As for the IUPAC name of this compound, it would be Benzenecarbonitrile, or simply, benzonitrile.

This compound has the molecular formula C₆H₅CN. It receives this name over the other ones, because this is a aromatic cyanide, and aromatic cyanide has nitriles behavior, (Strength of the nitrogen over the carbon), therefore, this group has priority over the whole molecule. As the principal chain is the benzene, the carbonitrile is the radical of the group, and when cyano groups are as radical and not as principal chains, it has a behavior of nitrile, and it's named as nitrile. therefore, option b, cannot be the name.

phenylcyanide is naming the phenyl as radical, but the benzene is the principal group in this compound, and the cyanide is the radical. It happens the same thing with cyanophenyl.

Hope this helps

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A reaction produces 0.809 moles of h2o. how many molecules are produced?
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Use Avogadro's number which is numerically equal to ...

( 6.02 multiply by 10 to the power of negative 23) 


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0.809 mole contains....(6.02* 10^-23) (0.809)
which is equal to [4.951* 10 ^-23]

if I made no mistake in calculation...it must be correct
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A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using this information, calculate Ka for HA.
slavikrds [6]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant ({ \alpha}) is :

\qquad \sf  \dashrightarrow \:  \alpha =  \frac{dissociation \:  \: percentage}{100}

\qquad \sf  \dashrightarrow \:  \alpha =  \frac{9.4}{100}  = 0.094

Now, at initial stage :

  • \textsf{ Conc of HA = 0.200 M}

  • \textsf{Conc of H+ = 0 M}

  • \textsf{Conc of A - = 0 M}

At equilibrium :

  • \textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M}

  • \textsf{Conc of H+ = 0.094(0.200)  = 0.0188 M}

  • \textsf{Conc of A - = 0.094(0.200)  = 0.0188 M}

Now, we know :

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}}

( big brackets represents concentration )

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}}

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}}

\qquad \sf  \dashrightarrow \: { K_a \approx 0.00195 }

\qquad \sf  \dashrightarrow \:  {K_a \approx 1.9 × {10}^{-3} }

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