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andrew-mc [135]
3 years ago
9

Can someone please help me out !!!

Chemistry
1 answer:
lakkis [162]3 years ago
5 0

Explanation:

Volume = FL - SL

= 39 - 33

= 6

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Answer:

valence electron

Explanation:

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3 years ago
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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2 years ago
An atom that normally has __________ electrons in its outer (valence) shell tends not to form chemical bonds with other atoms.
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Explanation:

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3 years ago
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