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Tju [1.3M]
3 years ago
15

How many atoms in 10.0 g of Ar?

Chemistry
1 answer:
NARA [144]3 years ago
6 0

Answer:

1.51 x 10^23 atoms of Ar

Explanation:

Start out with what you're given.

10.0 g of Ar/1 * 1 mol/39.948 g Ar * 6.022*10^23 atoms/1 mol

(It's easier if you write it out.)

You always want to start out w/ what you're given and continue your unit conversions from there. Since you started out with grams of Argon, you want grams of Argon on the bottom to cancel it out. Once you found the unit conversion: 1 mol/39.948 g Ar (molar mass of Ar), now it's time to convert to atoms. Since 1 mol is up on top, put 1 mol on the bottom. Considering you're looking for atoms, use Avagradro's number: 6.022*10^23.

Once simplified, you should get 1.51 x 10^23 atoms of Ar.

***I hope this helps! I also suggest using a mole map, if you're interested.***

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Answer:

<u>One lone-Pair is present in Ammonia</u>

<u></u>

Explanation:

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3 years ago
What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?
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Answer:

\% diss = 50\%

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

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Thus, it is possible to find x given the pH as shown below:

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So that we can calculate the initial concentration of the acid:

\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\

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\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%

Best regards!

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