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WINSTONCH [101]
4 years ago
11

I recently started a science experiment and need help coming up with a catchy title. Can you help me? Here is my problem stateme

nt:
Does salt concentration effect the boiling temperature of an egg?

Please let me know what I can title my project if you have any ideas. I would appreciate it. :)
Chemistry
2 answers:
kirill115 [55]4 years ago
7 0
I think Salty Boiling would be a good one lol :)
Anna71 [15]4 years ago
5 0
"Egg-saline"
Kind of like "eggcellent." Maybe "egg-saline eggs" would work.

Saline means "impregnated with salt."
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2 Na + 1 Cl2 → 2 Naci<br> How many grams of NaCl are created from 23,2 grams Cl2?
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In a 0.100 m hf solution, the percent dissociation is determined to be 9.5%. calculate the ka for hf based on this data.
Elden [556K]

Answer : The dissociation constant (Ka) = 9.025 × 10^{-4}

Solution :  Given,

                Concentration HF solution = 0.100 M

                 % Dissociation = 9.5 %

The equation for dissociation of HF is :

                  HF \rightleftharpoons H^{+}+ F^{-}

The Ka expression for HF is :

Ka=\frac{[H^{+}][F^{-}]}{[HF]}        ............. (1)

Step 1 : we find the [H^{+}] by using the concentration and % dissociation.

 [H^{+}] = Concentration HF solution ×  % Dissociation

 [H^{+}] = 0.100 M × \frac{9.5}{100} = 9.5 × 10^{-3} M

Step 2 : For [F^{-}] , the concentration of  [F^{-}] is equal to the  [H^{+}]. From the above equation the stoichiometry of   [F^{-}] and [H^{+}] is 1:1.

Therefore,

[F^{-}] =  [H^{+}]  = 0.100 M × \frac{9.5}{100} =    =  9.5 × 10^{-3} M

         

Now, put all the values in equation (1), we get

Ka=\frac{(9.5\times10^{-3})\times(9.5\times10^{-3})}{(0.1)}

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