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Snezhnost [94]
3 years ago
7

Describe the solubility of water compared to ethanol?

Chemistry
1 answer:
Neko [114]3 years ago
4 0
As you know ethanol is a an alcohol and alcohol is a hydrocarbon. Alcohol is made up of a carbon chain which Is always non polar and a OH group which is polar. According to the solubility rule like substances dissolves like substance. Using ethanol chemical formula. Ethanol has a 2 carbon chain and a OH group. water is polar so it will be attracted to the OH group. Carbon chain on the other hand is nonpolar so it will be repelled from the water.


Therefore the Solubility of alcohols is determined by the stronger of the two forces. The strength of the attraction of the OH group, and the amount of water they dissolve in.
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1 - 2Na + O2 ➔ Na2O Balanced Not Balanced
Mrrafil [7]

Answer:

1 not

2 not

3 not

4 balanced

5 not

6 not

7 balanced

Explanation:

the amount of elements must be equal in the reactant and products

5 0
3 years ago
Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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