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Mrrafil [7]
3 years ago
8

Water is dissolved into n-butanol (a polar liquid). Which is the second step at the molecular level as water dissolves into n-bu

tanol?
A. n-Butanol mixes with water.

B. Water molecules are carried into n-butanol.

C. n-Butanol molecules surround water molecules.

D. n-Butanol molecules are attracted to the surface of the water molecules.
Chemistry
2 answers:
hodyreva [135]3 years ago
5 0
Keeping it nice and simple, it is D- Water molecules surrounded by water molecules who are then carried into n-butanol

Hope this helps

the real answer is D n butanol molecules are attracted to the surface of the water molecules

Have a good day and if you have an ig use the tag Whataboutshe to gain more followers

plz dont report or flag. love you
elena55 [62]3 years ago
4 0

Answer:

C. n-Butanol molecules surround water molecules.

Explanation:

Because both liquids are polar, they can be mixed (polar dissolves polar, and nonpolar dissolves nonpolar). N-Butanol is the solvent, so it has a large amount in the solution, and water is the solute. When a liquid-liquid solution is placed, the solvent surrounds the molecules of the solute, so the solute is dissociated in the solution.

Then, the n-Butanol molecules surround the water molecules.

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How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?
givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

H2SO3\leftrightarrow H^{+}+HSO3^{-} --------(1)

where,  Ka1 = 1.5 x 10–2  or pKa1 = 1.824

HSO3^{-}\leftrightarrow H^{+}+SO3^{2-} --------(2)

where,  Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

i.e. \ 5.13\  moles/L*0.385L = moles\  NaOH\\therefore, \ moles\  NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L

Step 2:

For the second equivalence point setup an ICE table:

                  HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}

Initial           1.98                    ?                                       0

Change      -x                       -x                                       x

Equil           1.98-x                 ?-x                                    x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles

Volume of NaOH required is:

\frac{0.634\ moles}{0.615 moles/L}=0.389L

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0
3 years ago
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