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tiny-mole [99]
3 years ago
9

Question: How will soap or alcohol affect the hydrogen bonds between different water molecules? Hypothesis: Materials: Penny, wa

ter, soap, pipette, paper towel, 70% rubbing alcohol
Chemistry
2 answers:
leonid [27]3 years ago
8 0

Answer:

soap and alcohol will make the hydrogen bonds weak, and weaker

Explanation:

A water molecule is polar as the oxygen atom is strongly electronegative and draws the electrons it shares with the hydrogen atoms in the water molecule to itself. This creates a partial negative charge on the oxygen atom and partial positive charges on the hydrogen atoms. The attraction of neighboring oxygen to hydrogen atoms in a water molecule results in hydrogen bonding.

However, a detergent molecule which has a polar head and a hydrophobic tail, when added to water significantly interferes with the hydrogen bonding between water molecules. The polar head is attracted to water molecules whereas the hydrophobic tail of each molecule are not. The detergent molecules form clusters known as micelles with the polar heads attracted to water and the hydrophobic tails directed towards the center of the micelle, away from water molecules. This weakens the hydrogen bonding between water molecules

An alcohol contains a polar -OH group in its molecules. However, the hydrogen bonding between alcohol molecules are weaker than that in water because the other end of the alcohol molecule has a C-O bond which is less polar than the O-H bond. Therefore, when an alcohol is added to water, it weakens the hydrogen bonding in water.

This weakening of the hydrogen bonding in water also has the effect of lowering the surface tension of water.

Stella [2.4K]3 years ago
7 0

Answer:

A soap or alcohol will interfere with the hydrogen bonding between water molecules making it weaker

Explanation:

A water molecule is polar as the oxygen atom is strongly electronegative and draws the electrons it shares with the hydrogen atoms in the water molecule to itself. This creates a partial negative charge on the oxygen atom and partial positive charges on the hydrogen atoms. The attraction of neighboring oxygen to hydrogen atoms in a water molecule results in hydrogen bonding.

However, a detergent molecule which has a polar head and a hydrophobic tail, when added to water significantly interferes with the hydrogen bonding between water molecules. The polar head is attracted to water molecules whereas the hydrophobic tail of each molecule are not. The detergent molecules form clusters known as micelles with the polar heads attracted to water and the hydrophobic tails directed towards the center of the micelle, away from water molecules. This weakens the hydrogen bonding between water molecules

An alcohol contains a polar -OH group in its molecules. However, the hydrogen bonding between alcohol molecules are weaker than that in water because the other end of the alcohol molecule has a C-O bond which is less polar than the O-H bond. Therefore, when an alcohol is added to water, it weakens the hydrogen bonding in water.

This weakening of the hydrogen bonding in water also has the effect of lowering the surface tension of water.

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At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration
wlad13 [49]

Answer:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.03901 mol/liter
  • [Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

  • Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
  • Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

  • Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
  • Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

            H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

I            0.0481      0.0326            0

C              -x                 -x              +x

E          0.0481-x    0.0326-x         x

<u />

<u>4. Equilibrium expression</u>

       K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}

     0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}

<u />

<u>5. Solve:</u>

            x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0

Use the quadatic formula:

x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}

The positive result is x = 0.00909

Thus the concentrations are:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
  • [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

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