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aksik
2 months ago
6

Behold a mixture of oily/hydrophobic (yellow) and water (purplish) molecules. recall that hydrophobic regions are ones with few

or no partial charges--so there's nothing for water's hydrogen interaction donors or acceptors to 'play with'. so! how many of the water molecules are 'stuck' contacting at least part of a hydrophobic molecule in this arrangement?

Chemistry
1 answer:
Tems11 [2.7K]2 months ago
7 0

Answer:

  around 40

Explanation:

The included diagram shows 5 hydrophobic molecules, each surrounded by 9 water molecules. Therefore, there are "around 40" water molecules that are in contact with the hydrophobic molecules.

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You can test it by combining it with vinegar. The acetic acid present in vinegar reacts with sodium bicarbonate (baking soda) to produce carbon dioxide, resulting in an intense bubbling reaction that is harmless. Baking powder, however, won’t produce this effect.
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2 months ago
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a 0.5678 of KHP required 26.64cm³ of NaOH to complete neutralization.calculate the molarity of the NaOH solution​
lions [2927]

Answer:

Explanation:

0.5678 G        X GRAMS

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 MOL               1 MOL

0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4

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4 0
2 months ago
Using the equation pH = -log [H+], determine the pH of a solution with a hydrogen ion concentration, or [H+], of 1x10-5. HINT: l
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Response:

The result is a pH of 7. I’m uncertain about the mechanics of Hydrochloric acid reactions

8 0
1 month ago
A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74
KiRa [2933]

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product (CH3COOH)^{2} gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol (CH3COOH)^{2} / 2 mol CH3COOH)

= 0.005 mol (CH3COOH)^{2}

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^{2}

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

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4 0
2 months ago
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