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zhannawk [14.2K]
2 years ago
11

If two solutions with concentrations of 0.4 M sugar and 0.7 M sugar respectively are separated by a semipermeable membrane, duri

ng osmosis there is a net flow of Group of answer choices sugar molecules from the dilute to the concentrated solution sugar molecules from the concentrated to the dilute solution water molecules from the concentrated to the dilute solution water molecules from the dilute to the concentrated solution
Chemistry
1 answer:
user100 [1]2 years ago
5 0

Answer: Water molecules from the dilute to the concentrated solution

Explanation:

During Osmosis if a solution is separerated by a semipermeable membrane, the solvent (typically water) from the less concentrated solution in terms of solute goes through the semipermeable membrane to the solution with the higher concentration so that the concentrations between the solutions can be balanced.

With the above solutions therefore, water molecules would move from the solution of 0.4M of sugar to the solution with a 0.7M of sugar through the semipermeable membrane.

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Physical properties of a bag of microwaveable popcorn are the mass of it, the color of it, the size of it, and the weight of it. Two chemical properties of a bag of microwavable popcorn are it changed from seeds to popcorn and it popped.
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What could be done to change this carbide ion to a neutral carbon atom? A) remove 2 electrons B) add 2 electrons C) remove 4 ele
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Explanation:

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Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0
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Answer:

pH = 7.1581

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The equilibrium of NaHSO₃ with Na₂SO₃ is:

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<em>Where K of equilibrium is the Ka2: 6.5x10⁻⁸</em>

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HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

HSO₃⁻: 0.252 moles

SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

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SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

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pH = 7.187 + log [0.189] / [0.202]

<h3>pH = 7.1581</h3>

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2 years ago
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The disadvantages of each of the given model of electron configuration have been mentioned below:

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Answer:

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