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trasher [3.6K]
3 years ago
13

Write the balanced

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
4 0
It’s c I just did this so yea c
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Calculate the osmotic pressure associated with 50.0 g of an enzyme of molecular weight 98 g/mol dissolved in water to give 2600
andrew-mc [135]

Answer:

π = 4,882 atm

Explanation:

To calculate the osmotic pressure (π), the <em>Van´t Hoff equation</em> must be used, which is:

π x V = n x R x T

<em>Where: </em>

• π: Osmotic pressure, which is the difference between the levels of the solution and the pure solvent through a semipermeable membrane, which allows the passage of the solvent but not the solute

• V: Volume of the solution, in liters unit

• n: Number of moles of solute

• R: Constant of ideal gases, equal to 0.08206 L.atm / mol.K

• T: Absolute temperature, in Kelvin degrees

With the data you provide you can calculate the osmotic pressure by clearing it from the equation, we would be equal to:

π = (n x R x T) / V

However, all data must first be converted to the corresponding units in order to replace the values ​​in the equation.

<em>Solution volume ⇒ go from mL to L: </em>

1000 mL of solution ____ 1 L

2600 mL of solution _____ X = 2.6 L

Calculation: 2600 mL x 1 L / 1000 mL = 2.6 L

<em>Temperature ⇒ Go from ° C to K </em>

T (K) = t (° C) + 273.15 = 30.0 ° C + 273.15 = 303.15 K

<em>Number of moles of solute ⇒</em> <em>It can be calculated since we have the mass of the enzyme and its molecular mass: </em>

98.0 g of enzyme ____ 1 mol

50.0 g of enzyme _____ X = 0.510 moles

Calculation: 50.0 g x 1 mol / 98.0 g = 0.510 moles

Now, you can replace the values ​​in the Van´t Hoff equation and you will get the result:

 π = (n x R x T) / V

π = (0.510 mol x 0.08206 L.atm / mol.K x 303.15 K) / 2.6 L = 4.882 atm

Therefore, <em>the osmotic pressure will be 4,882 atm</em>

3 0
3 years ago
A sample of potassium chlorate (15.0 g) is dissolved in 201 g of water at 70 °c with precautions taken to avoid evaporation of a
zlopas [31]
Answer is: unsaturated.
Solubility of potassium chlorate on 70°C is approximately 30 grams in 100 grams of water.
Solubility of potassium chlorate on 70°C is approximately 10 grams in 100 grams of water.
So if dissolve 15 g of potassium chlorate in 201 g of water, there is less salt than it solubility and solution is unsaturated.
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Its either A or B, 50-50 chance?
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