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iren [92.7K]
3 years ago
10

Given that you added 5 ug of a purified acid phosphatase to tube a, calculate the amount of acid phosphatase (in ug) that was pr

esent in the 400ul of the wheat germ extract in tube b
Chemistry
1 answer:
Leona [35]3 years ago
5 0

solution:

Amount of purified acid phosphatase added to tube A = 5\mu g(micrograms)\\Amount of acid phosphatase present in 400\mu g of wheat germ extract in tube B=1/100=5x/400\\100x=(0.5)(400)\\X=(0.5)(400)/100=2\mu g\\The amount of acid phosphatase present in 400\mu g of the wheat germ extract in tube B is 2\mu g

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How many molecules are contaiined in 125 grams of water, H2O?
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Answer:

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List the official SI units
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Read 2 more answers
The following solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentrations.
lions [1.4K]

Answer:

1M MgCl₂ > 1M KCl > 1M C₁₂H₂₂O₁₁

Explanation:

The osmotic pressure (π) is the pressure needed to impede the osmose, it means that it's the necessary pressure to prevent the solvent to go through a membrane.

It can be calculated by:

π = M*R*T*i

Where M is the molarity of the solution (mol/L), R is the ideal gas constant, T is the temperature, and i the van't Hoff factor.

This factor is a way to correct the number of particles that are dissolved in a solute, and it can be calculated by:

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Because all the substances have the same molarity (1 M) and are at the same temperature, let's analyze the value of i, which is directly proportional to π.

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Both KCl and MgCl₂ are soluble salts and will dissociate completely (α = 1), but MgCl₂ will have 3 particles (Mg²⁺ + 2Cl⁻), and KCl only one particle (K⁺ and Cl⁻), so qMgCl₂ > 1KCl, and so πMgCl₂ > πKCl, which will be higher than 1.

1M MgCl₂ > 1M KCl > 1M C₁₂H₂₂O₁₁

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3 years ago
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