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Len [333]
3 years ago
7

A 1-L soda bottle contains dissolved carbon dioxide gas. Which bottle would have the highest concentration of dissolved gas? a b

ottle that is sealed a bottle that was opened 1 minute ago a bottle that was opened 1 hour ago a bottle that was opened 1 day ago
Chemistry
2 answers:
Svetlanka [38]3 years ago
8 0

Answer:

A. A bottle that is sealed

Explanation:

lesya [120]3 years ago
3 0

Answer:  The correct answer would be : "A bottle that is sealed".

I hope that this helps you !

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An inhibitor is added to an enzyme-catalyzed reaction at a concentration of 26.7 μM. The Vmax remains constant at 50.0 μM/s, but
fomenos

Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

<h3>What is the Ki for the inhibitor?</h3>

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

  • Kma = Km/(1 + [I]/Ki)

Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

where:

  • Kma is the apparent Km due to inhibitor
  • Km is the Km of the enzyme-catalyzed reaction
  • [I] is the concentration of the inhibitor

Solving for Ki:

where

[I] = 26.7 μM

Km = 1.0

Kma = (150% × 1 ) + 1 = 2.5

Ki = 26.7 μM/{(2.5/1) - 1)

Ki = 53.4 μM

Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

Learn more about enzyme inhibition at: brainly.com/question/13618533

5 0
2 years ago
Help me please!! What is the name of the alkyne molecule
juin [17]

Answer:

ans is (2) 2,4- hexadiene

5 0
3 years ago
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

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You measure the mass of a model car to be 230 grams.The actual mass is 218 grams .what is your percent error
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Your percent is 5.50458716
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what would be the ph of an aqueous solution of sulphuric acid which is 5×10^_5 mol l^_1 in concentration​
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the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature

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