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Elena L [17]
2 years ago
6

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

alculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20C and an atmospheric pressure of 682 TM
Chemistry
1 answer:
netineya [11]2 years ago
4 0

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

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The km of an enzyme is 5. 0 mm. Calculate the substrate concentration when this enzyme operates at one‑quarter of its maximum ra
Makovka662 [10]

The substrate concentration of the enzyme operating at one‑quarter of its maximum rate is = 0.333.

Relationship between Km and substrate concentration is -

Km is the concentration of substrate.It allows the enzyme to achieve half Vmax. High Km enzyme requires a higher concentration of substrate to get Vmax. Since, Km is a constant. If the substrate concentration is increased, it has no effect on it.

An enzyme with a high Km has a low affinity for its substrate. The substrate concentration Km corresponds to the substrate concentration.

The substrate concentration at which the reaction rate of the enzyme-catalyzed reaction is half of the maximum reaction rate Vmax.

The equation is:

                <em>V₀ = </em><u><em>Vmax [S] </em></u>

<em>                          [S] + Km</em>

Here,

V₀ is initial rate,

Km is the dissociation constant between the substrate and the enzyme,

Vmax is the maximum rate, and

S is the concentration of substrate.

taking fraction of V₀ and Vmax :

<u><em>    V₀    </em></u><em> = </em><u><em>     [S</em></u><em>]</em><u><em>     </em></u><em>  </em>

<em>Vmax      [S] + Km</em>

<u><em>   </em></u><u>V₀   </u> =   <u>   0.5Km   </u>  = 0.333

Vmax       1.50 + Km

Therefore, the substrate concentration of this enzyme operating at one‑quarter of its maximum rate is = 0.333.

To learn more about substrate concentration,

brainly.com/question/18237939

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3 0
1 year ago
Please help!
Grace [21]
The most common reaction that causes spoilage isn't a reaction at all. Molds and Bacteria are attracted to the easily found presence of water in the fruit. They find a natural place to reproduce and what they do causes spoilage.

Very few sources talk about the chemical changes that take place. If you put fruit in a refrigerator it slows the spoiling process down. That means that the chemical reaction has to be endothermic (it requires heat to occur)

The process of spoilage is speeded up by bananas for example, giving up Ethylene gas. You do not want to put a banana with tomatoes, because tomatoes are very sensitive to Ethylene. (It's OK to eat them together. They make a terrific salad. Yum).

I cannot find a definitive source that connects all this together, but the conduct of the fruit in refrigerators confirms what I am saying.

Spoilage is a very complex reaction and interaction with the environment. I have given you a hint of what happens but you should search it out to convince yourself of the outcome.
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