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2 years ago

15

Describe the relationship between an enzyme, substrate, and active site.

1 answer:

alexdok [17]2 years ago

8 0

So what I know is that enzyme and substrate are like lock and key meaning that when the active site of the enzyme changes, the enzyme will not fit to the substrate which will lead the enzyme to denature. Hope this helps.

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The number of protons in every atom of an element

ANTONII [103]

Answer: The atomic number is the number of protons in an atom of an element. In our example, krypton's atomic number is 36. This tells us that an atom of krypton has 36 protons in its nucleus.

Explanation:

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2 years ago

A molecule contains 24.36 g of nitrogen and 62.64g of sliver

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Answer:

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2 years ago

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Methane's chemical formula is CH. Is there a bond between any of the hydrogen atoms? Why or why not? (1 point)

irinina [24]

Answer:

No, there is not because it would form H2 instead of methane if hydrogen bonded with itself.

Explanation:

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2 years ago

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djyliett [7]

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Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of

shusha [124]

Answer:

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}

Explanation:

1. Write down the balanced chemical reaction:

$2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}$

2. Find the limiting reagent:

- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

For the hexane:

$70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}$

For the oxygen:

$81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}$

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

$\frac{0.81}{2}=0.41$

For the oxygen:

$\frac{2.54}{19}=0.13$

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

$81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}$

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3 years ago