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mixas84 [53]
3 years ago
14

9.Competitive inhibition occurs when a O A. substrate binds to an enzyme in the active site and activates the enzyme B. molecule

binds to an enzyme in the active site and prevents the substrate from binding C. molecule binds to an enzyme at a site other than the active site and inhibits the substrate from binding O D. molecule binds to an enzyme but doesn't change the shape of the active site Mark for review (Will be highlighted on the review page) << Previous Question Next Question >>​
Chemistry
1 answer:
alexgriva [62]3 years ago
5 0

Answer:

.Competitive inhibition occurs when a O A. substrate binds to an enzyme in the active site and activates the enzyme B. molecule binds to an enzyme in the active site and prevents the substrate from binding C. molecule binds to an enzyme at a site other than the active site and inhibits the substrate from binding O D. molecule binds to an enzyme but doesn't change the shape of the active site Mark for review (Will be highlighted on the review page) << Previous Question Next Question >>

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D

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Write a balanced equation for the following neutralization reaction HBr+LiOH&lt;-&gt;
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How many grams of Ag2S2O3 form
Art [367]

Taking into account the reaction stoichiometry, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgBr + Na₂S₂O₃ → Ag₂S₂O₃ + 2 NaBr

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • AgBr: 2 moles
  • Na₂S₂O₃: 1 mole
  • Ag₂S₂O₃: 1 mole
  • NaBr: 2 moles

The molar mass of the compounds is:

  • AgBr: 187.77 g/mole
  • Na₂S₂O₃: 158 g/mole
  • Ag₂S₂O₃: 327.74 g/mole
  • NaBr: 102.9 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • AgBr: 2 moles ×187.77 g/mole= 375.54 grams
  • Na₂S₂O₃: 1 mole ×158 g/mole= 158 grams
  • Ag₂S₂O₃: 1 mole ×327.74 g/mole= 327.74 grams
  • NaBr: 2 moles ×102.9 g/mole= 205.8 grams

<h3>Mass of Ag₂S₂O₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 375.54 grams of AgBr form 327.74 grams of Ag₂S₂O₃, 125 grams of AgBr form how much mass of Ag₂S₂O₃?

mass of Ag_{2} S_{2} O_{3} =\frac{125 grams of AgBrx327.74 grams of Ag_{2} S_{2} O_{3}}{375.54 grams of AgBr}

<u><em>mass of Ag₂S₂O₃= 109.09 grams</em></u>

Then, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

4 0
2 years ago
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