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Vladimir [108]
3 years ago
13

A competitive inhibitor of an enzyme-catalyzed reaction - cannot bind to the active site. - always interferes with product relea

se. - is usually structurally similar to the substrate. - binds to an allosteric site. - inhibits to the same extent at all substrate concentrations.
Chemistry
1 answer:
Brut [27]3 years ago
4 0

Answer:

is usually structurally similar to the substrate.

Explanation:

Competitive inhibitors resemble normal substrate and binds to enzyme at the active site usually and prevents substrate from binding.

Active sites are main location for the substrate-enzyme binding. These sites involve weak as well as reversible bonds between the substrate and the enzyme. These inhibitors bind to the active sites and form weak and  reversible bonds. Competitive inhibitors can be dissociated from active site by increasing concentration of the substrates. Substrates has to compete for active site and displace the bound competitive inhibitors.

<u>Hence, correct option is - is usually structurally similar to the substrate.</u>

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You carefully weigh out 11.00 g of caco3 powder and add it to 44.55 g of hcl solution. you notice bubbles as a reaction takes pl
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8 0
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An unknown compound is processed using elemental analysis and found to contain 117.4g of platinum 28.91 carbon and 33.71g nitrog
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Answer:

1 mole of platinum

Explanation:

To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.

The empirical formula for the compound can be obtained as follow:

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Divide by their molar mass

Pt = 117.4 / 195 = 0.602

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N = 33.71 / 14 = 2.408

Divide by the smallest

Pt = 0.602 / 0.602 = 1

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N = 2.408 / 0.602 = 4

The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.

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How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

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4 years ago
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