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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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When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain
yKpoI14uk [10]

<u>Answer:</u>

<u>For a:</u> The number of molecules of nitrogen dioxide is 4.52\times 10^{23}

<u>For b:</u> The mass of nitric acid formed is 54.81 grams

<u>For c:</u> The mass of nitric acid formed is 206 grams

<u>Explanation:</u>

The given chemical reaction follows:

3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)

  • <u>For a:</u>

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with \frac{3}{1}\times 0.250=0.75mol of nitrogen dioxide

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.75 moles of nitrogen dioxide will contain 0.75\times 6.022\times 10^{23}=4.52\times 10^{23} number of molecules

Hence, the number of molecules of nitrogen dioxide is 4.52\times 10^{23}

  • <u>For b:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = \frac{2}{3}\times 1.304=0.870 moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g

Hence, the mass of nitric acid formed is 54.81 grams

  • <u>For c:</u>
  • <u>For nitrogen dioxide:</u>

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol

  • <u>For water:</u>

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = \frac{1}{3}\times 4.90=1.63mol of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce \frac{2}{3}\times 4.90=3.27mol of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g

Hence, the mass of nitric acid formed is 206 grams

7 0
2 years ago
Which option can be classified as a pure substance
Zina [86]
I have done this question before and even though you didn’t provide the options. I think the correct option is elements
4 0
3 years ago
Read 2 more answers
Write a balanced equation formed when the following elements react with oxygen:a)Zinc
Ainat [17]

Explanation:

a) when zinc burnt in oxygen.

2Zn + O2 -----∆-----> 2ZnO(black residue)

b) when carbon burnt in oxygen.

C+O2----∆---> CO2.

c) when sulphur burnt in oxygen.

S+O2-----∆-----> SO2.

d) when Calcium burnt in oxygen.

2Ca+O2-----∆-----> 2CaO(black residue)

e) when Magnesium burnt in oxygen.

2Mg+O2-----∆----> 2MgO.

f) when sodium burnt in oxygen.

4Na+O2----∆-----> 2Na2O.

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What do these images model ?
jek_recluse [69]
The answer is A

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3 years ago
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Explain why a city that receives 10 hours more light each day has colder average temperatures. Explain
Anestetic [448]

Explanation:

This would be similar to how the poles work in the North and South. Due to their placement, they recieve much more sun light than the rest of the Earth since we rotate horizontally. However, their placement also makes them farther from the sun than the rest of Earth, therefore not receiving as much heat.

3 0
3 years ago
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