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Irina-Kira [14]
3 years ago
9

Predict the effect of adding a non competitive inhibitor to the reaction mixture on the rate of reaction at a high substrate con

centration. Support your prediction by describing how a non competitive inhibitor affects the structure and function of an enzyme.
Chemistry
1 answer:
Darya [45]3 years ago
8 0

Answer:

A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time

Explanation:

this is because It changes the conformation of an enzyme as well as its active site, which makes the substrate unable to bind to the enzyme effectively so that the efficiency of the enzyme decreases. A noncompetitive inhibitor binds to the enzyme away from the active site, altering/distorting the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively and most of the time also the inhibitor is reversible

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When a metal reacts with oxygen gas what is the product formed
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Reika [66]

Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

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