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Ilya [14]
3 years ago
13

The product formation was decreased when a substance b was added to an enzyme reaction. more substrate being added did not incre

ase the amount of produce formed. based on this we assume that substance b could be
Chemistry
1 answer:
borishaifa [10]3 years ago
7 0
When the product formation is decreased if a substance B is added to an enzyme reaction and more substrate being added would not increase the amount of produce formed, then we assume that substance b could be a noncompetitive inhibitor. This type of inhibitor would be one that would bind to the enzyme with or without the presence of a substrate in different sites at the same time. It would change the conformation of the enzyme and also the active sites. As a result, the substrate would not be able to bind to the enzyme more effectively than the usual. The overall efficiency would decrease.
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3 years ago
15.3 g of nano3 were dissolved in 100g of water in a calorimeter. The temperature of the water drops from 25.00°c to 21.56°c. Ca
Arturiano [62]

Answer:

0.259 kJ/mol ≅ 0.26 kJ/mol.

Explanation:

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = ??? J).

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c is the specific heat of water (c of ice = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).

<em>∵ Q = m.c.ΔT</em>

∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.

<em>∵ ΔH = Q/n</em>

n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.

∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.

3 0
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  - www.eduplace.com › science › hmxs › pdf

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Will Mark Brainliest <br>Aswer for<br>box 1 and box 2<br>On top of the oval says <br>A-U-G-G-C-A
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