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vagabundo [1.1K]
3 years ago
10

A researcher claims that increasing temperature by 10 degrees will double the rate of an enzyme-catalyzed reaction. To test this

, he designed an experiment. For each tube, the researcher will measure the rate of product formation
Which best justifies the inclusion of tube 5 as a control?


It will provide a measurement of product formation in the absence of the substrate

It will provide a measurement of product formation in the presence of a denatured enzyme

It will show the effect of doubling the amount of substrate on the rate of product formation

It will show the effect of increased enzyme activity on the rate of product formation
Chemistry
1 answer:
BigorU [14]3 years ago
8 0

Answer:

It will provide a measurement of product formation in the presence of a denatured enzyme

Explanation:

For any experimental research study, it is a common practice to have sets of experimental runs as well as a control. The main essence of the control is to investigate the effects of a given substance such as an enzyme on the experimental conditions. This is done by comparing the experimental results in the presence of an enzyme with the results in the absence of an enzyme.

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2 years ago
Classify each process as an endothermic or exothermic process. drag the appropriate items to their respective bins. helpreset ex
myrzilka [38]
Answers:

1) <span>Breaking Solvent-Solvent Attractions is an Endothermic Process.

2) </span><span>Breaking Solute-Solute Attractions is an Endothermic Process.

3) </span><span>Forming Solute-Solvent Attractions is an Exothermic Process.

Explanation:
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7 0
3 years ago
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Which one of the following formulas represents an aldehyde? 
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2 years ago
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Makovka662 [10]

Answer:

K = 0.5

Explanation:

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The equilibrium constant, K, is defined as:

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<em>Where P represent the pressure at the equilibrium for each one of the gases involved in the equilibrium.</em>

<em />

As:

P PCl₅ = 1.0atm

P PCl₃ = 1.0atm

P Cl₂ = 2.0atm

K = 1.0atm / 1.0atm * 2.0atm

<h3>K = 0.5</h3>
7 0
2 years ago
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