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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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10-kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the con
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Answer:

Temperature = 0.605°C

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Checking the table for 134a pressure table. It is given that the specific volume of saturated liquid and the specific volume of the saturated vapor of 280kpa is 0.0007699 m^3/kg and 0.072352 m^3/kg respectively.

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300 - 280/ 320 - 280 = n - 0.072352/ (0.063604 - 0.072352).

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Specific volume = 14 × 10^-3/ 10 = 0.0014 m^3/kg.

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300 - 280/ 320 - 280 = b(I) - 199.54/ (196.7 - 199.54).

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Hence, b = [ 300 - 280/ 320 - 280 = j - 50.18 / 55.16 - 50.18] + [ ( 0.0014 - 0.00077735) / 0.067978 - 0.00077735] × 198.125.

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300 - 280/ 320 - 280 = T + 1.25 / 2.46 - 1.25.

Temperature = 0.605°C.

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Total enthalpy at 600kpa = 846.45 kJ.

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