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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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d.have a mass of 1 amu..

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The average strength of a hydrogen bond is approximately what percent of the average strength of a covalent bond?
lozanna [386]

<span>Hydrogen bonds are approximately 5% of the bond strength of covalent bonds, for example (C-C or C-H bonds).
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You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the
Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate  reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

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Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

8 0
3 years ago
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