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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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Nookie1986 [14]
7 is atomic symbol 9 is atomic density 8 is the atomic number
4 0
3 years ago
An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

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Answer:

A continent is a large distinct landmass.

There are seven continents on Earth.

Continents cover two-thirds of Earth’s surface.

Explanation:

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The rate in m/s is 5.2 * 10^-4 m/s.

<h3>What is the rate in m/s?</h3>

We know that the speed is given as the ratio of the distance covered to the time taken. In this case, we have been told that the rate at which the tide rises is 6.08 ft per hour. We would now need to convert the rate from  6.08 ft per hour to m/s.

Now;

We know that;

1 foot/hour = 8.5 * 10^-5 m/s

6.08 ft per hour = 6.08 ft per hour * 8.5 * 10^-5 m/s/1 foot/hour

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