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Mazyrski [523]
3 years ago
10

BIOCHEMISTRY HELP! You and a lab partner are performing on several molecular analogs to determine Km. One is a substrate analog,

one is a transition state analog, and the last is a product analog. The data you obtain shows that the three Km's are 0.0013 M, 0.025 M, and 0.0045 M (in no particular order). Given what you know about the induced-fit model of catalysis, which Km would you hypothesize to belong to which analog, and why?
Chemistry
1 answer:
Alina [70]3 years ago
5 0

Answer:

Michaelis constant is known as km which is the substrate concentration that encourages the compound to work at half maximum velocity represented by Vmax/2. Michaelis constant is inversely related to the substrate and the affinity of the enzyme.  

Induced fit model: The premise of the purported induced fit hypothesis, which expresses that the attachment or association of a substrate or some other atom to an enzyme causes an adjustment to the enzyme in order to fit or restrain its activity.  

In substrate, analog Km or Michaelis constant will be high as the substrate will stay because of analogs inhibit activity.

In the transitional state, analog Km will be in the middle of the substrate and product analogs. Progress state analogs are synthetic mixes with a structure catalyzed reaction that looks like the progressing condition of a substrate atom in a compound enzyme.  

In item simple thus Km is the least.  

0.0013 M = product ananlog,

0.025 M=Transition state, and

0.0045 M = Substrate analog

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Answer:

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3 years ago
Freon−12 (CF2Cl2), widely used as a refrigerant and aerosol propellant, is a dangerous air pollutant. In the troposphere, it tra
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Answer:

141g of CCl₄

Explanation:

First, we have to write the balanced equation.

CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)

We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.

V = 14.9 dm³ = 14.9 L

T = 21°C + 273.15 = 294.15 K

P = 1.48 atm

R = 0.08206 atm.L/mol.K

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.48atm.14.9L}{0.08206\frac{atm.L}{mol.K}.294.15K }=0.914mol

We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.

0.914molCF_{2}Cl_{2}.\frac{1molCCl_{4}}{1molCF_{2}Cl_{2}} .\frac{154gCCl_{4}}{1molCCl_{4}} =141gCCl_{4}

7 0
3 years ago
2. What is the weight of the hydrochloric acid that fills a 144.5 mL container? The density of
hoa [83]
I don’t understand the question sorry
8 0
3 years ago
Dinitrogen oxide (N₂O) gas was generated from the thermal decomposition of ammonium nitrate and collected over water. The wet ga
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Answer:

126.73 mL

Explanation:

The total pressure of the gas mixture is the sum of the vapor pressure of its constituents. So, the vapor pressure of N₂O(p) can be calculated:

750 = 18.85 + p

p = 750 - 18.85

p = 731.15 torr

It means that for 731.15 torr, N₂O occupied 130 mL. For the general gas equation, we know that

\frac{p1V1}{T1} = \frac{p2V2}{T2}

Where <em>p</em> is the pressure, <em>V</em> is the volume, <em>T</em> is the temperature, 1 is the initial state, and 2 the final state. For the same temperatue (21ºC), the equation results on Boyle's law:

p1V1 = p2V2, so:

731.15x130 = 750xV2

750V2 = 95049.5

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4 0
3 years ago
A sample of argon gas occupies 105 mL at 0.871 atm. If
qwelly [4]

Answer:

1. final pressure = 0.259atm

2. 196.84mmHg

Explanation:

Using Boyle's law of equation

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (mL)

V2 = final volume (mL)

According to the information given in this question:

V1 = 105mL

V2 = 352mL

P1 = 0.871atm

P2 = ?

Using P1V1 = P2V2

P2 = P1V1/V2

P2 = 0.871 × 105/352

P2 = 91.455/352

P2 = 0.2598

P2 = 0.259atm

To convert 0.259atm of the gas into mmHg, we multiply the value in atm by 760.

Hence, 0.259 × 760

= 196.84mmHg

6 0
3 years ago
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