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mixas84 [53]
3 years ago
14

9.Competitive inhibition occurs when a O A. substrate binds to an enzyme in the active site and activates the enzyme B. molecule

binds to an enzyme in the active site and prevents the substrate from binding C. molecule binds to an enzyme at a site other than the active site and inhibits the substrate from binding O D. molecule binds to an enzyme but doesn't change the shape of the active site Mark for review (Will be highlighted on the review page) << Previous Question Next Question >>​
Chemistry
1 answer:
alexgriva [62]3 years ago
5 0

Answer:

.Competitive inhibition occurs when a O A. substrate binds to an enzyme in the active site and activates the enzyme B. molecule binds to an enzyme in the active site and prevents the substrate from binding C. molecule binds to an enzyme at a site other than the active site and inhibits the substrate from binding O D. molecule binds to an enzyme but doesn't change the shape of the active site Mark for review (Will be highlighted on the review page) << Previous Question Next Question >>

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1-<br> the<br> Nalt and Gil-<br> Cu2+<br> ad out- <br> what are the spectator ions in this reaction
Mrac [35]

Answer:

cu2 is 5he correct answer

3 0
2 years ago
What is the velosity of a 72.3 kg jogger with a kinetic energy of 1080.0 J?
Svet_ta [14]

Answer: 5.47m/s

Explanation:

Mass = 72.3kg

K.E = 1080.0J

V =?

K.E = 1 /2MV^2

V^2 = 2K.E /M = (2x1080)/72.3

V = sqrt [(2x1080)/72.3]

V = 5.47m/s

7 0
3 years ago
The modern periodic table is organized by increasing ______________ ___________.
professor190 [17]

Answer:

atomic number.

Explanation:

5 0
2 years ago
Read 2 more answers
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M
lukranit [14]
I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.
5 0
3 years ago
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