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Nookie1986 [14]
3 years ago
13

Imagine that you are observing an enzyme-catalyzed reaction in lab. every time you add more enzyme, the reaction rate increases

proportionally until the reaction rate suddenly levels off. no other chemicals were added, and no modifications were made in the experimental setting. why do you think the reaction rate stopped increasing?
Chemistry
1 answer:
tester [92]3 years ago
6 0

Enzymes catalyze the chemical reactions, they act upon the reaction substrates and speed up the reaction. Enzymes have active sites, the places where the reaction substrates interact with the enzyme bringing about the conversion of substrates to products. So, as the enzyme concentration increases the rate of reaction increases till a point where the rate is leveled off. The rate does not further increase, as the substrate might have become limiting at that point. All the available amount of substrate would have been associated with the active sites of the enzymes. So, at that point although there is enough catalyst, lack of substrate would limit the rate of reaction.

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The decomposition of hydrogen peroxide (H2O2(aq)) is proposed to follow a reaction mechanism of: Step 1: H2O2(aq) + Br–(aq)→H2O(
lys-0071 [83]

Answer:

Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without

Explanation:

Let us consider the equation below:

Step 1:

H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)

Step 2:

BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)

From the above equation, we can see that Br– is unchanged.

This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.

8 0
2 years ago
Can someone please help me
Mrrafil [7]
I think the answer would be A because O is oxygen and it has 7. Although it’s in parentheses and has a 2 on the outside of those parentheses, so you would multiply and 7 x 2 = 14. 14 is larger than the other ones.
Hopefully I’m right and hopefully that helps.
8 0
3 years ago
A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution wa
finlep [7]

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

<em>Change in heat:</em>

q = m*S*ΔT

<em>Where q is heat in J,</em>

<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

<em>Molar heat of solution:</em>

3830.87J/0.1048 moles KNO3 =

36554J/mol =

<h3>36.55kJ/mol</h3>

<em />

6 0
2 years ago
Which resonance structure would most likely create the infrared (IR) spectrum for the equilibrium geometry calculation of SCN −
rjkz [21]

Answer:

Resonance Structures for SCN-:[S-C N]-

Resonance StructureEnergy (kJ/mol)[S-C N]--23.00[S=C=N]

3 0
3 years ago
LiOH+HBr---&gt; LiBr +h20 If you start with 10.0 grams of lithium hydroxide, how many grams of lithium bromide will be produced?
kicyunya [14]
<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
7 0
3 years ago
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