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Vesnalui [34]
3 years ago
13

The first chart shows the effect of the normal breakdown of material from reactants to products. The second chart shows the same

reaction but with an added modification.
Which statement best describes the difference between the charts?

Reaction A has an enzyme.
Reaction B has an enzyme.
Reaction A has an inhibitor.
Reaction B has an inhibitor.

Chemistry
2 answers:
Reika [66]3 years ago
8 0

Answer:

B

Explanation:

Reaction B has an enzyme

sashaice [31]3 years ago
5 0

Answer:

Reaction B has an enzyme.

Explanation:

In both charts, we can see the energy versus the progress of the reaction. In both charts, reactants and products have the same energy, but chart B has a lower activation energy (The difference in energy between the reactants and the activated complex). This is probably due to the action of an enzyme. Enzymes lower the activation energy in order to increase the reaction rate as we can see in the Arrhenius equation.

k=A.e^{-Ea/R.T}

where,

k is the rate constant

A is the collision factor

Ea is the activation energy

R is the ideal gas constant

T is the absolute temperature

The higher the rate constant, the higher the reaction rate.

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1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules
Ivanshal [37]

Answer:

1. 2.510kJ  

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

1cal=4.184J\\\\1kJ=1000J

Then, we perform the conversion as follows:

600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ

2. Here, we use the general heat equation:

Q=mC(T_2-T_1)

And we plug in the given mass, specific heat and initial and final temperature to obtain:

Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ

Regards!

7 0
2 years ago
you are running a transformation reaction, and in your microfuge tube currently is your plasmid and your insert fragments (both
ElenaW [278]

A foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction.

<h3>How to explain the reaction?</h3>

With the aid of two enzymes, ligase and restriction enzymes, a foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction. Each enzyme detects a target DNA sequence and cuts it nearby, while ligase aids in connecting the DNA. When two bits of DNA have complimentary bases, it facilitates their joining.

Plasmid and the insert fragment are both present in the microfuge tube, and they both have compatible sticky ends. However, the ligase has been denatured and is no longer active because the prior student left it outside rather than freezing it; despite this, we had already put the ligase into the tube. Ligase aids in binding the plasmid and insert fragments together, but because it is denatured in this instance, it will no longer be able to do so. As a result, no transformation process will take place. And since ligase links DNA fragments together by catalyzing the development of connections between the nearby nucleotides, the two fragments will not be able to unite.

Learn more about reactions on:

brainly.com/question/11231920

#SPJ1

4 0
1 year ago
One scientific investigation may result in
Paladinen [302]

Answer:

a new discover a new theory

4 0
2 years ago
Read 2 more answers
A cylinder, with a piston pressing down with a constant pressure, is filled with 2.10 moles of a gas (n1), and its volume is 46.
olga2289 [7]

Answer:

The volume is  V_2 =15.33 \  L

Explanation:

From the question we are told that

    The number of moles of the gas is  n_1 = 2.10 \  moles

      The volume of the gas is  V  =  46.0 \  L

        The  of moles of the gas that leaked is  n_2 =  0.700 \  mol

Generally from the ideal gas law

      PV  =  nRT        

Generally at constant pressure and temperature

        \frac{V}{n}  =  constant

So   \frac{V_1 }{n_1}  = \frac{V_2 }{n_2}

=>    V_2 = \frac{ V_1 *  n_2 }{ n_1}

=>    V_2 = \frac{  46.0 *  0.700  }{ 2.10 }  

=>    V_2 =15.33 \  L

             

     

5 0
2 years ago
What would happen to a weak base dissociation equilibrium if more products
Elina [12.6K]

Answer:

Both B and D are correct.

Explanation:

B + H₂O ⇌ BH⁺ + OH⁻

If you add more products, the position of equilibrium will shift to the left to decrease their concentrations (Le Châtelier's Principle). The concentration of reactants will increase, but the equilibrium concentrations of products will also be higher than they were initially.

A is wrong. The equilibrium constant is a constant. It does not change when you change concentrations.

C is wrong. Per Le Châtelier's Principle, the concentrations must change when you ad a stress to a system at equilibrium.

(This is a poorly-worded question. "They" are probably expecting answer D.)

6 0
3 years ago
Read 2 more answers
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