1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
babunello [35]
4 years ago
6

A catalyst can increase the rate of a reaction ________. Select one: a. by increasing the overall activation energy (Ea) of the

reaction b. by providing an alternative pathway with a lower activation energy c. by lowering the activation energy of the reverse reaction d. by changing the value of the frequency factor (A) e. All of these are ways that a catalyst might act to increase the rate of reaction.
Chemistry
1 answer:
makvit [3.9K]4 years ago
5 0

Answer:

by providing an alternative pathway with a lower activation energy

Explanation:

Cataysts are the substances that neither consumed nor produced during a chemical reaction, but increase the rate of the reaction.

So, catalysts remain chemically unchnaged after the completion of the reaction.

For a reaction to occur, reactants molecule must collide and successful collisions lead to product formation.

A certain minimum amount of energy is required for successful collision. This minimum amounr of energy is called activation energy.

One way to increase speed of reaction is to provide energy to the reactants  and other way to increase speed is to provide alternate pathway with lower activation energy.

Catalyst increases a rate of reaction by providing an alternate pathway with lower activation energy.

Therefore, the correct option is option b 'by providing an alternative pathway with a lower activation energy'

You might be interested in
An unknown liquid has a mass of 4.25 × 108 mg and a volume of 0.250 m3. what is the density of the liquid in units of g/ml?
mrs_skeptik [129]
Density= mass/volume
             step  one :
  convert  m3  to  ml
1m^3  =1000000ml
0.250m^3 x1000000=250000ml
         
           step  two:  convert  mg   to  g
 1mg=0.001g,  therefore 4.25 x108mg=0.459g
density  is  therefore= 0.459g/250000=1.836 x10^-6g/ml
8 0
4 years ago
Read 2 more answers
This answer please :)
Artemon [7]

Answer:

ithink it is c but i dont my ghess

Explanation:

i want to do thos for the points tbh

5 0
3 years ago
For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)
mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is 1.70\times 10^{15}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = 151.2 kJ = 151200 J

\Delta S^o = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(151200J)-(328.0K\times 169.4J/K)

\Delta G^o=95636.8J=95.6kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = 95636.8 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k

k=1.70\times 10^{15}

Therefore, the value of equilibrium constant for this reaction at 328.0 K is 1.70\times 10^{15}

3 0
3 years ago
An inhibitor is added to an enzyme-catalyzed reaction at a concentration of 26.7 μM. The Vmax remains constant at 50.0 μM/s, but
fomenos

Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

<h3>What is the Ki for the inhibitor?</h3>

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

  • Kma = Km/(1 + [I]/Ki)

Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

where:

  • Kma is the apparent Km due to inhibitor
  • Km is the Km of the enzyme-catalyzed reaction
  • [I] is the concentration of the inhibitor

Solving for Ki:

where

[I] = 26.7 μM

Km = 1.0

Kma = (150% × 1 ) + 1 = 2.5

Ki = 26.7 μM/{(2.5/1) - 1)

Ki = 53.4 μM

Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

Learn more about enzyme inhibition at: brainly.com/question/13618533

5 0
3 years ago
How many molecules are there in 5H20?<br> A.5<br> B.7<br> C.10<br> D.11
Ad libitum [116K]

Answer:

the answers B which is 7

5 0
3 years ago
Other questions:
  • A partially-enclosed coastal region where sea water mixes with freshwater is called a(n) _______.
    14·1 answer
  • In an atom:
    7·1 answer
  • Athe gas that siports burning is
    5·1 answer
  • How do I solve this?? ok having a hard time BRAINLIST​
    10·1 answer
  • How can you make hard water into soft water ?
    14·2 answers
  • Please hlep me on this 2 questions
    9·2 answers
  • Ok help it’s for homework
    15·2 answers
  • Help please<br> Protein produced from mutated strand?<br><br> And effect of mutation??
    9·1 answer
  • Hypothesis and Observations
    13·1 answer
  • Photosynthesis makes (__________) and (____________) for an ecosystem.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!