Chemistry, 03.07.2019 15:00 annie1799

What factors could be used to increase reaction rate? check all that apply. using an inhibitor using a catalyst increasing temperature increasing pressure decreasing concentration

Answers: 1

Show answers

Answers

Answer from: alexsysantiago54

The rate of a reaction can be increased by:

-using a catalyst- it lowers the activation energy and leads to less energy required to break bonds of reactants. This lower activation energy leads to more products formed with less time, hence an increase reaction rate.

-increasing temperature- the particles of a molecule move faster and undergo more collision which leads to an increase in the speed of the reaction.

-increasing pressure- means there is more particles of reactants in a reduced volume. The particles do not need to move long distances to find another particle to react with, hence the rate of reaction increases.

Using an inhibitor will not increase the rate of the reaction because, an inhibitor binds to the active site where a catalyst is supposed to act. This means a higher activation energy and thus a decrease in reaction rate.

Likewise, decreasing the concentration implies few particles available to collide with each other and that slow down the speed of the reaction.

Answer from: Marisajordon16

The correct answers are all of the middle ones or-

using a catalyst increasing temperature increasing pressure

Explanation:

I just finished doing the assignment on edge 2020

Answer from: Quest

i have know clue i guess aluminum

i think lol

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

Another question on Chemistry

Chemistry, 22.06.2019 09:20

What will most likely happen when two bromine atoms bond together?

Answers: 3

Answer

Chemistry, 22.06.2019 09:50

Although respiratory organs vary across different organisms, they all contain respiratory surfaces that have a large surface area and are extremely thin. explain why having an extremely thin respiratory surface with a large surface area is advantageous for the process of gas exchange

Answers: 1

Answer

Chemistry, 22.06.2019 14:40

Which statement best describes the function of enzymes?

Answers: 1

Answer

Chemistry, 22.06.2019 18:10

Given the following at 25c calculate delta hf for hcn (g) at 25c. 2nh3 (g) +3o2 (g) + 2ch4 (g) > 2hcn (g) + 6h2o (g) delta h rxn= -870.8 kj. delta hf=-80.3 kj/mol for nh3 (g), -74.6 kj/mol for ch4, and -241.8 kj/mol for h2o (g)

Answers: 1

Answer

You know the right answer?

What factors could be used to increase reaction rate? check all that apply. using an inhibitor usin...

Questions