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devlian [24]
3 years ago
7

What is the smallest part of an element?

Chemistry
2 answers:
vova2212 [387]3 years ago
6 0

Answer:electrons

Explanation: I looked it up

Ainat [17]3 years ago
4 0
An atom is the smallest part of an element
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What is the empiracle formula of amylose?
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Answer:

(C6H10O5)n

Explanation:

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Consider the conversion of succinate to fumarate in the Citric Acid Cycle (reaction below). This reaction is endergonic under st
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Succinate oxidation to fumarate The following reactions transform succinate to regenerate oxalacetate. The first of these reactions is carried out by an oxidation catalyzed by succinate dehydrogenase. The hydrogen acceptor is FAD, since the free energy change is insufficient to allow NAD to interact. The final product is fumarate.

Explanation:

The condensation reaction of GDP + Pi and the hydrolysis of Succinyl-CoA involve the H2O necessary to balance the equation.

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How many major reservoirs does carbon move between?<br> A. 3<br> B. 4<br> C. 5<br> D. 2
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Which two conditions can limit the usefulness of the kinetic-molecular theory in describing gas behavior?
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Read 2 more answers
Calculate the energy for the transition of an electron from the n = 5 level to the n = 6 level of a hydrogen atom. E = Joules Is
kramer

Answer:

For an electron to move from a lower energy level to a higher energy , that electron needs to absorb energy sufficient enough to excite it to make the transition. Hence it is an absorption process. The required energy of transition  E = 2.665 x 10⁻²⁰J

Explanation:

Using the Rydberg's equation we can calculate the wavelength of the photon of energy transition as follows:

1/λ = R . (1/nf² - 1/ni²)

where

λ is the required wavelength of the photon needed to be absorbed to excite the electron to transit from level 5 to 6.  

(Note that for the electron to transit to from energy level 5 to 6, the photon would have to fall from level 6 to 5 in order to emit the required energy to excite the electron)

R is the Rydberg's constant 1.097 x 10⁷ m⁻¹

nf is the final level of the photon

ni is the initial level of the photon

1/λ = 1.097 x 10⁷ m⁻¹ (1/5² - 1/6²)

1/λ = 1.3407 x 10⁵ m⁻¹

λ = 7.458 x 10⁻⁶ m

This implies that that is the wavelength of the photon required to excite the electron to transit from energy level 5 to 6. Using the equation below, we can calculate the energy of transition as

E = h.c/λ

where

E is the required energy of transition

h is the Planck's constant (6.626 x 10⁻³⁴ Js)

c is the speed of light (3 x 10⁸ms⁻¹)

λ is the wavelength calculated above

E = 6.626 x 10⁻³⁴ Js  x  3 x 10⁸ms⁻¹/ 7.458 x 10⁻⁶ m

E = 2.665 x 10⁻²⁰J

3 0
3 years ago
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