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The primitive atmosphere of Earth may have favored the abiotic synthesis of organic molecules because it **had close to no oxygen, hence, it was powerfully reducing. It had pure energy sources too in form of ultraviolet rays from the sun (it was extremely hot! Too hot for water in liquid form to exist in fact!), outgassing from the numerous volcanoes and lightning**.

Abiotic synthesis means a formation reaction where the reactants are charged to become products, almost all on their own, with no catalyst especially.

So, these factors would have helped the organic molecules to form as most of what is needed to overcome the energy barrier and become products.

7 0

3 years ago

A crystal that absorvd water from air is (blank)

raketka [301]

The answer is c hygroscopic

7 0

4 years ago

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

ololo11 [35]

Hence the order of the transition in order of increasing frequency of the photon absorbed or emitted will be : d < a < c < b

$E = -13.6$ × $Z$ ²/ $n$ ²

where,

$E$ = energy of orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

$E$ ₁ = -13.6× 1²/1² = -13.6eV

Energy of n = 2 in an hydrogen atom:

$E$ ₂ = -13.6 × 1²/2² = -3.40eV

Energy of n = 3 in an hydrogen atom:

$E$ ₃ = -13.6× 1²/3² = -1.51eV

Energy of n = 4 in an hydrogen atom:

$E$ ₄ = -13.6× 1²/4² = -0.85eV

Energy of n = 5 in an hydrogen atom:

$E$ ₃ = -13.6× 1²/ 5² = -0.54eV

a) n = 2 to n = 4 (absorption)

Δ $E$ ₁ = E₄ - E₂ = -0.85- (- 3.40) = 2.55eV

b) n = 2 to n = 1 (emission)

Δ E₂ = E₁ - E₂ = -13.6 - (- 3.40) = -10.2eV

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

ΔE₃ = E₅ - E₂ = -0.54 -( -3.40) = -2.85eV

d) n = 4 to n = 3 (emission)

ΔE₄ = E₃ - E₄ = -1.51 - (- 0.85) = - 0.66eV

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

$E = hv$

h = Planck's constant

$v =$ frequency of the wave

So, the increasing order of magnitude of the energy difference :

$E$ ₄< E₁ <E₃ <E₂

The H atom electron transitions in order of increasing frequency of the photon absorbed or emitted will be d < a < c < b

: d < a < c < b

To learn more about transitions visit the link:

brainly.com/question/28304182?referrer=searchResults

The question is incomplete , complete question is:

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

(a) n = 2 to n = 4

(b) n = 2 to n = 1

(c) n = 2 to n = 5

(d) n = 4 to n = 3

#SPJ4

6 0

2 years ago