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polet [3.4K]
3 years ago
7

The retina of a human eye can detect light when radiant energy incident on it is at least 4.00 × 10−17 J. For light of 535−nm wa

velength, how many photons does this energy correspond to?
Chemistry
1 answer:
VLD [36.1K]3 years ago
7 0

Answer:

1080 photons

Explanation:

ΔE = hc/λ => Energy per photon

h = 6.63 x 10⁻³⁴ j·s

c = 3 x 10⁸ m/s

λ= 535 nm = 5.35 x 10⁻⁷ m

ΔE/photon = (6.63 x 10⁻³⁴ j·s)(c = 3 x 10⁸ m/s)/(5.35 x 10⁻⁷ m) = 3.71 x 10⁻¹⁹ j/photon

For #photons in 4.00 x 10⁻⁷ j =  (4.00 x 10⁻⁷ j) / (3.71 x 10⁻¹⁹ j/photon) = 1076 photons ≅ 1080 photons (3 sig. figs.)

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3 years ago
The standard enthalpy of formation of Ca2+ (g) (∆H = 1934 kJ/mol)
Alex73 [517]

Answer:

see explanation

Explanation:

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3 years ago
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They achieve stable structures by sharing their single, unpaired electron.
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3 years ago
Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2
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Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

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In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

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And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

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And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

<em />

7 0
3 years ago
Read 2 more answers
Necesito saber esto porfavor, ¿ cuales son las ventajas que representan los espectroscopia en cuanto al estudio de los elementos
igomit [66]

Answer:

Por favor,no ponga exactamente lo que pongo,Trate de parafrasear.

La espectroscopia es el estudio de la interacción entre la materia y la radiación electromagnética en función de la longitud de onda o frecuencia de la radiación. La espectroscopia puede ser muy útil para ayudar a los científicos a comprender cómo un objeto como un agujero negro, una estrella de neutrones o una galaxia activa produce luz, qué tan rápido se mueve y de qué elementos está compuesto. Se pueden producir espectros para cualquier energía de la luz, desde ondas de radio de baja energía hasta rayos gamma de muy alta energía.

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5 0
3 years ago
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