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3 years ago

If the frequency of the photon increases what happens to the wavelength and energy?

1 answer:

8 0

The energy per photon is proportional to the frequency of the radiation when considered as waves, ie inversely proportional to the wavelength. Double the wavelength, halve the photon energy. This means that long wavelength radiation (radio waves) has low photon energy and so does not penetrate matter.

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15. D is correct, exothermic reactions release heat

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3 years ago

Which of these is a chemical property? boiling point odor ability to rust color

NISA [10]

Answer:

ability to rust

Explanation:

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3 years ago

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vagabundo [1.1K]

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3 years ago

What best describes the conditions to becoming a fossil

Ede4ka [16]

Hello

I don't quite understand the question your asking and if there is any answers provided but I will still give you mine.

Fossils are formed in a number of different ways, but most are formed when a plant or animal dies in a watery environment and is buried in mud and silt. Soft tissues quickly decompose leaving the hard bones or shells behind. Over time sediment builds over the top and hardens into rock.

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3 0

3 years ago

Calculate Ho298 for the process

Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$ $\Delta H^0_1 = -314 kJ$ ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$ $\Delta H^0_2 = -80kJ$ ..............(2)

The final reaction is as follows:

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$ $\Delta H^0_3 = ?$ .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

$\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$

= -314 kJ + (-80) kJ

= -394 kJ

Thus, we can conclude that $H^{o}$ at 298 K for the given process is -394 kJ.

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3 years ago