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4 years ago

Compare a photon and quantum. How are they alike? How are they different?

1 answer:

3 0

Quantum is deemed to be a distinct container with energy kept in, whereas photon is considered as an elementary particle. Photon and quantum are two very significant principles talked about in modern physics. These principles are broadly used in quantum chemistry, quantum physics, electromagnetic theory, particle physics, optics, etc. These principles are also applied in a lot of actual world applications such as high resolution microscopy, LASERs, measurements of molecular distances, photochemistry, and quantum cryptography. It is highly important to have knowledge and a good understanding in photon and quantum so that we could excel in the aforesaid fields. A photon is an elementary particle, however, a quantum is not an elementary particle. A quantum does not have the properties of both wave like and particle like, but a photon does. A photon is not about a measure of quantity and can be defined as a quantum of energy, on the contrary, a quantum can be described as a measure of quantity.

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The process of breaking food is to release energy is called _____

just olya [345]

Answer:

The process by which food is broken down to release energy is called respiration

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3 years ago

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When Rutherford performed his experiment, only 1 in 20,000 alpha particles bounced straight back or were

Anna007 [38]

Answer:

when rutherford performed his experiment, only 1 in 20,000 alpha particles bounced straight back or were deflected greatly. the rest went straight through the gold foil. e) based on this evidence, what is in atom's center? positively charged particles.

Explanation:

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3 years ago

How a convection current can enable a hawk or eagle to soar upward without flapping its wing

Maurinko [17]

The convection current is one of hot air rising, in this example, and then cooling and returning down. The rising air pushes the eagle up higher into the air. It just has to spread its wings and let the air push it into the air

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3 years ago

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You react 2.43 grams of magnesium with oxygen from the air according to the following reaction. The final mass of the product, m

Leona [35]

Answer:

The mass of oxygen that reacted is approximately 1.6 grams of oxygen

Explanation:

The given information are;

The mass of magnesium in the reaction = 2.43 grams

The final mass of the magnesium oxide = 4.12 grams

The chemical equation for the reaction is given as follows;

2Mg + O₂ → 2MgO

From which we see that two moles of magnesium, Mg, reacts with one mole of oxygen gas molecule, O₂, to produce two moles of magnesium oxide, MgO

The number of moles of magnesium present, $n_{Mg}$ , is given as follows;

$n_{Mg} = \dfrac{Mass \ of \, magnesium}{Molar \ mass \ of \, magnesium} = \dfrac{2.43}{24.305} \approx 0.1 \ moles$

By the given chemical equation, 2 moles of magnesium reacts with one mole of O₂, therefore;

1 mole of magnesium will react with 1/2 moles of oxygen which is 0.5 moles of oxygen also;

0.1 mole of magnesium will react with 0.05 moles of oxygen

The mass of one mole of oxygen = The molar mass of oxygen = 32 g/mol

The mass of oxygen in the reaction = The number of moles of oxygen × The molar mass of oxygen

The mass of oxygen in the reaction = 0.05 × 32 = 1.6 grams

Also from the molar mass of MgO which is 40.3044 g/mol, we have;

The mass fraction of oxygen = 16/40.3044 = 0.39698 ≈ 0.4

The mass of oxygen = 0.39698 × 4.12 = 1.636 g

Therefore, the mass of oxygen in the reaction is approximately 1.6 grams.

8 0

3 years ago

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K

Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

$\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S$