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almond37 [142]
3 years ago
15

Which of the following quantities contains the greatest number of particles?

Chemistry
1 answer:
Nadya [2.5K]3 years ago
3 0
I believe the correct answer would be the last option. All of the quantities given above contain the same number of particles. We determine this by using the avogadro's number. It represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. 

2 moles of carbon atoms ( 6.022 x 10^23 particles / mole  ) = 1.20 x10^24 particles
<span>
2 moles of carbon dioxide molecules </span>( 6.022 x 10^23 particles / mole  ) = 1.20 x10^24 particles<span>

2 moles of diatomic oxygen molecules </span>( 6.022 x 10^23 particles / mole  ) = 1.20 x10^24 particles

As you can see, no matter what is the gas as long as they have the same number of moles, they would also have same number of particles<span />
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3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

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True or False: Cohesion is the attraction between particles of the same<br> substance
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Explanation:

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3 years ago
Calculate the de broglie wavelength of a subatomic particle that is moving at 351 km/s if its mass is 9.11 à 10â31 kg. λ = hmuh
gregori [183]

The de Broglie wavelength of a subatomic particle is 2.09 nm.

 λ = h m v = h

momentum : wherein 'h' is the Plank's steady. This equation pertaining to the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated the use of this relation is de Broglie wavelength.

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Wavelength is the gap between the crests of waves or a person's fashionable mind-set. An instance of wavelength is the gap between the crest of two waves. An instance of wavelength is while you and some other character share the equal standard attitude and might for that reason speak properly.

calculation is given in the image below

de Broglie wavelength λ = h/mv

                                          = (6.626 * 10^-34)/9.1 * 10^-31 *351 *10^3

                                          = 2.07 *10^-9

                                 Hence, = 2.op nm    

Learn more about de Broglie wavelength here:-brainly.com/question/16595523

#SPJ4

6 0
2 years ago
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