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tekilochka [14]
3 years ago
5

What is the wavelength of a 264-Hz sound wave when the speed of sound is 345 m2 1.31 m A. 0.77 m C. 6.09 m D. 0.11 m​

Chemistry
1 answer:
antoniya [11.8K]3 years ago
8 0

The wavelength : 1.31 m

<h3>Further explanation  </h3>

Waves are vibrations that travel  

There are several parts of the wave namely:  

<em>frequency, period, wavelength and wave velocity  </em>

Frequency (f): the number of waves that occur in one second  

Period (T): The time required to travel one wave  

Wavelength (λ): the distance from one crest to another, or from one trough to another (on transverse waves) and the distance between two consecutive compressions or between two consecutive rarefactions (on longitudinal waves)  

Fast wave propagation (v): the distance traveled by waves per unit of time  

The formula:

 \tt v=\lambda\times f

The speed of a sound wave 345 m, and the frequency = 264 Hz , so

v = 345 m

f = 264 Hz,  

then the wavelength :

\tt \lambda=\dfrac{v}{f}=\dfrac{345}{264}=\boxed{\bold{1.31~m}}

 

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Cu is in FCC structure. We know that even at close to melting temperature, vacancies are still sparsely distributed, i.e., it is
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Explanation:

FCC is face centered cubic lattice. In FCC structure, there are eight atoms at the eight corner of the cubic unit cell and one atom centered in each of the faces. FCC unit cells consist of four atoms, (8/8) at the corners and (6/2) in the faces.

Given that, Cu has FCC structure and it contains a vacancy at origin (0, 0, 0). And there is no other vacancy directly adjacent to the vacancy at the origin. So, all the adjacent positions contain Cu atoms. Hence, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy.

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5 0
2 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

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R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

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3 years ago
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Vsevolod [243]

Answer:

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The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.

In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.

Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.

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