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2 years ago

14

Fragilidad de metales

1 answer:

Lemur [1.5K]2 years ago

7 0

Answer: La fragilidad es la propiedad de algunos metales de no poder experimentar deformaciones plásticas, de forma que al superar su límite elástico se rompen bruscamente. La acritud es la propiedad de un metal para aumentar su dureza y su resistencia por el efecto de las deformaciones.

Explanation: y estas en el Brianly de ingles no el de español.

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Would it be C..? <br> More NO2 and SO2 would form..help..please and thank you

Scorpion4ik [409]

Answer:

C. More NO2 and SO2 will form

Explanation:

Le Chatelier's Principle : It predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc

It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.

In given Reaction SO3 is introduced(increased) .

So equilibrium will shift in the direction where SO3 should be consumed(decreased)

Hence the equilibrium will go in backward direction , i.e

$NO + SO_{3} \rightarrow NO_{2} + SO_{2}$

So more and more Of NO2 and SO2 will form

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3 years ago

Model B: In this model the atom is a solid sphere that cannot be divided up into smaller

vivado [14]

Answer:

They are called corpuscles

Explanation:

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3 years ago

Hey guys, about to submit my draft for business and chemistry.

drek231 [11]

Answer:

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and give it to ur teacher

8 0

3 years ago

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. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

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3 years ago

What is the enthalpy of reaction?

erma4kov [3.2K]

Answer:

Enthalpy is the amount of heat released or absorb. The total enthalpy can be found by subtracting the products' enthalpy by the reactants' enthalpy

4 0

2 years ago

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