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kramer
3 years ago
5

If particles of a solid move slow and particles of

Chemistry
1 answer:
Nana76 [90]3 years ago
3 0

Answer:

<u>The</u><u> </u><u>gas</u><u> </u><u>particles</u><u> </u><u>would</u><u> </u><u>have</u><u> </u><u>higher</u><u> </u><u>Kinetic</u><u> </u><u>energy</u><u>.</u>

Explanation:

From the formular of kinetic energy:

{ \sf{KE =  \frac{1}{2}m {v}^{2}  }}

KE is the kinetic energy possessed by particle.

KE is the kinetic energy possessed by particle.m is the particle mass.

KE is the kinetic energy possessed by particle.m is the particle mass.v is the velocity attained by a particle

{ \sf{KE =  \{ \frac{1}{2} m \}} {v}^{2} }

keeping the mass, m constant:

{ \sf{KE = k {v}^{2} }}

Therefore, KE is directly proportional to v² :

{ \sf{KE \:  \alpha  \:  {v}^{2} }}

so, when velocity increases, KE also increases.

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2 years ago
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What is the average atomic mass listed for nitrogen in the periodic table?
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3 years ago
A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:
bekas [8.4K]

Answer: The standard enthalpy of formation  of H_2O(g) is  -252.1 kJ/mol.

Explanation:

The balanced chemical reaction is,

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]

Putting the values we get :

\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]

67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]

H_f{H_2O}=-252.1kJ/mol

Thus standard enthalpy of formation  of H_2O(g) is  -252.1 kJ/mol.

3 0
3 years ago
If someone trys to lur u into a car with candy...say NO. and run away
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Answer:

as u should. candy ain't even that tempting >^<

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Most of the earth's volcanoes:
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