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3 years ago

How does a sample of hydrogen at 10 °C compare to a sample of hydrogen at 350 K?

Chemistry

2 answers:

Aleks04 [339]3 years ago

8 0

Answer: -

The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Explanation: -

Temperature of the hydrogen gas first sample = 10 °C.

Temperature in kelvin scale of the first sample = 10 + 273 = 283 K

For the second sample, the temperature is 350 K.

Thus we see the second sample of the hydrogen gas more temperature than the first sample.

We know from the kinetic theory of gases that

The kinetic energy of gas molecules increases with the increase in temperature of the gas. The speed of the movement of gas molecules also increase with the increase in kinetic energy.

So higher the temperature of a gas, more is the kinetic energy and more is the movement speed of the gas molecules.

Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Pie3 years ago

3 0

The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K would be the answer to this question.

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Teniendo en cuenta los siguientes fenómenos: ebullición del agua- movimiento de un cuerpo- disolución de sal en agua- combustión

miv72 [106K]

Answer:

Las siguientes son reacciones químicas;

combustión de leña

oxidación del hierro

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Explanation:

Una reacción química da como resultado la formación de una (s) sustancia (s) nueva (s), mientras que un cambio físico no conduce a la formación de una sustancia nueva.

Las siguientes son reacciones químicas;

combustión de leña: la combustión de madera implica la oxidación del carbono según la reacción; C (s) + O2 (g) -------> CO2 (g)

oxidación del hierro: La oxidación del hierro conduce a la formación de óxidos de hierro. Como; 2Fe (s) + O2 (g) ----> 2FeO (s)

descomposición del agua en hidrógeno y oxígeno: esta es una reacción química en la que el agua se descompone de la siguiente manera; 2H2O (l) -----> 2H2 (g) + O2 (g)

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3 0

3 years ago

The bacteria can chemically combine nitrogen with hydrogen to form ammonia (NH3). This combining process is called nitrogen fixa

nikklg [1K]

Answer:

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Explanation:

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3 years ago

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

3 years ago

Which of the following observations is usually not evidence of a chemical change ?

Nookie1986 [14]

Answer:

Explanation:

Have a nice day

5 0

4 years ago

Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20

KengaRu [80]

Answer:

$m_{Ag}=2,265.9g$

Explanation:

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In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

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