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meriva
3 years ago
15

Nuclear changes differ from normal chemical changes in that all nuclear changes

Chemistry
1 answer:
kogti [31]3 years ago
8 0

Answer:

(1) Nuclear reactions entail a transition in the nucleus of an atom, which normally results in the formation of a new substance. Chemical reactions, on the other hand, only involve electron rearrangement and do not involve nuclei modifications. (4) Nuclear reactions are unaffected by the element's chemical form.

Explanation:

Hope this helps!

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what elements would you expect to lose electrons in chemical changes? Potassium, sulfur, fluorine, barium or copper.
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Potassium, Barium, and Copper because they are metals, and most metals tend to lose electrons when forming an ion.
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Equation below is used to determine the heat flux for convection
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4 years ago
5. A gas has a pressure of 310 kPa at 237 degrees C.<br>What will its pressure be at 23 degrees C?)​
slavikrds [6]

Answer:

The pressure of the gas at 23 C is 179.92 kPa.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

In short, when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

Studying two states, one initial 1 and the other final 2, it is satisfied:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 310 kPa
  • T1= 237 C= 510 K (being 0 C= 273 K)
  • P2= ?
  • T2= 23 C= 296 K

Replacing:

\frac{310 kPa}{510 K} =\frac{P2}{296 K}

Solving:

P2=296 K*\frac{310 kPa}{510 K}

P2= 179.92 kPa

<u><em>The pressure of the gas at 23 C is 179.92 kPa.</em></u>

5 0
3 years ago
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