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GREYUIT [131]
3 years ago
13

Suppose that a substance in a beaker is heated over a burner in a science lab. Which observation would most likely indicate that

a chemical change has occurred in the substance?
If the substance is a liquid or solid, an increase in temperature would indicate a chemical change.
If the substance is a liquid, a change of some of the liquid to gaseous form would indicate a chemical change.
If the substance is a solid, a change of some of the solid to liquid form would indicate a chemical change.
If the substance is a liquid or solid, production of an odor would indicate a chemical change.
Chemistry
2 answers:
saul85 [17]3 years ago
7 0

^^

Otherwise known as D.

(this is for people who are just looking for letters)

loris [4]3 years ago
5 0

Answer:

If the substance is a liquid or solid, production of an odor would indicate a chemical change.

Explanation:

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Warmer atoms expand. There is also more movement. This is also explained when putting it into terms with solid vs. gas. in a solid, the atoms vibrate, but they can't move very far because there is not a lot of space between them. In a gas, on the other hand, the atoms are spread apart, and they move very fast. Putting it into context with water, the steam (gas) version is hot, and the ice cube (solid) version is cold.
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3 years ago
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2 years ago
A helium-filled weather balloon has a volume of 793 L at 16.9°C and 759 mmHg. It is released and rises to an altitude of 4.05 km
mariarad [96]
<h3>Answer:</h3>

1082.96 L

<h3>Explanation:</h3>

We are given;

  • Initial volume of helium gas, V1 = 793 L
  • Initial temperature, T1 = 16.9°C

            But, K = °C + 273.15

  • Thus, initial temperature, T1 is 290.05 K
  • Initial pressure, P1 = 759 mmHg
  • New pressure at 4.05 km, P2 = 537 mmHg
  • New temperature at 4.05 km, T2 = 7.1 °C

                                                             = 280.25 K

Assuming we are required to calculate the new volume at the height of 4.05 km

We are going to use the combined gas law.

  • According to the combined gas law;

\frac{P1V1}{T1}=\frac{P2V2}{T2}

  • Rearranging the formula;

V2=\frac{P1V1T2}{P2T1}

V2=\frac{(759mmHg)(793L)(280.25K)}{(537mmHg)(290.05K)}

V2=1082.96L

Therefore, the new volume of the balloon at the height of 4.05 km is 1082.96 L

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Answer:

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