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loris [4]
4 years ago
11

A scientist is trying to identify an unknown substance. The answer to which of the following questions would give the scientist

information about a chemical property of the substance?. . Is it possible to change the shape of the substance? . . Does the substance corrode when left in water? . . What color is the substance? . . At what temperature does the substance melt? .
Chemistry
2 answers:
borishaifa [10]4 years ago
7 0
The one question that would give the scientist information about a chemical property of the substance is "Does the substance corrode when left in water?" The correct answer is B.
Drupady [299]4 years ago
5 0

Explanation:

Chemical property is defined as the property of a substance which is observed during a reaction where the chemical composition identity of the substance gets changed.  For example: color, reactivity etc.

Physical property is defined as the property which can be measured and whose value describes the state of physical system. For Example: State, density etc.

The question of which will give us information regarding chemical property of the substance will be:

Does the substance corrode when left in water?

This is because corrosion is the type of chemical reaction in which damaging of the surface of the substance takes place due to its surroundings.Here identity of the substance changes.Hence answering this question will give us an information regarding chemical property of the substance.

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SO2 + H2O = H2SO3 and not H2SO4?
kramer
It's because the Law of Conservation of Mass WILL NOT be followed if, from SO2 and water, sulfuric acid is formed. Making a balanced chemical reaction is not possible, no matter what you do. To obtain or make sulfuric acid, SO2 still needs to be oxidized further to SO3.
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3 years ago
An Equilibrium with Cobalt Complex Ions.Investigate the equilibrium between two different complex ions of cobalt. The reaction i
Paladinen [302]
  • The change in color from blue to pink of the cobalt complexes here has been the basis of cobalt chloride indicator papers for the detection of the presence of water. It is also used in self-indicating silica gel desiccant granules.
  • Pink cobalt species + chloride ions ⇌ Blue cobalt species + water molecules

<u>Explanation</u>:

  • The adjustment in color from blue to the pink of the cobalt complexes here has been the premise of cobalt chloride indicator papers for the detection of the presence of water. It is likewise utilized in self-demonstrating silica gel desiccant granules.  

Pink cobalt species + chloride particles ⇌ Blue cobalt species + water molecules  

  • The response of [Co(H2O)6]2+(aq) + 4Cl–(aq) → [CoCl4]2–(aq) + 6H2O(l) is endothermic. In this manner, as per Le Chatelier's rule, when the temperature is raised, the situation of the balance will move to one side, shaping a greater amount of the blue complex particle at the expense of the pink species.  
  • Including concentrated hydrochloric raises the chloride particle fixation, making the equilibrium move to one side, as per Le Chatelier. Including water brings down the chloride particle fixation, moving the equilibrium the other way.
  • As an extension, it is conceivable to show that it is the Cl–particles in the hydrochloric acid that move the balance by including a spatula of sodium chloride rather than the pink arrangement. This delivers a bluer color, however, this may take some time because the salt is delayed to dissolve.
5 0
3 years ago
Ammonia is produced by the Haber process. The equation is shown.
choli [55]

Answer:

option D

Explanation:

Increasing the temperature increases the yield of ammonia and speeds up the reaction as chemical reaction is affected by temperature.

7 0
3 years ago
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Answer:

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Explanation:

8 0
3 years ago
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
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