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vovikov84 [41]
3 years ago
10

An unknown gas at standard temperature and pressure

Chemistry
1 answer:
Dvinal [7]3 years ago
7 0

Answer

V2 = 24.872

Explanation:

If 3.7 moles of propane are at a temperature of 28°C and are under 154.2 kPa of pressure, what volume does the sample occupy? ... 145 g of carbon dioxide have a volume of 34.13 dmº What is the temperature ... a pressure of 3.11 atm, a 276 g sample of

an unknown noble gas occupies 13.46L

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The correct answer is:Some substances keep the same molecular structure when they break down, and others do not.

Some substances change their form when they dissolve and some do not. Dissolution can sometimes be regarded as a sort of reaction between a chemical substance and water.

Usually, we can consider dissolution of a substance in water as a sort of chemical reaction for some substances. For instance, an ionic substance interacts with water to form ions. similarly, some salts become hydrolysed in water and give acidic/basic solutions as  result of that.

However, some substances do not interact with water upon dissolution. They rather remain as molecular entities because they are not composed of ions.

We can see that some substances keep the same molecular structure when they break down, and others do not keep the same molecular structure  when they dissolve hence it is difficult to classify dissolving as a physical or a chemical change.

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Dehydrohalogenation of tert-pentyl bromide at higher temperatures will produce 2-methyl-1-butene as the chief product when
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Find the final equilibrium temperature when 10.0 g of milk at 10. c is added to 1.60 x 10^2g of coffee with a temperature of 90.
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The concept that can be used in order to answer this item is that of the conservation of heat among the system. We let T be equal to the final temperature. The equation that would allow us to relate the initial and final conditions of both substances is as follows,

     m₁cp₁(T - T₁)  = m₂cp₂(T₂ - T)

The first entity, 1, is the milk and the second entity, 2, is the coffee. We are given that the specific heats of both substances are just equal so we can eliminate them from the equation. Substituting the known values,

   (10 g)(T - 10°) = (1.60 x 10^2 g)(90° - T)

The value of T from the equation is 85.29°C.

Answer: 85.29°C
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3 years ago
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