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nikitadnepr [17]
3 years ago
11

Why can it be difficult to demonstrate the Law of Conservation of Mass in action when a chemical reaction takes place in an open

system?
Chemistry
1 answer:
rjkz [21]3 years ago
3 0

Answer:

In simple words, in an open system, the system that has external interactions. Such interactions can take the form of information, energy, mass etc. So keeping this definition in mind, the law of conservation of mass in open system is difficult because when chemical reaction takes places things change form and some things are released as energy, That energy than escapes the system and interacts with other surrounding. This is why it is difficult to demonstrate Law of Conservation of Mass because the changes in mass due to chemical reaction cannot be accounted for properly since the lost mass (in form of energy) was lost to environment.

Whereas, in closed system, all components occur inside a closed container or something where external interactions are minimal to none so the energy lost due to chemical reaction does not escape the system thus it can be accounted for properly to demonstrate the law of conservation of Mass.

Explanation:

Too many words but should get the point across hopefully!

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EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?
AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq).

The equilibrium constant for this reaction is called the self-ionization constant of water:

K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}].

Note that water isn't part of this constant.

The value of K_w at 25 °C is 10^{-14}. How to memorize this value?

  • The pH of pure water at 25 °C is 7.
  • [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}
  • However, [\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3} for pure water.
  • As a result, K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14} at 25 °C.

Back to this question. [\text{H}^{+}] is given. 25 °C implies that K_w = 10^{-14}. As a result,

\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}.

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The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.

Mole of 960 grams SO2 = 960/64 = 15 moles

Equivalent mole of sodium sulfite that reacted = 15 moles

Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams

More on stoichiometric problems can be found here: brainly.com/question/14465605

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3 0
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The symbol for the hydroxide ion is _____. <br> H +<br> H 3O - <br> H(OH) <br> OH -
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Explanation:

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