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Vedmedyk [2.9K]
3 years ago
5

What are sports examples that involve transfer of momentum

Chemistry
2 answers:
Alona [7]3 years ago
7 0

Answer:  

In sports some , examples are  using a heavier bat or racket and increasing running speed or hand speed.

Explanation:

MOMENTUM: is the  mass and the velocity, you can increase momentum by increase either of these elements

Elena-2011 [213]3 years ago
6 0
Baseball could be one
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We will know that caramelizing sugar is a chemical change because it changes the sugar's color. A chemical change involves the forming of new substances from specific substances called reactants. Caramelizing sugar releases volatile chemicals that cause its caramel flavor.

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A rock face at the top of a mountain is composed of layers in which seashells are embedded. This type of rock is ​
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Sedimentary rock.

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When limestone form as ocean water evaporates and leaves calcium carbonate behind, they are then deposited in the seabed. Thus, the type of rock that can be formed this way is a sedimentary rock.

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Consider the reaction: P(s) + 3/2 Cl2(
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K = Ka/Kb

<h3>Further explanation</h3>

<u>Given:</u>

  • \boxed{ \ P_{(s)} + \frac{5}{2} Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_a \ }
  • \boxed{ \ PCl_3_{(g)} + Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_b \ }

<u>Question:</u>

Write the equilibrium constant for this reaction:

\boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = ? \ }

in terms of the equilibrium constants, Ka and Kb

<u>The Process:</u>

Let us solve the problem above.

\boxed{ \ P_{(s)} + \frac{5}{2} Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_a \ } ... (Equilibrium-1)

\boxed{ \ PCl_3_{(g)} + Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_b \ } ... (Equilibrium-2)

Consider the reaction: \boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = ? \ }

Equilibrium-2 is reversed to match the target reaction (Hint: PCl₃). Then the two equilibrium reactions are added together.

\boxed{ \ P_{(s)} + \frac{5}{2} Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_a \ } ... (Equilibrium-1)

\boxed{ \ PCl_5_{(g)} \rightleftharpoons PCl_3_{(g)} + Cl_2_{(g)} \ \ \ \ \ \frac{1}{K_b} \ } ... (Equilibrium-2, after reversed)

---------------------------------------------------------------- (+)

\boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = K_a \times \frac{1}{K_b} \ }

<u>Notes:</u>

  • \boxed{PCl_5} is eliminated.
  • \boxed{\frac{5}{2}Cl_2 \ subtracted \ by \ Cl_2 \ equal \ to \ \frac{3}{2}Cl_2}.
  • If the two equilibrium reactions are added together, the equilibrium constants are multiplied by each other.

Thus, the equilibrium constant for the target reaction in terms of the equilibrium constants, Ka and Kb, is

\boxed{\boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = \frac{K_a}{K_b} \ }}

<h3>Learn more</h3>
  1. Conservation of mass  brainly.com/question/9473007
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Keywords: consider the reaction, P, Cl2, PCl3, PCl5, Ka, Kb, K, write, the equilibrium constant, for this, in terms, reverse, multiply

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15 g

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