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Rasek [7]
3 years ago
7

Which of the following changes can alter the form of a substance but does not necessarily change it to an other substance?

Chemistry
1 answer:
Goshia [24]3 years ago
8 0

Answer:

  • The physical changes can alter the form of a substance but does not necessarily change it to an other substance

Explanation:

  • The physical properties of matter is one that does not depend on composition.It just change the form of matter into another form with changing into composition.
  • Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others.
  • The three states of matter are: solid, liquid, and gas. The melting point and boiling point are related to changes of the state of matter. All matter may exist in any of three physical states of matter.

Example:

  1. When water is boiled then it change into vapors. Now the form of water is change but its composition is same like water
  2. When water freezes into ice, now phase are different i.e water is liquid while ice is solid but both have same composition.
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Consider the reaction to produce methanolCO(g) + 2H2 (g) &lt;-----&gt; CH3OHAn equilibrium mixture in a 2.00-L vessel is found t
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Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

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Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

7 0
3 years ago
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