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4 years ago

In which phase(s) do the molecules take the shape of the container?

Chemistry

2 answers:

Katen [24]4 years ago

6 0

Answer:

Liquid and gas

Explanation:

Liquid and gas are the phases of matter that take the shape of container.

This is very simple to imagine, if we have a piece of rock and we put it in a container, it will not take the shape of container as it already has a definite shape and volume. Liquid when put in a container takes the shape of container but varies in volume as per the container.

However, the gas phases is the phase of matter that perfectly takes the shape of container and occupies all the volume of container as well.

If we recall Dalton's Law of Partial Pressures we can see that the pressure exerted by the gas components in a container is same like the pressure exerted by the gas alone. These partial pressures of the component of gas combine in such a way that they exert total pressure equal to the constituents' pressure on the container. This way gases occupy all the volume of a container and take the shape of a container they're placed in.

Hope it help!

strojnjashka [21]4 years ago

5 0

In the liquid and gas phase molecules take the shape of the container.

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Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

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3 years ago

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makvit [3.9K]

Use the formula E=mc^2
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Answer:

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