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alisha [4.7K]
1 year ago
13

A+gas+at+250k+and+15atm+has+a+molar+volume+12%+less+than+the+perfect+gas+law+character. +calculate+the+compressibility+factor

Chemistry
1 answer:
shutvik [7]1 year ago
6 0

The compression factor based on the information given is 0.083.

<h3>How to calculate the compression factor?</h3>

The way to calculate the compression factor goes this:

V(ideal) = RT/P

= (0.0821 × 250)/15

= 1.368

V(real) will be:

= V(ideal)/12

= 1.368/12 = 0.114

The compression factor will be:

= 1.368/(12 × 1.368)

= 0.083

Therefore, the compression factor is 0.083.

Learn more about compression factor on:

brainly.com/question/24261456

#SPJ12

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"How does land change temperature compared to water?"
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It’s A. Land heats up and cools quickly then water
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Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come
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Answer:

B

Explanation:

For solving this we need a heat balance

Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}

By changing the corresponding relations, we have

m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\

By cancelling similar factor, we obtain

\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\

Which means that the change of temperature in A is twice the change of B

3 0
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At 700 K, the reaction 2SO2(g) + O2(g) &lt;====&gt; 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f
nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

  • If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
  • If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
  • If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

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3 years ago
HELPPPPP PLEASE!!!
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C I think it’s C I’m semi guessing
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