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Deffense [45]
3 years ago
8

Consider the following reversible reaction

Chemistry
1 answer:
kogti [31]3 years ago
7 0

Answer:

The right choice is  K(eq) = ([CO₂]) / [O₂]

Explanation:

The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium

For a system undergoing a reversible reaction described by the general chemical equation

  • a A + b B → x X + n N

An equilibrium constant can be expressed as following

K(eq) = ([X]ˣ*[N]ⁿ) / ([A]ᵃ*[B]ᵇ)

where [] is the equilibrium concentration of each component.

and the concentration of solid is conventionally considered as 1

<u>in your example: </u>

The reversible reaction

  • C ₍s₎ + O₂ ₍g₎  ↔ CO₂ ₍g₎

taking in consideration that C is a solid and the concentration of solid is conventionally considered as 1

So,  the equilibrium constant for this reversible reaction can be expressed as following

  • K(eq) = ([CO₂]) / [O₂]

∴ The right choice is  K(eq) = ([CO₂]) / [O₂]

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explain why samples of gold and copper can have the same extensive properties, but not the same intensive properties
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Extensive properties, as volume and mass, depend on the amount of material. So, you can have a sample of gold and a sample of copper with the same volume as long as you have different amount of each one.

On the other hand, intensive properties do not depend on the amound of material but on the chemical constitution of the material. Density is an intensive property, so gold and copper have different densities. That is why you can use intensive properties to characterize different materials.
3 0
3 years ago
if the pressure of a gas at constant volume is 3.5 atm at 100°c, what will the pressure be if the tempature is changed to 250°c?
Degger [83]
Sorry don't know this one

6 0
3 years ago
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
3 years ago
Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Leokris [45]
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ. 
6 0
3 years ago
I need help with the first question what the answer?
Margaret [11]

Answer:

Density is 2.7 grams

Explanation:

formula for density is p= mass over volume [p being density]. So all you have to do is divide

8 0
3 years ago
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