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daser333 [38]
3 years ago
14

As the volume of a 1 mole sample of gas increases, with the temperature remaining constant, the pressure exerted by the gas:

Chemistry
1 answer:
Vinvika [58]3 years ago
7 0
I believe the correct answer would be decreases. At constant temperature, the volume and pressure are inversely related that is when one value increases the other decreases or as one decreases the other increases. Hope this helps. Have a nice day.
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Which statement is true of the particles that make up a substance?
victus00 [196]

Particles in a gas have more energy than particles in a liquid. Because in gaseous state particles are free to move around due to which kinetic energy of molecule or gas increases and hence overall energy increases

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2 years ago
A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3. Determine the equilibrium pressure of NO2 if Kp for the reac
dlinn [17]

Explanation:

Reaction equation for this reaction is as follows.

     NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)

It is given that K_{p} = 0.0118.

According to the ICE table,

              NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)

Initial:           0.86      0.86               0            0

Change:          -x          -x                 +x           +x

Equilibrium:  0.86 - x   0.86 - x         x           x

Hence, value of K_{p} will be calculated as follows.

           K_{p} = \frac{P_{NO_{2}} \times P_{SO_{3}}}{P_{NO} \times P_{SO_{3}}}

         0.0118 = \frac{x \times x}{(0.86 - x)^{2}}

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4 0
3 years ago
Ethanol undergoes combustion in oxygen to produce carbon dioxide gas and liquid water. The standard heat of combustion of ethano
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Answer:

-277.6\ \text{kJ/mol}

Explanation:

Heat of combustion of C_2H_5OH(l) = -1366.8 kJ/mol

Heat of formation of CO_2(g) = -393.5 kJ/mol

Heat of formation of H_2O(l) = -285.8 kJ/mol

The reaction is

C_2H_5OH(l)\rightarrow 2CO_2(g)+3H_2O(l)

Assuming x is the standard enthalpy of formation of ethanol

-1366.8=2\times (-393.5)+3\times (-285.8)-x\\\Rightarrow x=2\times (-393.5)+3\times (-285.8)+1366.8\\\Rightarrow x=-277.6\ \text{kJ/mol}

The standard enthalpy of formation of ethanol is -277.6\ \text{kJ/mol}.

5 0
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Alex17521 [72]

Answer:

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Explanation:

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Answer:

Explanation:

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b.Metric unit

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