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lukranit [14]
4 years ago
13

Which of the following does NOT involve a change of state? a. pouring water into a vacuum-insulated bottle b. sublimation of dry

ice c. freezing water d. vaporization of alcohol e. melting ice
Chemistry
1 answer:
Elena L [17]4 years ago
6 0
<h2> The correct option is (a).</h2>

Explanation:

  • There is no change of state in pouring water into a vacuum insulated bottle as the vacuum created inside the bottle only reduces the time limit for the transfer of heat by conduction and convection.
  • It helps the content filled in the bottle to be hot and cool for a longer time.

Hence, option (a) is correct.

Rest of the options (b), (c), (d), (e) is not correct because of the following reasons:

  • In the sublimation of dry ice, the dry ice (solid-state) is changed to the vapour form.
  • In the freezing of water, the liquid state is changed to solid-state.
  • During the vaporization of alcohol, the liquid state is changed to the vapour form.
  • During the melting of ice, the solid-state is changed to the liquid state.

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2) If I have an unknown quantity (# of moles) of gas at a pressure of 32 atm, a volume of 70 L, and a temperature of 25˚C. How m
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Answer:

91.5 mol

Explanation:

Volume of gas = 70 L

Temperature = 25°C

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Moles of gas  = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

25+273.15 = 298.15 K

By putting values,

32 atm × 70 L = n ×0.0821 atm.L /mol.K × 298.15 K

2240 atm.L = n ×24.48 atm.L /mol

n = 2240 atm.L / 24.48  atm.L /mol  

n = 91.5 mol

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Read 2 more answers
Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe
baherus [9]

Answer:

The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

3 0
3 years ago
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