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hichkok12 [17]
3 years ago
12

Does a solution have to involve a liquid? Explain your answer

Chemistry
2 answers:
FromTheMoon [43]3 years ago
7 0
No becasue a solution is only in a linear equation
Ulleksa [173]3 years ago
3 0
Not actually because a solution in math also involves the answer as well as physics. 
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shusha [124]
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OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial  pressure of  methanol is  P_{CH_3OH_{(g)}} =0.077 \  bar

The partial  pressure of carbon monoxide is  P_{CO} = 0.382 \ bar

The partial  pressure at  hydrogen is  P_H =  O \  bar

Explanation:

From the question we are told that

  The volume of the  flask is  V_f = 10.0 \  L

   The initial pressure of carbon monoxide gas is  P_{CO} = 0.461 \ bar

   The initial  temperature of carbon monoxide gas is T_{CO} = 22.0^oC

   The volume of the hydrogen gas is  V_h  =  200 mL = 200 *10^{-3} \  L

    The initial  pressure of the hydrogen is P_H  =  7.10 \  bar

    The initial temperature of the hydrogen  is  T_H = 271 \  K

The reaction of  carbon monoxide and  hydrogen is  represented as

         CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}

Generally from the ideal gas equation the initial number of moles of carbon monoxide is  

        n_1  =  \frac{P_{CO} *  V_f }{RT_{CO}}

Here R is the gas constant with value  R  = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K

=>     n_1  =  \frac{0.461  *  10 }{0.0821 * (22 + 273)}

=>     n_1  = 0.19

Generally from the ideal gas equation the initial number of moles of Hydrogen  is  

       n_2  =  \frac{P_{H} *  V_H }{RT_{H}}

      n_2  =  \frac{ 7.10 *  0.2 }{0.0821 * 271 }

=> n_2  =  0.064

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CO

=>      0.064 moles of  hydrogen gas will react with  x  mole of  CO

So

          x = \frac{0.064}{2}

=>       x = 0.032 \ moles \ of  \  CO

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CH_3OH_{(g)}

=>      0.064 moles of  hydrogen gas will react with  z  mole of  CH_3OH_{(g)}

So

          z = \frac{0.064}{2}

=>       z = 0.032 \ moles \ of  \ CH_3OH_{(g)}

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is  

          n_k = n_1 - x

=>       n_k = 0.19  - 0.032

=>       n_k = 0.156

So at the end of the reaction , the partial pressure for  CO is mathematically represented as

      P_{CO} = \frac{n_k  *  R *  T_{CO}}{V}

=>    P_{CO} = \frac{0.158   *  0.0821 *  295}{10}

=>    P_{CO} = 0.382 \ bar

Generally the partial pressure of  hydrogen is  0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial  pressure of the methanol is mathematically represented as

         P_{CH_3OH_{(g)}} = \frac{z  *  R *  T_{CO}}{V_f}

Here  T_{CO} is used because it is given the question that the   temperature  returned to 22.0 degrees C

So

      P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 *  295}{10}

     P_{CH_3OH_{(g)}} =0.077 \  bar

6 0
3 years ago
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Methanol (CH3OH) has a heat of fusion of 3.16 kJ/mol. Which of the following is the heat of solidification that occurs when 64 g
Alexandra [31]
To determine the heat dissipated when a substance freezes, we multiply the heat of fusion of the substance to the mass of the substance that freezes. We calculate as follows:

Heat = -3.16 (64/32.06) = - 6.32 kJ

Hope this answers the question.
5 0
3 years ago
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Will a double replacement reaction occur if a potassium hydroxide solution is mixed with a lead(II) nitrate solution? Briefly ex
ohaa [14]
No - a precipitation will occur though. Potassium nitrate is soluble in water, so the potassium and nitrate ions will remain spectator ions and stay in solution. Lead (II) hydroxide is not soluble, and will precipitate out of solution to form a solid product.
3 0
3 years ago
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The vapor pressure of chloroform is 173.11 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in chloroform is aspirin
Over [174]

Answer:

VP (solution) = 171.56 mmHg

Explanation:

Vapor pressure of pure solvent(P°) - Vapor pressure of solution (P') = P° . Xm

Let's replace the data:

173.11 mmHg - P' = 173.11 mmHg . Xm

Let's determine the Xm (mole fraction for solute)

Mole fraction for solute = Moles of solute / Total moles

Total moles = Moles of solute + moles of solvent.

Let's determine the moles

Moles of solvent → 623.4 g / 119.4 g/mol = 5.22 moles

Moles of solute → 9.322 g / 180.1 g/mol = 0.052 moles

Total moles = 0.052 + 5.22 = 5.272 moles

Xm = 0.052 moles / 5.272 moles = 0.009 → 9/1000

173.11 mmHg - P' = 173.11 mmHg . 9/1000

P' = - (173.11 mmHg . 9/1000 - 173.11 mmHg)

P' = 171.56 mmHg

4 0
3 years ago
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