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miss Akunina [59]
3 years ago
8

Can someone help me solve this please 2.150 x 10 ^-12 dm / 5.02 x 10 ^-5dm^3

Chemistry
1 answer:
zavuch27 [327]3 years ago
4 0

4.3 x 10⁻⁸dm⁻²

Explanation:

Problem:

                   \frac{2.150 x 10^{-12} dm }{5.02  x 10 ^{-5} dm^{3} }

 To solve this problem, we have apply the right rule of indices operation.

If we have an indices in the form;

 \frac{a^{n} }{b^{k} } = (\frac{a }{b } )^{n-k}

  Solving this problem:

   = \frac{2.15}{5.02}   x  10^{-12 -(-5})

   = 0.43 x 10⁻⁷

   = 4.3  x 10⁻¹ x 10⁻⁷

   = 4.3 x 10⁻⁸

for the units:

 \frac{dm}{dm^{3} }   = \frac{dm}{ dm x dm x dm} = \frac{1}{dm^{2} }  = dm^{-2}

  The solution is:  4.3 x 10⁻⁸dm⁻²

   

learn more:

long division brainly.com/question/1747117

#learnwithBrainly

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    <u><em>Explanation</em></u>

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<h3><u><em> </em></u>Empirical formula  calculation</h3>

 Step 1:  find  the  moles   CO2  and  H2O

moles =mass/molar mass

moles   of CO2 =  6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol =  0.15  moles

Step 2: Find the moles  ratio  of Co2:H2O  by diving  each mole by smallest mole(0.15)

that  is  for  CO2 = 0.15/0.15  =1

              For H2O = 0.15/0.15 =1

therefore  the mole ratio  of Co2 : H2O = 1:1  which  implies that 1 mole of Co2  and 1  mole of H2O is  formed  during combustion reaction.


From the  the law of mass conservation the number  of atoms in reactant side  must  be equal to  number of  atoms  in product side

therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

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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

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

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Explanation:

From the question we are told that

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   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

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        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}

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Generally the partial  pressure of the methanol is mathematically represented as

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

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