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IgorLugansk [536]
3 years ago
11

How many grams of NaHCO3 are needed to prepare 250 mL of 0.50 M NaHCO3?

Chemistry
1 answer:
forsale [732]3 years ago
6 0

Answer:

10.5g

Explanation:

First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:

Volume = 250mL = 250/1000 = 0.25L

Molarity = 0.5M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.5 x 0.25

Mole = 0.125 mole

Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:

Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol

Number of mole of NaHCO3 = 0.125 mole

Mass of NaHCO3 =?

Mass = number of mole x molar Mass

Mass of NaHCO3 = 0.125 x 84

Mass of NaHCO3 = 10.5g

Therefore, 10.5g of NaHCO3 is needed.

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Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

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

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

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