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Nastasia [14]
3 years ago
13

A student weighs out a 6.64 g sample of , transfers it to a 500. mL volumetric flask, adds enough water to dissolve it and then

adds water to the 500. mL tick mark. What is the molarity of cobalt(II) fluoride in the resulting solution
Chemistry
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

the molarity of cobalt(II) fluoride in the resulting solution is = 0.137  M

Explanation:

Given that :

a student dissolves 6.64 g of CoF₂  into 500 mL of water

volume of the solution(water) = 500 mL = 0.50 L

The standard  molar mass of CoF₂ is 96.93 g/mol

number of moles of CoF₂ = mass of CoF₂/molar mass of CoF₂

number of moles of CoF₂ = 6.64 g/96.93 g/mol

number of moles of CoF₂ = 0.0685 mol

The molarity of any given substance is known to be as the number of moles of solute dissolved in one litre of solution.

Thus ;

Molarity of  cobalt(II) fluoride CoF₂ in the resulting solution is = number of moles / Volume in (L)

Molarity of  cobalt(II) fluoride CoF₂  = 0.0685 mol/ 0.50 L

= 0.137  M

Thus ; the molarity of cobalt(II) fluoride in the resulting solution is = 0.137  M

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In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

4 0
3 years ago
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