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Volgvan
3 years ago
6

What determines how ionic bonds will form?

Chemistry
1 answer:
Harrizon [31]3 years ago
8 0

Answer:

I think the answer should be B. Number of valence electrons

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A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp
vichka [17]

Answer:

The balanced chemical equation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = T_1= 21.36^oC

Final temperature of the calorimeter = T_2= 26.37^oC

Heat absorbed by calorimeter = Q

Q=C\times \Delta T

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol

3 0
3 years ago
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
trasher [3.6K]

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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