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dimulka [17.4K]
3 years ago
12

HCl can be produced by the exothermic reaction H2(g)+2ICl(g)→2HCl(g)+I2(g) . A proposed mechanism for the reaction has two eleme

ntary steps as shown below.
Step 1: H2(g)+ICl(g)→HI(g)+HCl(g) (slow)

Step 2: HI(g)+ICl(g)→HCl(g)+I2(g) (fast)
(d) Write a rate law for the overall reaction that is consistent with the proposed mechanism.
(e) On the incomplete reaction energy diagram below, draw a curve that shows the following two details.

The relative activation energies of the two elementary steps
The enthalpy change of the overall reaction

Chemistry
1 answer:
xenn [34]3 years ago
8 0

Answer:

Explanation:

The solutions and the answers can be found in the attachment

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Masja [62]

Answer:

B. Angle BOA

Hope this helped! :)

8 0
3 years ago
Need help will<br> Give 55 points
CaHeK987 [17]

D, L, A, L,?, j, b, ,, ,, ,,

4 0
3 years ago
If an experiment produces 5 g but should have made 500 g, what is the percent<br> yield?
hoa [83]

Answer:

Percentage of yield = 1%

Explanation:

Given:

Amount of yield = 5g

Total amount of product = 500 gram

Find:

Percentage of yield = ?

Computation:

⇒ Percentage of yield = [Amount of yield / Total amount of product]100

⇒ Percentage of yield = [5g / 500g]100

⇒ Percentage of yield = [0.01]100

⇒ Percentage of yield = 1%

7 0
4 years ago
The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this
azamat

Answer : The value of K_{sp} for BaCO_3 is 19.36\times 10^{-10}mole^2/L^2.

Solution : Given,

Solubility of BaCO_3 in water = 4.4\times 10^{-5}mole/L

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

                            BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3

Initially                   -                   0        0

At equilibrium       -                   s         s

The Solubility product will be equal to,

K_{sp}=[Ba^{2+}][CO^{2-}_3]

K_{sp}=s\times s=s^2

[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L

Now put all the given values in this expression, we get the value of solubility constant.

K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2

Therefore, the value of K_{sp} for BaCO_3 is 19.36\times 10^{-10}mole^2/L^2.

3 0
4 years ago
The solubility of cadmium cyanide is 1.70 g/100 mL of water at 25°C. Which of the following solutions is unsaturated?
hram777 [196]

Answer:

The answer to your question is: letter B is correct

Explanation:

Calculate the Molarity of each solution an compare it to the solubility

Solubility

MW Cd(CN)2    = 252 g

                                         252 g      ----------------  1 mol

                                           1.7 g   ------------------    x

                                        x = (1.7 x1 ) / 252

                                        x = 0.006 mol

            Molarity = 0.006 / 0.1 = 0.067

a.97.0 g of Cd(CN)2 in 5.00 mL

                                          252 g      ----------------  1 mol

                                             97 g --------------------   x

                                          x = (97 x 1) / 252

                                          x = 0.38

            Molarity = 0.38 / 0.005 =75              Saturated

b. 325.5 g of Cd(CN)2 in 20.5 L

                                          252 g      ----------------  1 mol

                                        325.5 g     ----------------    x

                                        x = (325.5 x 1) / 252

                                        x = 1.29 mol

            Molarity = 1.29 / 20.5 = 0.063          Unsaturated

c. No right answer.

d. 4.25 g of Cd(CN)2 in 250.0 mL

                                            252 g      ----------------  1 mol

                                             4.25 g     ---------------   x

                                           x = (4.25 x 1) / 252

                                           x = 0.017

               Molarity = 0.017 / 0.25 = 0.067       Saturated

e. 5.750 g of Cd(CN)2 in 330.0 mL                

                                             252 g      ----------------  1 mol

                                              5.75 g ------------------    x

                                           x = (5.75 x 1) / 252 = 0.023

              Molarity = 0.023 / 0.33 = 0.07        Saturated

   

4 0
4 years ago
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