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Blababa [14]
3 years ago
8

ICS AND EQU Identifying intermediates in a reaction mechanism _ Consider the following mechanism for the formation of iodine: H2

(g) + ICI(9) → HI(g) + HCI(g) (1) HI(g) + ICI(g) → 12(g) + HCl(g) (2) Write the chemical equation of the overall reaction: Are there any ○yes intermediates in this no mechanism? If there are intermediates, write down their chemical formulas. Put a comma between each chemical formula, if there's more than one. Check Explanation
Chemistry
1 answer:
Mama L [17]3 years ago
7 0

Explanation:

Mechanism for the formation of iodine:

Step 1 :

H_2(g) + ICI(g) \rightarrow  HI(g) + HCI(g)

Step 2 :

HI(g) + ICI(g) \rightarrow I_2(g) + HCl(g)

By adding both reaction we will get the overall reaction of formation of iodine:

H_2(g)+2ICl(g)\rightarrow I_2(g)+2HCl(g)

Intermediate are defined as short lived chemical species which are formed during the course of a chemical reaction. They are formed in one step and in an other step they serve as a reactant.

The intermediate formed between the course of mechanism : HI

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Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO;?
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Answer:

K₂CO₃    

Explanation:

Given parameters:

Number of moles of K = 0.104mol

Number of moles of C = 0.052mol

Number of moles of O = 0.156mol

Method

From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.

Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.

  • Divide the given moles through by the smallest which is C, 0.052mol.
  • Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
  • This is the empirical formula

Solution

Elements                             K                       C                    O

Number of moles            0.104                0.052            0.156

Dividing by the

smallest                       0.104/0.052     0.052/0.052  0.156/0.052

                                            2                           1                     3

The empirical formula is K₂CO₃      

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Carbon atoms are extremely small and are one of the only atoms that are structurally stable enough to form various different kinds of macromolecules.
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