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ikadub [295]
3 years ago
4

Roberto examines two molecular models. He sees that the atoms are the same in each model and are attached in the same order in e

ach model. However, he cannot make the models appear the same without breaking a double bond. What do the models represent?
two structural isomers that have the same molecular formula
two geometric isomers that have the same molecular formula
two organic compounds that have the same structural formula
two models of a single organic compound

Chemistry
2 answers:
Dafna11 [192]3 years ago
8 0
<span>B) Two geometric isomers that have the same molecular formula </span>
mylen [45]3 years ago
3 0
I have drawn the scenario as discussed in statement. The bond color shows their position before and after change. Structure 1 and 2 are those two models which are in hands of Roberto. These two structures are not identical to each other, as the blue and green bonds are not same (still the molecular formula of both compound is same). When structure 2 is broken at double bond, rotated at 180° degree and re assembled it forms the structure (3) exactly equal and same to structure 2.

Explanation:
                   It means that the positions of substituents are having different positions in space (3D). And we know well that the double bond can't be rotated as single bond. So, due to double bond rotation restriction we are having two isomers (such isomers which have different position in 3D are called Stereoisomers) and this stereoisomer which takes place due to double bond restriction is called Geometrical Isomerism or cis trans Isomerism.

Result:
           These two models (1 and 2) represent two geometric isomers that have the same molecular formula.

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4 0
4 years ago
Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)     \Delta H_1=-241.8kJ

(2) X(s)+2Cl_2(g)\rightarrow XCl_4(s)    \Delta H_2=+461.9kJ

(3) \frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)    \Delta H_3=-92.3kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) H_2O(g)\rightarrow H_2O(l)    \Delta H_5=-44.0kJ

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) 2H_2O(g)\rightarrow 2H_2(g)+O_2(g)     \Delta H_1=2\times 241.8kJ=483.6kJ

(2) XCl_4(s)\rightarrow X(s)+2Cl_2(g)    \Delta H_2=-461.9kJ

(3) 2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)    \Delta H_3=4\times -92.3kJ=-369.2kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) 2H_2O(l)\rightarrow 2H_2O(g)    \Delta H_5=2\times 44.0kJ=88.0kJ

The expression for enthalpy of main reaction will be:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5

\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)

\Delta H=-1048.6kJ

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0
3 years ago
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