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dangina [55]
3 years ago
14

Consider the following equilibrium: 2SO2(g) + O2(g) 2SO3(g) Which of the following is a correct equilibrium expression? Keq=[SO3

][SO2][O2] Keq=[SO3]2[SO2]2[O2] Keq=[SO2][O2][SO3] Keq=[SO2]2[O2][SO3]2
Chemistry
1 answer:
Ne4ueva [31]3 years ago
7 0
Answer : Option B) K_{eq}  =  \frac{[SO_{3}]^{2}  }{[SO_{3}]^{2} . [O_{3}]^{2}}

Explanation : The equilibrium constant is often expressed in terms of the molar concentration of products divided by the product of molar concentrations of reactants.

So, K_{eq} =  \frac{[products]}{[reactants]}.

Now, here in the given reaction of  

2SO_{2}  + O_{2} ----\ \textgreater \  2SO_{3}

So, we have the equation for equilibrium as K_{eq} = \frac{[SO_{3}]^{2} }{[SO_{3}]^{2} . [O_{3}]^{2}}
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Draw structures from the following names, and determine which compounds are optically active:(c) 1-bromo-1-chlorobutane
kirill [66]

The given compound 1-bromo-1-chlorobutane is an optically active compound .

Because this compound does not have plane of symmetry (POS) and center  of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .

It has  chiral center.

Here , the chiral carbon is bonded to 4 distinct groups such as : chlorine , bromine , hydrogen and propyl .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group  is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

Learn more about chiral center here:

brainly.com/question/9522537

#SPJ4

7 0
2 years ago
What is the bond hybridization for an oxygen in CO32-?
il63 [147K]

Answer:

sp2 hybridization

Explanation:

The central carbon in the carbonate ion, CO32-, has sp2 hybridization. Carbonate has a central carbon atom bonded to three oxygen atoms.

3 0
3 years ago
Consider the reaction below to answer the following question: This reaction is an example of: A. A Michael reaction B. A Robinso
Anestetic [448]

The reaction is missing, so i have attached it.

Answer:

C. An intramolecular aldol condensation

Explanation:

From the attached image showing the reaction, we can see that the left hand side of the reaction has 2 carbonyl groups which are the double bonds attached to the oxygen atoms.

Now, on the right hand side, we can see that a six member ring has been formed.

This 6 member ring is produced because one of the carbonyls on the left hand side was deprotonated at the alpha position thereby serving as a nucleophile, which then attacks the carbon in the other carbonyl.

This process is possible because it underwent the processes of deprotonation, intramolecular aldol addition, proton transfer and elimination to yield the right hand side product which is ɑ,β-unsaturated carbonyl compound.

Thus, the correct answer is option C

4 0
3 years ago
11. Propane (C3Hg) is a fuel commonly used in gas grills.
Mrac [35]

Answer:

81.71%

Explanation:

One mole of propane contains 3 moles of carbon atoms and 8 moles of hydrogen atoms, as seen from the molecular formula of C_3H_8. In order to calculate the percent of carbon in propane by mass, we need to remember that %w/w (or percent mass) formula states that:

\omega=\frac{m_{component}}{m_{total}}\cdot100\%

That is, we need to divide the mass of the component of interest by the total mass of the compound and multiply by 100 to obtain the percentage.

For simplicity, let's take 1 mole of propane and find the mass of 1 mole (hence, we'll be finding the molar mass of propane). To do that, we add the 3 molar masses of carbon and 8 molar masses of hydrogen to obtain a total of:

M_{C_3H_8}=3M_C+8M_H=3\cdot12.011 \frac{g}{mol}+8\cdot1.00784\frac{g}{mol}=44.096 \frac{g}{mol}

Now that we have the molar mass of propane, we also need to find the total mass of carbon in 1 mole of propane. We know that we have a total of 3 moles of carbon which corresponds to:

M_C=3\cdot12.011 \frac{g}{mol}=36.033 \frac{g}{mol}

Dividing the mass of carbon present by the total mass of the compound will yield the mass percentage as defined by the formula we introduced:

\omega_C=\frac{36.033\frac{g}{mol}}{44.096 \frac{g}{mol}}\cdot 100\%=81.71\%

5 0
3 years ago
If volume remains the same while the mass of a substance ________, the density of the substance will______________
victus00 [196]
Black and blank sorry idk
4 0
4 years ago
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