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2 years ago

Describe polar covalent bonds using water as an example

1 answer:

3 0

A covalent bond is a bond formed by atom sharing.
In water molecule, there are twice the number of hydrogen atoms than the oxygen atoms. Its structure is H-O-H. The electronegative difference between the H and O allows them to be polar because on side there is positive charge and on another side there is negative charge.

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jasenka [17]

Answer:

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Explanation:

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2 years ago

In the diagram shown, what is occurring at the section marked 4?

astra-53 [7]

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3 years ago

Which of the following choices represents elements with the most similar

matrenka [14]

Answer:

Explanation:

F Cl Br belongs to the Same group

5 0

3 years ago

21+ Cl2 → 12+201 -

Alina [70]

Answer:

$\rm 2\; I^{-} + Cl_2 \to I_2 + 2 \; Cl^{-}$ .

Start color: yellowish-green.

End color: dark purple.

Assumption: no other ion in the solution is colored.

Explanation:

In this reaction, chlorine gas $\rm Cl_2$ oxidizes iodine ions $\rm I^{-}$ to elemental iodide $\rm I_2$ . At the same time, the chlorine atoms are converted to chloride ions $\rm Cl^{-}$ .

Fluorine, chlorine, bromine, and iodine are all halogens. They are all found in the 17th column of the periodic table from the left. One similarity is that their anions are not colored. However, their elemental forms are typically colored. Besides, moving down the halogen column, the color becomes darker for each element.

Among the reactants of this reaction, $\rm I^{-}$ is colorless. If there's no other colored ion, only the yellowish-green hue of $\rm Cl_2$ would be visible. Hence the initial color of the reaction would be the yellowish-green color of $\rm Cl_2$ .

Similarly, among the products of this reaction, $\rm Cl^{-}$ is colorless. If there's no other colored ion, only the dark purple hue of $\rm I_2$ would be visible. Hence the initial color of the reaction would be the dark purple color of $\rm I_2$ .

5 0

2 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago