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sweet-ann [11.9K]
3 years ago
14

The electronegative oxygen that is central to a water molecule is __________ bound to two hydrogen atoms. These hydrogens are 'b

ent' to form a 105° angle because _________
a.) ionically, four of the outer e- about oxygen are shared with hydrogens

b.) covalently, four of the outer e- about oxygen are shared with hydrogens

c.) covalently, of hydrogen bonding with other water molecules

d.) ionically, of hydrogen bonding with other molecules
Chemistry
1 answer:
iragen [17]3 years ago
7 0

Answer: Option (b) is the correct answer.

Explanation:

When an atom is attached to another atom through sharing of electrons then bond formed between the atoms is known as a covalent bond. And, a bond formed by transfer of electrons from one atom to another is known as an ionic bond.

For example, electronic configuration of hydrogen is 1s^{1} and electronic configuration of oxygen is 1s^{2}2s^{2}2p^{4}. So, in order to attain stability hydrogen needs 1 more electron and an oxygen atom needs two electrons.

Therefore, two hydrogen atoms need to covalently bond through an oxygen atom leading to the formation of H_{2}O.

Thus, we can conclude that the electronegative oxygen that is central to a water molecule is covalently bound to two hydrogen atoms. These hydrogens are 'bent' to form a 105^{o} angle because four of the outer e- about oxygen are shared with hydrogens.

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The correct answer would be B: 4.

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2 years ago
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

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Answer:

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