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3 years ago

How many electrons will each element gain or lose in forming an ion?

1 answer:

8 0

It depends on the number of valence electrons required to make octet or duplet( in case of H). For example, Nitrogen(atomic number = 7) has electronic configuration(2,5) which means nitrogen has 5 valence electrons and requires 3 more electrons to complete its octet. After gaining 3 electrons from atoms of an element with less electronegativity than N, it forms nitride ion ( $N ^{-3}$ ). Hope this helps.

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How many moles of carbon are in a sample of 21.45 moles of heptane(C7H16)???????

saw5 [17]

The moles of carbon that are in the sample of 21.45 moles of heptane(C₇H₁₆) is 150.15 moles

<u><em>calculation</em></u>

moles of carbon = moles of heptane × number of C atom

number of C atom in heptane = 1 ×7 = 7 atoms

moles is therefore = 21.45 moles × 7 =150.15 moles

3 0

3 years ago

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(02.03 LC)

OLga [1]

The age of earth I think

3 0

4 years ago

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What is the difference between constitutional isomers and confirmational isomers?

babymother [125]

Answer:

<h3>Constitutional Isomers are two molecules with the same composition but different constitution. Confirmational Isomers are two molecules with the same configuration but different confirmation.</h3>

Explanation:

4 0

4 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.

6 0

3 years ago

Read 2 more answers

Can a new element be discovered and be placed in between copper and nickel on the periodic table? Explain your answer.

laila [671]

Answer:

Yes and no

Explanation:

If a new element is discovered and needed to be put into the table of elements, there would have to be a process to find a spot for it. The table is used heavily in chemistry and it revolves around how the elements are placed. So there would have to be a developed spot for the new element.

3 0

3 years ago