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nikitadnepr [17]

2 years ago

7

Anew fictitious element was discovered. It is a metal named jolmium, J,and it has three valence electrons. If it combineswith io

dine it forms jolmium iodide. Answer the following questions.a.Is this compound ionic or covalent
a. Is this compound ionic or covalent?
b. What is the formula of Jolmium iodide?
c. What is the charge on the metal J in this compound?
d. Explain how you know that is the charge?

1 answer:

cluponka [151]2 years ago

3 0

Answer:

A new fictitious element was discovered.

It is a metal named jolmium, J, and it has three valence electrons.

If it combines with iodine it forms jolmium iodide. Answer the following questions.

a.Is this compound ionic or covalent

b. What is the formula of Jolmium iodide?

c. What is the charge on the metal J in this compound?

d. Explain how you know that is the charge?

Explanation:

It is given the element jolmium J is a metal with three valence electrons.

Since metals are highly electropositive, they lose electrons easily and form cations (Ions with a positive charge).

$J^{3+}$ ion will be formed.

b) The formula of jolmium iodide is shown below:

$J^{+3} I^{-1}\\The formula becomes:\\JI_{3}$

a) The compound is ionic in nature.

Since the compound formed between a metal atom and a nonmetal atom will take place by transfer of electrons.

Hence, it is ionic in nature.

c) The charge on the metal J in this compound is +3.

d) Given J has three valence electrons.

That means it can lose three electrons to form a bond.

So, its valency is three.

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With the information given we can assume that is:

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By the ICE (initial, change, equilibrium) analysis:

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C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

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Missing in your question :

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Ka1 for (H2CO3) = 4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

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2) equation 2 for Ka1 = 4.7 x 10^-7

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3) equation 3 for Ka2 = 5.6 x 10^-11

HCO3-(aq) + H2O(l) → CO3-2 (aq) + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l) Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq) + CO3-2(aq) Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq) ↔ Ca2+ (aq) + HCO3-(aq) K= 80.55

from the overall equation:

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when we have [Ca2+] = [HCO3-] so we can assume both = X

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when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
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so, by substitution on K expression:

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