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GuDViN [60]
4 years ago
8

Calcum chloride contains only calcum and chloride What is the fomula for this compound? fou need the Peodie Table for thia quest

ion to detemine he proups of these elemente and thus determine ther onzaon deo for the Podic Table of he Elements in he Introduction to the queton) You might ao choope to acpes he Express your answer as a chemical formule View Available His) C. AE4 CaCI
Chemistry
1 answer:
mafiozo [28]4 years ago
3 0

Answer:

CaCl_2

Explanation:

Calcium is the element of the group 2 and period 4 which means that the valence electronic configuration is [Ar]4s^2.

Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is [Ne]3s^23p^5.

Thus, calcium losses 2 electrons to 2 atoms of chlorine and these 2 atoms of chlorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.

Ca         Cl

2            1

Cross multiplying the valency, We get, CaCl_2

Thus, the formula of calcium chloride is CaCl_2.

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Using the bond energy data from your text (or the internet), determine (show calculations for) the approximate enthalpy change ,
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Answer:

∆H=  <u>438 KJ/mol</u>

Explanation:

First, we have to find the <u>energy bond values</u> for each compound:

-) Cl-Cl = 243 KJ/mol

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-) F-Cl = 193 KJ/mol

If we check the reaction we can calculate the <u>number of bonds</u>:

Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)

In total we will have:

-) Cl-Cl = 1

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-) F-Cl = 6

With this in mind. we can calculate the <u>total energy for each bond</u>:

-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

-) F-F = (3*159 KJ/mol) = 477 KJ/mol

-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the <u>products</u> and the <u>reagents</u>:

Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a <u>subtraction</u> between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = <u>438 KJ/mol</u>

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I hope it helps!

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