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Answer:

The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.

Explanation:

Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Thus, the overall reaction becomes;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?

According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;

Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

Answer:

Explanation:

In KCl, the two elements that combine to create KCl are potassium (K) and chlorine (Cl).

Potassium, as a Group 1 element, possesses one valence electron in its outermost shell which it readily donates during bonding. Every element aims to achieve a stable electron configuration, typically with 2 or 8 electrons in its outer shell. Potassium is characterized by its lower electronegativity and higher ionization energy, making it more likely to donate its electron than to accept one. On the other hand, chlorine belongs to Group 17 and has 7 electrons in its outer shell, requiring just one additional electron to complete its octet. Chlorine’s higher electronegativity and lower ionization energy facilitate its tendency to accept an electron rather than donate it.

The bond between potassium and chlorine that results in KCl is termed an electrovalent bond.

Reaction equation:

K + Cl → KCl

Answer:

The temperature difference is 293.15 Kelvin.

Explanation:

The provided information:

The temperature difference between the two matters is 20°C

We need to determine this difference in Kelvin =?

To solve this;

Using the formula:

0°C +273.15

Now substituting the values in place of 0.

20°C + 273.15 = 293.15 K

Hence, the temperature differential between the two samples is 293.15 K.