Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
Ionization energy increases from left to right because the left wants to lose elctrons and the right wants to gain electron
As you go a group it is easier lose lose because the electrons are farther away from the nucleus and there is less attraction from the positive charges.
It should be 3p3. the p level can hold 6 electrons
Rubidium group 1, 1 valence electrons very reactive
Mg2,2 very reactive
Al 3, 3 reactive
Answer:
I think B
Explanation:
There are more negative ions than positive ions
<h2>
Answer:
atoms</h2>
Explanation:
The given formula of the compound is 
The formula says
Every mole of contains 
moles of atoms of hydrogen.
Given that number of moles of compound is 
So,the number of moles of hydrogen atoms present is 
Since each mole has 
atoms,
moles has 
atoms.
