Hydrogen fusion is the answer. It is also known as <span>hydrogen burning. </span>
The electron configuration that belongs to the atom with the lowest first ionization energy is francium.
<h3>What is ionization energy? </h3>
Ionization energy is defined as the minimum amount of energy required to remove the most loosely electron present in outermost shell.
<h3>Ionization energy across period</h3>
Ionization energy increase as we move from left to right in the period. This can be explained as when we move from left to right along period new electron is added to the same shell which increase the nuclear charge. Hence results int he decrease in size. Due to this decrease in size more energy is required to remove electron from outermost shell.
<h3>Ionization energy along group</h3>
Ionization energy decrease as we move from top to bottom along group. This can be explained as we move from top to bottom new electron is added to new shell. Due to addition of new shell the size of atom increases which results in the decrease in the nuclear charge. Due to this less amount of energy is needed to remove an electron.
Thus, we concluded that the electron configuration that belongs to the atom with the lowest first ionization energy is francium.
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Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0
b) To calculate the constant we have the following expression:
Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K
c) The equilibrium pressure of O₂ over M is:
1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs
2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams
Answer: 40.1%
Explanation: The mass of calcium in this compound is equal to 40.1 grams because there's one atom of calcium present and calcium has an atomic mass of 40.1 . The molar mass of the compound is 100.1 grams. Using the handy equation above, we get: Mass percent = 40.1 g Ca⁄100.1 g CaCO3 × 100% = 40.1% Ca.
