Removal of the loosely bounded electron is defined by the ionization energy. Chlorine needs less energy to add electrons and has higher ionization power. Thus, option A is correct.
<h3>What is ionization energy?</h3>
The ability of an electron to accept or give electrons to another element in a chemical reaction by forming and creating bonds and the positive and negative charges has been defined by the ionization energy.
Magnesium has lower ionization energy than chlorine due to its large size and smaller nuclear charge. On the other hand, chlorine can easily add electrons to the valence shell which is almost full.
Therefore, option A. chlorine has higher ionization energy and can add electrons with minimum energy.
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Answer:
moles H = 2 x 0.0649=0.130
ions = 0.130 x 6.02 x 10^23=7.81 x 10^22
Explanation:
It would be 1.55x10 to the 9th
hope that helps you
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
<span>0.70 mol/0.250 L = 2.8 M</span>