The trends and exceptions to the trends in ionization energy observed includes;
B, ionization energy tends to increase across a period because the nuclear charge increases.
C, ionization energy tends to increase across a period because electrons are added to the same main energy level.
E, The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron, in the p orbital from the nucleus.
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example to remove an electron from a neutral fluorine atom to form a positively charged ion. <span />
Answer:
99°C
Explanation:
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Number of moles = 5 x 10^24 / 6.02 x 10^23 = 8.305 moles. Volume= moles x 22.4 = 186.032 liters. Hope this helps!
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
% yield = 80.719
<h3>Further explanation</h3>
Given
22.0 g of Mgl₂
25.0 g of Mg
25.0 g of l₂
Required
The percent yield
Solution
Reaction
Mg + I₂⇒ MgI₂
mol Mg = 25 g : 24.305 g/mol = 1.029
mol I₂ = 25 g : 253.809 g/mol = 0.098
Limiting reactant = I₂
Excess reactant = Mg
mol MgI₂ based on I₂, so mol MgI₂ = 0.098
Mass MgI₂ (theoretical):
= mol x MW
= 0.098 x 278.114
= 27.255 g
% yield = (actual/theoretical) x 100%
% yield = (22 / 27.255) x 100%
% yield = 80.719
