A. Linear would be the answer
Good luck
The more electronegativity atom attracts electrons more strongly and gains a slightly negative charge, the less electronegative atom has a slightly positive charge.
Which of the following equations can be used to determine the change in enthalpy of a system
ΔHreaction=ΔHproducts−ΔHreactants
Answer:
0.190L of hydrogen may be produced by the reaction.
Explanation:
Our reaction is:
3Mg + 2H₃PO₄ → Mg₃(PO₄)₂ + 3H₂
We need to determine the limting reactant. Let's find out the moles of each:
5.159×10²¹ atoms . 1 mol / 6.02×10²³ atoms = 0.00857 moles of Mg
55.23 g . 1 mol / 97.97 g = 0.563 moles of acid
2 moles of acid react to 3 moles of Mg
0.563 moles of acid may react to: (0.563 . 3) /2 = 0.8445 moles of Mg
Definetely the limting reactant is Mg.
As ratio is 3:3, 3 moles of Mg can produce 3 moles of hydrogen
Then, 0.00857 moles of Mg must produce 0.00857 moles of H₂
At STP, 1 mol of any gas occupies 22.4L
0.00857 mol . 22.4L / 1mol = 0.190L
will be largest for 
.
Explanation: Ionization energy is the energy to knock off an electron from a gaseous atom of ion. First ionization energy or is the energy required to remove 1 loosely held electron from 1 mole of gaseous atoms to produce 1 mole of gaseous ion carrying (+)1 charge.
The electrons are filled according to Aufbau's rule and the orbitals which are strongly held to the nucleus follows the order 
.
Electron is released from the outermost shell that is from the electrons which are loosely held to the nucleus, this follows the pattern 
.
In configurations, 
The loosely held orbital is 4s, therefore electron will be lost from that easily.
Now, in 3p orbital, one configuration has 5 electrons and one has 1 electron.
The configuration having 5 electrons will be more tightly held by the nucleus because it has more electrons that the one having only 1 electron. Hence, the electron will be lost easily from the configuration having 
as the valence shell.
Therefore, the configuration will the largest .
