What is the Kc for the following reaction at equilibrium at 500∘C if [N2]= 0.41 M , [H2]= 0.41 M , and [NH3]= 2.2 M ? N2(g)+3H2(
1 answer:
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Explanation:
(a) When we move across a period then there occurs a decrease in atomic size of the elements due to increase in force of attraction between the protons and electrons of an atom.
On the other hand, when we move down a group then there will occur an increase in atomic size of the elements due to increase in number of electrons which will also increase number of shells.
For example, Mg and Ca are group 2 elements, Br is a group 17 element and Ar is a group 18 element.
Hence, these are arranged in order of increasing radius as Ar < Br < Mg < Ca.
(b) Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period but decreases along a group.
Hence, order of increasing ionization energy for the given species is as follows.
Na < O < = Ne
(c) Electronegativity is defined as the ability of an atom to attract electrons towards itself. When we move across a period then there occurs an increase in electronegativity of the atoms.
And, when we move down a group then there occurs a decrease in electronegativity of the atoms.
Therefore, for the given elements order of increasing electronegativity is as follows.
Al < H < O F
Hello!
The reaction between HBr and KOH is the following:
HBr+KOH→H₂O + KBr
To calculate the amount of HBr left after addition of KOH, you'll use the following equations:
![HBr_f=HBr_i-KOH=([HBr]*vHBr)-([KOH]*vKOH) \\ \\ HBr_f=(0,25M*0,64L)-(0,5M*0,32L)=0 mol HBr](https://tex.z-dn.net/?f=HBr_f%3DHBr_i-KOH%3D%28%5BHBr%5D%2AvHBr%29-%28%5BKOH%5D%2AvKOH%29%20%5C%5C%20%20%5C%5C%20HBr_f%3D%280%2C25M%2A0%2C64L%29-%280%2C5M%2A0%2C32L%29%3D0%20mol%20HBr)
That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be neutral, close to 7.
Have a nice day!
The reaction between HBr and KOH is the following:
HBr+KOH→H₂O + KBr
To calculate the amount of HBr left after addition of KOH, you'll use the following equations:
That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be neutral, close to 7.
Have a nice day!