Explanation:
From the knowledge of law of multiple proportions,
mass ratio of S to O in SO:
mass of S : mass of O
= 32 : 16
= 32/16
= 2/1
mass ratio of S to O in SO2:
= mass of S : 2 × mass of O
= 32 : 2 × 16
= 32/32
= 1/1
ratio of mass ratio of S to O in SO to mass ratio of S to O in SO2:
= 2/1 ÷ 1/1
= 2
Thus, the S to O mass ratio in SO is twice the S to O mass ratio in SO2.
I don't know which one I should answer
Answer:
NH4+(aq) → NH3(aq) + H+(aq)
Explanation:
Following arrhenius, an acid can be defined as:
An Arrhenius acid is a substance that, when added to water, increases the concentration of H+ ions in water.
NH4+(aq) → NH3(aq) + H+(aq)
The ammonium ion acts as a weak acid in aqueous solution, dissociating into ammonia and a hydrogen ion.
An Arrhenius base is a substance that, when added to water, increases the concentration of OH- ions in water.
NH4+(aq) will not dissciate in OH- ions. So it's not a base, but an acid.
Answer: -
IE 1 for X = 801
Here X is told to be in the third period.
So n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.
Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
I think the answer is a but I am not for sure