Answer:
Along period electronegativity and ionization energy increases.
Along group electronegativity and ionization energy decreases.
Explanation:
Along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. Thus the attraction of the atoms for valance electrons increases. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required, and electronegativity also increases.
Along group:
As we move from top to bottom in periodic table the atomic sizes increases.The electrons are added in next energy level in every next element. Thus the valance electrons farther away from the nucleus and hold of nucleus becomes weaker, because of weak nuclear attraction atomic radii increases and electronegativity and ionization energy decreases.
Answer:
A. Cu^+2(aq)cathode ---> Cu^+2(aq)anode
Explanation:
Electrolysis is the process in which current is passed through a solution thereby causing a chemical change at the anode and cathode. Copper is being purified using electrolysis by using impure copper at the anode and pure copper at the cathode. This pure and impure copper are placed in a copper(ii)sulfate electrolyte solution and dc current is made to pass through it. The resulting changes at the anode and cathode are given by the equation:
cathode: Cu²⁺ + 2e⁻ ⇒ Cu
anode: Cu ⇒ Cu²⁺ + 2e⁻
To determine what gas is this, we use Graham's Law of Effusion where it relates the rates of effusion of gases and their molar masses. We do as follows:
r1/r2 = √(M2 / M1)
Let 1 be the the unkown gas and 2 the H2 gas.
r1/r2 = 0.225
M2 = 2.02 g/mol
0.225 = √(2.02 / M1)
M1 = 39.90 g/mol
From the periodic table of elements, most likely, the gas is argon.
I believe the correct answer from the choices listed above is the third option. The <span>oxidation number of Nitrogen in HNO2 would be +3. It is calculated as follows:
1 + x + (-2)(2) = 0
x = +3
Hope this answers the question. Have a nice day.</span>
