Density = mass / volume
= 69g / 23 ml
= 3 g / ml.
Thus, the density of the sample is 3 grams per ml or 3g/ ml
Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend 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. 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.
Trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
PH= −log
10
[H
+
]
= −log
10
(0.001)
= −log
10
(10
−3
)
= −(−3)log
10
10
pH=3.
01
Answer:
When an atom attracts extra electrons it becomes a negative ion. The negative ion is larger than the original atom. The positive nucleus remains the same, with the same attractive force. However, this attractive force is now pulling on more electrons and therefore has less effect.
Or
Positive ions are formed by removing one or more electrons from the outermost region of the atom. The opposite is true of negative ions. When electrons are added to form an anion, the increased electron-electron repulsions cause the electrons to spread out more in space. Thus, anions are larger than their parent atoms.
Answer:
Explanation:
Reaction given
6 H⁺ + 2 MnO₄⁻ + 5 (COOH)₂ = 10CO₂ +8H₂O + 2 Mn⁺²
Oxidation number of Mn in MnO₄⁻
= x - 4 x 2 = -1
x = 8 -1
+ 7
Oxidation no of Mn in Mn⁺² = +2
So its oxidation no is decreased from + 7 to + 2 . Hence it is reduced.
