Yes, K+ is<span> a </span>potassium<span> ion, and Mg</span>2+ is<span> a magnesium ion. But However, when non-metallic elements gain the </span>electrons<span> to form anions, Yes the end of their name </span>is<span> changed to “-ide.” and yes the example, a fluorine </span>atom<span> gains </span>one electron<span> to </span>become<span> a yes fluoride ion (F</span>-<span>) sooo yeessyes</span>
Answer:
0.0177 L of nitrogen will be produced
Explanation:
The decomposition reaction of sodium azide will be:

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas
The molecular weight of sodium azide = 65 g/mol
The mass of sodium azide used = 100 g
The moles of sodium azide used = 
so 1.54 moles of sodium azide will give =
mol
the volume will be calculated using ideal gas equation
PV=nRT
Where
P = Pressure = 1.00 atm
V = ?
n = moles = 2.31 mol
R = 0.0821 L atm / mol K
T = 25 °C = 298.15 K
Volume = 
The correct answer is ClO, ClO3-, ClO- and ClO4-
Kossel and Lewis in 1916 developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electron in order to have an octet( 8 electron) in their shells. This is known as octet rule.
In ClO2-, oxygen contains 8 electrons in its valence shell and oxygen will share one electron with chlorine to complete the octet of Cl. In other four, we can clearly see that there are more or less than 8 electrons in the outer shell of oxygen so we can clearly say that ClO, ClO3-, ClO- and ClO4- are disobeying the octet rule.
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
<u>8:4 and 4:8</u> both have CCP structure............................... ᐛ