<span>It’s
easier to determine the elecron configurations for the p-block elements
in periods 1,2,3 than to determine the electrons configurations for the
rest of the p-block elements in the periodic table beacause
from period 4, specifically from the element 31 (Ga), the atoms start to fill the d orbitals, and the energy levels of the 3d orbitals ara quite similar to the energy levels of 4p orbitals. So, for the elements Cr and Cu the right configurations do not match the configurations predicted using Aufbau method and Hund rules. Those are not the only exceptions but the two first. All is due to the proximity of the energy of the d and p orbitals and the fact that the rearrangement of the electrons result in a lower energy level. </span>
Valence electrons is the answer to ur question
Correct answer: Option D, <span>
K = 5.04 × 10^52</span>
Reason:
We know that,
Ecell =
,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52
The coefficient for NaNO₃ = 6
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
AI(NO₃)₃ +Na₂SO₄ →
Al₂(SO₄) +
NaNO₃
give coefficient
aAI(NO₃)₃ +bNa₂SO₄ →
Al₂(SO₄)₃ +c
NaNO₃
Al, left=a, right=2⇒a=2
N, left=3a, right=c⇒3a=c⇒3.2=c⇒c=6
Na, left=2b, right=c⇒2b=c⇒2b=6⇒b=3
The equation becomes :
2AI(NO₃)₃ +3Na₂SO₄ →
Al₂(SO₄)₃ +6NaNO₃
Answer:
Option B.
Explanation:
As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.
Then 2 moles of ethane will produce 4 moles of CO₂
