Answer:
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O
Explanation:
Let's consider the reaction between acetic acid and strontium hydroxide. This is a neutralization reaction, in which an acid reacts with a base to form salt and water. The unbalanced equation is:
HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O
We have 1 acetate ion to the left and 2 to the right, so we will multiply HC₂H₃O₂(aq) by 2.
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O
Finally, we multiply water by 2 to get the balanced equation.
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O
The number of electrons it take to fill the mos formed from the combination of the 3s orbitals of two atoms simply is 14 electrons.
<h3>How electrons are distributed in the 3s orbitals.</h3>
The 3s orbital possess two different spherical nodes which is highly connected with the principal quantum number. In order words, it has 2 radial nodes. Also the shape of the 3s orbital is spherical in shape.
That being said, from the context of the above given task, the number of electrons which fill the mos formed from the combination of the 3s orbitals of two atoms is fourteen electrons.
However, the electron configuration of an atom simply is the arrangement of electrons in the electron shell or orbit of the atom of that element.
In conclusion, it can be deduced from above s orbital has a maximum of two electrons and this energy increases as the orbitals increases.
Read more on electron:
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setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :
Q<K⇒The reaction moved to the right (products)
Setup 2 :
Q=K⇒the system at equilibrium
Setup 3 :
Q>K⇒The reaction moved to the left (reactants)
Answer: C)The smaller the soil particles, the more water and air available for plant growth.
Explanation:
