Answer: the reaction contains significant amounts of products and reactants at equilibrium.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
is the constant of a certain reaction at equilibrium for gaseous reactants and products.
For reaction : 
![K_{eq}=\frac{[NH_3]\times [CO]}{[HCONH_2]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNH_3%5D%5Ctimes%20%5BCO%5D%7D%7B%5BHCONH_2%5D%7D)
![4.84=\frac{[NH_3]\times [CO]}{[HCONH_2]}](https://tex.z-dn.net/?f=4.84%3D%5Cfrac%7B%5BNH_3%5D%5Ctimes%20%5BCO%5D%7D%7B%5BHCONH_2%5D%7D)
The equilibrium lies far to the right when K is much greater than 1.
The equilibrium lies far to the left when K is much lesser than 1.
Thus as the value of is greater than 1, which means the reaction contains significant amounts of products and reactants at equilibrium.
Energy requires for removing one electron from the atom is first ionization energy & when another electron is removing than that energy is called 2nd ionization energy.
The answer is donating. This is because when an atom loses electron, it <em>donates</em> them to another atom.
Chemical reaction: Sr(OH)₂ + 2HNO₃ → Sr(NO₃)₂ + 2H₂O.
part A) m(Sr(OH)₂) = 21,78 g.
n(Sr(OH)₂) = m(Sr(OH)₂) ÷ M(Sr(OH)₂).
n(Sr(OH)₂) = 21,78 g ÷ 87,62 g/mol.
n(Sr(OH)₂) = 0,248 mol.
c(Sr(OH)₂) = n(Sr(OH)₂) ÷ V(Sr(OH)₂).
c(Sr(OH)₂) = 0,248 mol ÷ 0,075 L.
c(Sr(OH)₂) = 3,3 mol/L.
part B) V(Sr(OH)₂) = 34,21 mL = 0,03421 L.
n(Sr(OH)₂) = 0,03421 L · 3,3 mol/L = 0,112 mol.
From chemical reaction: n(Sr(OH)₂) : n(HNO₃) = 1 : 2.
n(HNO₃) = 0,225 mol.
V(HNO₃) = 0,225 mol ÷ 0,25 mol/L.
V(HNO₃) = 0,9 L = 900 ml.
