Answer:
Explanation:
Iso-electronic species have same number of electrons . Positive charged ions will have smaller size . As electrons add , size increases due to electronic repulsion .
Following species are isoelectronic .
Al³⁺ < Mg²⁺ < Na¹⁺ < Ne < F⁻¹ < O⁻² < N⁻³
__ KClO₃ → __ KCl + __ O₂
Left Side:
1 K
1 Cl
3 O
Right Side:
1 K
1 Cl
2 O
Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.
This gives us 2KClO₃ → __ KCl + 3O₂.
However, this equation is still not balanced.
Left Side:
2 K
2 Cl
6 O
Right Side:
1 K
1 Cl
6 O
In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.
2KClO₃ → 2KCl + 3O₂
Answer:
3)The reaction is not at equilibrium and willproceed to the right.
Explanation:
The reaction quotient of an equilibrium reaction measures relative amounts of the products and the reactants present during the course of the reaction at particular point in the time.
It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.
Q < Kc , reaction will proceed in forward direction.
Q > Kc , reaction will proceed in backward direction.
Q = Kc , reaction at equilibrium.
Given that:
Q = 
K = 
Since, Q < K , reaction is not at equilibrium and will proceed to right, in forward direction.
Answer:
they have the same molar solubility at 44.5 degrees celsius
Answer:
4 moles of SO3 will be produced from 6 moles of oxygen.
Explanation:
From the reaction given
S8 + 12 O2 ----> 8 SO3
12 moles of oxygen reacts to form 8 moles of SO3
if 6 moles of oxygen were to be used instead, it has been reduced to half of the original mole of oxygen used. Then the moles of SO3 will also be reduced to half.
6 moles of O2 will yield 4 moles of SO3
12 moles = 8 moles
6 moles = ?
? = 6 * 8 / 12
? = 48/ 12
? = 4 moles of SO3.
