Answer:
2 moles
Explanation:
The following were obtained from the question:
Molarity = 0.25 M
Volume = 8L
Mole =?
Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:
Molarity = mole of solute/Volume of solution.
With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:
Molarity = mole of solute/Volume of solution.
0.25 = mole of MgCl2 /8
Cross multiply to express in linear form
Mole of MgCl2 = 0.25 x 8
Mole of MgCl2 = 2 moles
Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution
Answer:
They reduce the bond angle to be slightly lower than the tetrahedral bond angle, approximately 104.45 degrees.
Explanation:
The unshared pair of electrons or lone pair electrons in order to have the minimum repulsion possible with each other pushes the other bonding pairs closer together making the bond angle smaller or bent.
The bond angle is slightly lower than the tetrahedral bond angle of 108 degrees, leaving the water molecule with a bent molecular geometry.
Since medals form cations
nonmedals form anions
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb