P orbitals have 3 sub-shells, each of which can hold one pair of electrons that have opposing spins. This leads to p orbitals holding a maximum of 6 electrons
Answer:
decant/ filter then decompose the electrolytes by electrolysis
Explanation:
The given question is incomplete. The complete question is as follows.
A 2.300×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.4 ml . the density of water at 20.0∘c is 0.9982 g/ml.
Calculate the molality of the salt solution.
Express your answer to four significant figures and include the appropriate units.
Explanation:
Molality is defined as the number of moles present in kg of a solvent.
Mathematically, Molality = 
Also,
Mole of solute = Molarity of solute x Volume of solution
= (0.0230 M) x (1.000 L) = 0.0230 mol of solute
Therefore, mass of solvent will be as follows.
= 997.7 g
= 0.9977 kg (as 1 kg = 1000 g)
Therefore, we will calculate the molality as follows.
Molality =
= 0.02306 mol/kg
thus, we can conclude that molality of the given solution is 0.02306 mol/kg.
Answer:
pt1
The half-life of the element is 1,600 years.
pt2
k = ln(2)/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1,600 years) = 4.33 x 10⁻⁴ year⁻¹.
kt = ln([A₀]/[A]),
where, k is the rate constant of the reaction (k = 4.33 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 312.0 g).
[A] is the remaining concentration of the sample ([A] = 9.75 g).
∴ t = (1/k) ln([A₀]/[A]) = (1/4.33 x 10⁻⁴ year⁻¹) ln(312.0 g/9.75 g) = [8,000 year].
-Hops
The theoretical yield is 0.0110 g H₂.
<em>Moles of H₂O</em> = 0.0987 g H₂O × (1 mol H₂O/18.02 g H₂O) = 0.005 477 mol H₂O
<em>Moles of H₂</em> = 0.005 477 mol H₂O × (2 mol H₂/2 mol H₂O) = 0.005 477 mol H₂
<em>Theoretical yield</em> of H₂ = 0.005 477 mol H₂ × (2.016 g H₂/1 mol H₂)
= 0.0110 g H₂