Answer:
B. s, p, d, f
Explanation:
These things are often referred to as suborbitals and you normally have s,p,d,f.
S has 1 two orbitals
P has 3 orbitals
D has 5 orbitals
F has 7 orbitals
and each orbital can house 2 electrons
Answer:
Number of moles of iron are 4 mol.
Explanation:
Given data:
Mass of iron = 223.4 g
Number of moles = ?
Solution:
Molar mass of iron = 55.85 g/mol
Formula:
Number of moles = mass/molar mass
Number of moles = 223.4 g/ 55.85 g/mol
Number of moles = 4 mol
Answer:
Rubidium= [Kr] 5s^1
Calcium= [Ar] 4s^2
Aluminium= [Ne] 3s^2 3p^1
Explanation:
A noble gas configuration begins with the elemental symbol of the last noble gas prior to the atom. The symbol is then followed by the remaining electrons.
Hope this helped! good luck :)
Answer:
0.147 billion years = 147.35 million years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of Potassium-40 is 1.25 billion years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
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/(1.25 billion years) = 0.8 billion year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
<em></em>
where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).
[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).
- At the time needed to be determined:
<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>
- If we start with 100% Potassium-40:
∴ The remaining concentration of Potassium-40 ([A] = 88.88%).
and that of argon-40 produced from potassium-40 decayed = 11.11%.
- That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).
∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.
