For radioactive decay, we can relate current amount, initial amount, decay constant and time using:
N = No x exp(-λt)
Half-life = ln(2)/λ
λ = ln(2) / 5730
N/No = 80% = 0.8
0.8 = exp( -ln(2)/5730 x t)
t = 1844 years
Here we have explain that the maximum possible electrons present in nitrogen valence shell is 8 whereas in phosphorous 12 valence electrons are present.
Although both nitrogen (N) and phosphorous (P) belongs to the same series there are several properties which are different between both the element. The number of electrons present in nitrogen is seven which are present in the -s and -p orbitals. The electronic configuration of nitrogen is 1s²2s²2p³. In which the outermost electrons are the valence electrons i.e. 5 valence electrons are present. The maximum orbitals are possible under the principal quantum number 2 are -s and -p orbitals. Now the maximum capacity of the p orbital to contain 6 electrons, as it is half filled in nitrogen another 3 electrons can be incorporated. Thus the maximum number of electrons can be present in nitrogen is 10 among which 8 is the valence electrons.
On the other hand there are 15 electrons in phosphorous the electronic configuration is 1s²2s²2p⁶3s²3p³. Now the principal quantum number 3 can have three orbitals -s, -p and -d. So another 13 electrons can be incorporated (3 in -p orbital and 10 in -d orbital) among which upto 12 electrons can be its valence electrons.
Answer:
2.42L
Explanation:
Given parameters:
V₁ = 1.8L
T₁ = 293K
P₁ = 101.3kPa
P₂ = 67.6kPa
T₂ = 263K
Unknown:
V₂ = ?
Solution:
To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;
All the units are in the appropriate form. We just substitute and solve for the unknown;
101.3 x 1.8 / 293 = 67.6 x V₂ / 263
V₂ = 2.42L
The mass number goes on top, and the atomic number goes on bottom.
Therefore, the answer is C:
40
K
19
