The electronic configuration of a ground-state Cr-atom :
1s²2s²2p⁶3s²3p⁶4s²3d⁴
<h3><u>What are electronic configurations?</u></h3>
The arrangement of an atom's or molecule's (or other physical structure's) electrons in their atomic or molecular orbitals is known as the electron configuration in atomic physics and quantum chemistry. For instance, the neon atom's electron configuration is 1s² 2s² 2p⁶, which means that 1, 2 and 6 electrons, respectively, are present in each of the 1s, 2s, and 2p subshells.
According to electronic configurations, each electron moves individually within an orbital while being surrounded by an average field produced by all other orbitals. Slater determinants or configuration state functions are used to mathematically characterize configurations.
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Answer: Please find answer in explanation column
Explanation:
During radioactive decay, the __unstable ________ isotope decays into a _stable ___________ isotope that has a different ____proton _______________ number
Or
During radioactive decay, the _ unstable parent nuclide ________ isotope decays into a _stable daughter nuclide ___________ isotope that has a different ____proton _______________ number.
There are 3 types of radioactive decay;alpha, beta and gamma, Of which the above clearly explains the beta decay. In beta decay, the unstable isotope having excess neutrons will undergo a beta decay emitting a beta particle.( ⁰₋₁e) causing the nucleus to loose a neutron but gain a proton.
Some heavy unstable isotopes which undergo radioactive (beta decay ) to become stable isotopes are phosphorus-32, strontium-90, iodine-131
Using Strontium 90 as an example , we have
⁹⁰₃₈St ----->⁹⁰₃₉Y + ⁰₋₁e
Strontium an unstable isotope undergoes a beta radioactive decay to form Yttrium.
Answer:
192.9
Explanation:
From the question,
Ke = [HCL]²/[H₂][CL₂].......................... Equation 1
Where Ke = Equilibrium constant.
Given: [HCL] = 0.0625 M, [H₂] = 0.0045 M, [CL₂] = 0.0045 M
Substitute these values into equation 1
Ke = (0.0625)²/(0.0045)(0.0045)
ke = (3.90625×10⁻³)/(2.025×10⁻⁵)
ke = 1.929×10²
ke = 192.9
Hence the equilibrium constant of the system = 192.9
The time required to reduce the concentration from 0.00757 M to 0.00180 M is equal to 1.52 × 10⁻⁴ s. The half-life period of the reaction is 9.98× 10⁻⁵s.
<h3>What is the rate of reaction?</h3>
The rate of reaction is described as the speed at which reactants are converted into products. A catalyst increases the rate of the reaction without going under any change in the chemical reaction.
Given the initial concentration of the reactant, C₀= 0.00757 M
The concentration of reactant after time t is C₁= 0.00180 M
The rate constant of the reaction, k = 37.9 M⁻¹s⁻¹
For the first-order reaction:
0.00180 = 0.00757 - (37.9) t
t = 1.52 × 10⁻⁴ s
The half-life period of the reaction:
Half-life of the reaction = 9.98 × 10⁻⁵s
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