This reaction is most likely to fall under SN2 because the thing called carbonication does not occur in SN1. The carbon forms a partial bond with the nucleophile during the intermediate phase and the leaving group. So for this question the reaction will fall under SN2.

3 0

2 years ago

a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter

ladessa [460]

Answer:

The concentration of chloride ion is $2.82\times10^{-3}\;mol/L$

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the $Cl^-$ content 100 ppm suggests the presence of 100 mg of $Cl^-$ in 1 L of solution.

The molar mass of $Cl^-$ is equal to the molar mass of Cl atom as the mass of the excess electron in $Cl^-$ is negligible as compared to the mass of Cl atom.

So, the molar mass of $Cl^-$ is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of $Cl^-$ present in 100 mg (0.100 g) of $Cl^-$ is calculated as shown below:

$N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol$

So, there is $2.82\times10^{-3}\;mol$ of $Cl^-$ present in 1 L of solution.

5 0

3 years ago

Effective nuclear charge, Zeff, is defined as:

Alborosie

Answer:

The effective nuclear charge for a valence electron in oxygen atom: $Z_{eff} = 4.55$

Explanation:

Effective nuclear charge $[Z_{eff}]$ is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.

It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s).

$\Rightarrow Z_{eff} = Z - s$

For an oxygen atom-

Electron configuration: (1s²) (2s² 2p⁴)

The atomic number (actual nuclear charge): Z = 8

The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:

⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45

Therefore, the effective nuclear charge for a valence electron in oxygen atom is:

$Z_{eff} = Z - s = 8 - 3.45 = 4.55$

Therefore, the effective nuclear charge for a valence electron in oxygen atom: $Z_{eff} = 4.55$

6 0

2 years ago