QUESTION 3

Since there are 3 Hydrogen atoms present, the formula mass of H is 1.0 × 3 = 3.0 g/mol. Therefore, by adding them up, the formula mass of ammonia is: [14.0 g/mol + 3.0 g/mol] = 17.0 g/mol.

7 0

2 years ago

P-fluoroanisole reacts with sulfur trioxide and sulfuric acid. Draw the major product of this substitution reaction; if applicab

Ipatiy [6.2K]

Answer: (Structure attached).

Explanation:

This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.

The aromatic ring in p-fluoroanisole has two sustituents, an halogen and a methoxy group, which are ortho-para directing substituents.

Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using sulfur trioxide as an electrophile (very reactive).

The reaction occurs in three steps:

The attack on the electrophile forms the sigma complex.
The loss of a proton regenerates an aromatic ring.
The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).

Normally, a mixture of ortho-para substituted products would be obtained. However, since both para positions are occupied, only the ortho substituted product is obtained here.

8 0

2 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

2 years ago