Chemistry is the study of all of the following EXCEPT

Compute the end-to-end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule with a number-

MArishka [77]

Answer:

5156 Â

Explanation:

Chains of monomers that are being linked together constitute what is called polymers. in this question; we are to compute the end to end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule.

Using the following approach:

The first step is to determine the repeating units in polymer

The number of repeating unit = average molecular weight/ atomic weight of PVC

Given that:

average molecular weight = 256,131 g/mol

atomic weight of PVC = 62.498 g

Then;

The number of repeating unit = 256,131 g/mol / 62.498 g = 4098.227143 mol

Now the distance is calculated by using the formula:

d = (C-C) × sin 109.5/2

C-C bond length = 1.54 angstroms

tetrahedral bond angle = 109.5°

Then d = (1.54) × sin 109.5/2

d = 1.258

Thus ; the end to end separation is :

4098.227143 × 1.258 = 5155.57 Â

The answer is 5156 Â (since no decimal should be included )

6 0

2 years ago

The heat of vaporization (∆Hvap) of the element mercury (Hg) is 59.0 kJ/mol. If the vapor pressure of Hg is 0.0017 torr at 25°C,

astra-53 [7]

Answer:

Tb Hg = 656.726 K

Explanation:

normal boiling point (Tb):

Clasius-Clapeyron's law:

Tb = [(RLn(Po)/ΔHv) + (1/To)]∧(-1)

∴ R = 8.314 J/K.mol

∴To = 25°C ≅ 298 K

∴ Po = 0.0017 torr = 2.24 E-6 atm

∴ ΔHv = 59.0 KJ/mol = 59000 J/mol

⇒ Tb = [(8.314 J/K.mol)Ln(2.24 E-6))/(59000 J/mol)) + (1/298 K)]∧(-1)

⇒ Tb = [- 1.833 E-3 K-1 + 3.355 E-3 K-1 ]∧(-1)

⇒ Tb = [1.523 E-3]∧(-1)

⇒ Tb = 656.726 K

6 0

2 years ago

when a volume of a gas is changed from __ L to 4.50 L, the temperature will change from 38.1 C to 15.0C

kow [346]

Assuming that the change of volumen was done at constant pressure and the quantity of gas did not change, you use Charles' Law of gases, which is valid for ideal gases:

V / T = constant => V1 / T1 = V2 / T2 => V1 = [V2 / T2] * T1.

Now plug in the numbers ,where T1 and T2 have to be in absolute scale.

T1 = 38.1 + 273.15 K = 311.25K

T2 = 15.0 + 273.15 K = 288.15K

V1 = 4.5L * 311.25K / 288.15 K = 4.86L.

Answer: 4.86

3 0

2 years ago

What is the maximum number of grams of ammonia, nh3, which can be obtained from the reaction of 10.0 g of h2 and 80.0 g of n2? n

Lelechka [254]

Answer: The mass of ammonia produced is 28.22 g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)