I need help, I don't understand the question

Chemistry

1 answer:

Oduvanchick [21]3 years ago

3 0

Remember this.

Ionic molecules has ionic bonds
Nonpolar molecules has dispersion (Van del Waals)
Polar molecules could either have hydrogen bonding or Dipole-Dipole. Hydrogen bonding is when you have F, O or N with H, every other polar molecule is dipole-dipole.

a. polar- dipole-dipole
b. polar- hydrogen bonding
c. nonpolar- dispersion
d. nonpolar- dispersion
e. polar- dipole-dipole
f. polar-dipole-dipole
g. nonpolar- dispersion
h. polar- hydrogen bonding.

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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig

kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

$v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s$

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

$K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J$

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

$\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K$

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, $\Delta S_{env} = -1.18 J/K$