Why are Noble gases inert?

Two positively charged objects (N pole) are separated by a large distance. One of the positively charged objects (N pole) is rep

Eddi Din [679]

Answer:

B

Explanation:

Applying law of electrostatic which states that like charges repel each other and unlike charges attract each other

N and S are unlike charges that turn and make the former repulsive force (due to two like charges N and N)to reduce and attractive force between N and S to increase

6 0

3 years ago

What is the acronym for ribonucleic acid?

Masteriza [31]

The acronym is RNA. Ribo Nucleic Acid

4 0

3 years ago

H2 H2O H2O2 CH4 All four examples represent

miv72 [106K]

It is covalent bonds

8 0

3 years ago

Read 2 more answers

Ammonia can be produced via the chemicalreaction

otez555 [7]

Answer:

3)The reaction is not at equilibrium and willproceed to the right.

Explanation:

The reaction quotient of an equilibrium reaction measures relative amounts of the products and the reactants present during the course of the reaction at particular point in the time.

It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.

Q < Kc , reaction will proceed in forward direction.

Q > Kc , reaction will proceed in backward direction.

Q = Kc , reaction at equilibrium.

Given that:

Q = $3.56\times 10^{-4}$

K = $6.02\times 10^{-2}$

Since, Q < K , reaction is not at equilibrium and will proceed to right, in forward direction.

3 0

3 years ago

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, $\Delta S_{env} = -1.18 J/K$

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.

Learn more: brainly.com/question/22655760

6 0

2 years ago