Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.
Answer: 43 L
Answer:
PBr3 - Molecule , Polar
N2H2 - Molecule , (Polar in E- form and Non- polar in Z form)
C2H2 - Molecule , Non- Polar
N2 - Molecule , Polar
NCl3 - Molecule , Polar
SiF4 - Molecule , Non- Polar
NH3 - Molecule , Polar
F - Not- Molecule (atom)
H2 - Molecule and Non- Polar
Explanation:
Molecule : these are group of two or more atoms joined by strong force of attraction.
H2 is non- polar because it is homoatomic molecule.(made up of same element)
N2 is non- polar because it is homoatomic molecule.
Answer:
A. More mass
C. Shorter distance between them
Explanation:
The two characteristics of a body experiencing greater gravitational force are that they have mass and a shorter distance between them.
This is conformity with Newton's law of universal gravitation.
The law states that "every object attracts one another with a force that is directly proportional to their masses and inversely proportional to the square of the distance between them".
This law implies that the more the mass of two bodies, the more the gravitational force of attraction. And that the shorter the square of the distance between them, the more the attraction.
Answer:
(a) 
(b) 
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

Thus, since the final pressure is 3.60 bar, we can write:

The moles of helium could be computed via solver as:

Or algebraically:

In such a way, the volume of the compartment B is:

Finally, he mole fraction of He is:

Regards.