Decide which element probably forms a compound with chlorine that has a chemical formula most and least similar to the chemical formula of the compound formed by chlorine and rubidium. Comparing chemical formula of compound formed with chlorine: phosphorus beryllium magnesium lithium most similar to rubidium least similar to rubidium

Most similar - Lithium

Least similar - Phosphorus

Explanation:

Rubidium is an element in group 1A of the periodic table. It is a metal and forms an ionic compound with chlorine. The formula of the compound is RbCl.

If we look at the options, Lithium is also a group 1A element and forms an ionic compound with chlorine having the formula LiCl which is very much similar to RbCl chemically.

Phosphorus is a nonmetal. Its compounds with chlorine, PCl3 and PCl5 are covalent and does not resemble RbCl in any way.

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.