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2 years ago

A gas made up of atoms escapes through a pinhole 2.04 times as fast as kr gas. write the chemical formula of the gas.

2 answers:

6 0

<span>The chemical formula for the unknown gas is Ne. Since we're looking for the rate at which a gas escapes through a small hole, we're dealing with effusion. For effusion, the rate is proportional to the velocity of the gas particles. Kinetic energy E = 0.5 mv^2 Since the kinetic energy of individual gas particles is the same if their temperatures are the same, we can create the following equality: 0.5 m1(v1)^2 = 0.5 m2(v2)^2 Double each side to make it simplier. m1(v1)^2 = m2(v2)^2 Divide both sides by m1 and by (v2)^2, giving (v1)^2/(v2^2) = m2/m1 And take the square root, giving (v1)/(v2) = sqrt(m2/m1) Now let's use the value 1 and the atomic weight of Kr for v1 and m1 1/(v2) = sqrt(m2/83.798) And for v2, we'll use the value 2.04 1/2.04 = sqrt(m2/83.798) Now solve for m2. 1/2.04 = sqrt(m2/83.798) 1/4.1616 = m2/83.798 83.798/4.1616 = m2 20.13600538 = m2 So the atomic weight of the unknown gas should be close to 20.136. Looking at a periodic table, I find that neon has an atomic weight of 20.18 which is quite close. Additionally, since neon is a noble gas, its gas particles consist of individual atoms. So the unknown gas is neon.</span>

GrogVix [38]2 years ago

6 0

Gas consisting of atoms exits through the eye of a needle 2.04 times faster than Kr gas. The chemical formula of the gas must be determined.

Rate1/Rate2 = square root M1/M2

$\frac{Rate1}{Rate2}$

<h2>Further explanation</h2>

Graham's law reveals the relationship between the level of effusion or diffusion and the molar mass of a gas. Diffusion describes the spread of gas throughout the second volume of gas, while effusion describes the movement of gas through a small hole into an open space.

Chemical Problems Graham Law

One way to apply Graham's law is to determine whether one gas will have an effect faster or slower than the others and to quantify the rate difference.

For example, if you want to compare the effusion rates of hydrogen gas (H 2) and oxygen gas (O 2), you use the molar mass of gas (2 for hydrogen and 32 for oxygen, which is the atomic mass multiplied by 2 because each molecule contains two-atom) and connect it in reverse:

rate H 2 / rate O 2 = 32 1/2 / 2 1/2 = 16 1/2 / 1 1/2 = 4/1

So, hydrogen gas molecules produce four times faster than oxygen molecules.

Another type of Graham legal problem might require you to find the molecular weight of a gas if you know the identity of one gas and the ratio between the effusion levels of two gases is known.

M 2 = M 1 Rate 1 2 / Rate 2 2

Learn more

Graham's Effusion Law brainly.com/question/3903505

Chemical Formula brainly.com/question/6951261

Details

Class: College

Subject: Physics

Keywords: Atoms, Chemical, Formula

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