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If two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is

Fed [463]

If the gases are at the same temperature and pressure, the ratio of their effusion rates is directly proportional to the ratio of the square roots of their molar masses:

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

Where

R₁ and R₂ are the rates of the two gas
M₁ and M₂ are the molar masses of the two gas

From the Graham's law equation, we can see that the ratio of the rates of effusion of the two gases is directly proportional to the square root of their molar masses

Learn more about Graham's law of diffusion:

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3 0

2 years ago

When ignited, solid ammonium dichromate decomposes in a fiery display. This is the reaction for a "volcano" demonstration. The d

Lerok [7]

Answer:

Your question is half unfinished, regarding the chromium III oxide the correct option that expresses the inorganic formula of said compound is "A"

Explanation:

In the reaction an initial salt reacts giving as product water vapor, nitrogen gas and an oxide that is chromium oxide.

Chromium oxide is an oxide that adopts the structure of corundum, compact hexagonal. It consists of an anion oxide matrix with 2/3 of the octahedral holes occupied by chromium. Like corundum, Cr2O3 is a tough, brittle material.

It is used as a pigment, green in color.

4 0

3 years ago

How have batteries gotten better over time

bekas [8.4K]

Answer:

Yes they have gotten better overtime

Explanation:

6 0

3 years ago

Read 2 more answers

Describe how thin layer chromatography is used in the isolation and extraction of lipids

BabaBlast [244]

Thin layer chromatography(TLC) works with the principle of separation through adsorption.

It is used in the isolation and extraction of lipids through the following steps:

apply the lipid samples spots in the bottom of the plate.
also apply sample solution to the marked spot
pour the mobile phase into the TLC chamber and use a moist filter paper to cover it. this is done to maintain equal humidity.
then place the plate in the TLC chamber and close it with a lid.
the plate is immersed into the solvent (mobile phase) for its development. this is done, keeping in mind that the sample spot should be above the solvent.
once the sample spots are developed, they are removed and dried.
this is later viewed using the UV light chamber to see the isolation of the lipid sample.

Learn more here:

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4 0

3 years ago

Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxid

kiruha [24]

Answer:

-471 Kj/mole acrylic acid

Explanation:

THIS IS THE COMPLETE QUESTION BELOW;

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?

The two equations from the reaction can be written as;

a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)

Δ H= -414Kj ........................ equation (a)

b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)

In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water

Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.

6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)

Δ H= -2484Kj.................. equation (c)

6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj

Then add equation (c) and equation(b) then we have

6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj

ΔH(net)= -2352Kj/5moles

=-471Kj/mole

therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid

6 0

2 years ago