Two identical containers, one red and one yellow, are inflated with different gases at the same volume and pressure. Both containers have an identically sized hole that allows the gas to leak out. It takes four times as long for the yellow container to leak out compared to the red container. If the red container is twice as hot as the yellow container, what is the ratio of the molar masses of the gases (Myellow / Mred)

Answers

Answer 1

Answer:

Answer:

Explanation:

Here we're dealing with the root mean square velocity of gases. We'll provide the formula in order to calculate the root mean square velocity of a gas:

$v_(rms)=\sqrt{(3RT)/(M)}$

Here:

$R = 8.314 (J)/(K mol)$ is the ideal gas law constant;

$T$ is the absolute temperature in K;

$M$ is the molar mass of a compound in kg/mol.

We know that the gas from the red container is 4 times faster, as it takes 4 times as long for the yellow container to leak out, this means:

$(v_(rms, red))/(v_(rms, yellow)) = 4$

We also know that the temperature of the red container is twice as large:

$(T_(red))/(T_(yellow)) = 2$

Write the ratio of the velocities and substitute the variables:

$(v_(rms, red))/(v_(rms, yellow))=\frac{\sqrt{(3RT_(red))/(M_(red))}}{\sqrt{(3RT_(yellow))/(M_(yellow))}}=4$

Then:

$\frac{\sqrt{(3RT_(red))/(M_(red))}}{\sqrt{(3RT_(yellow))/(M_(yellow))}}=\sqrt{(3RT_(red))/(M_(red))\cdot (M_(yellow))/(3RT_(yellow))}=\sqrt{(T_(red))/(T_(yellow))\cdot (M_(yellow))/(M_(red))}=4$

From here:

$16 = (T_(red))/(T_(yellow))\cdot (M_(yellow))/(M_(red))$

Then:

$(M_(yellow))/(M_(red)) = (16)/((T_(red))/(T_(yellow))) = (16)/(2) = 8$

Answer 2

Answer:

Final answer:

Considering Graham's Law of Effusion, and given that the temperature in the red container is twice that in yellow, the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.

Explanation:

The question is about comparing the molar masses of the gases based on the rate at which they escape or effuse from two different containers. The key to this problem lies in understanding Graham's Law of Effusion, which states that the rate at which a gas effuses is inversely proportional to the square root of its molar mass.

Firstly, note that it is given that the red container takes 1/4th the time as yellow to effuse completely, meaning the gas in the red container effuses 4 times faster than the gas in the yellow container. Hence, the ratio of rates of effusion is 4:1

It is also given that the temperature in the red container is twice that in the yellow. Given the gases are in the same volume and pressure, by Graham's law, the ratio of molar masses (Myellow / Mred) would be the square of the ratio of their effusion rates, however when different temperatures are considered, it's the square of [ratio of their effusion rates x (Tred / Tyellow)].

So the ratio of the molar mass of the yellow container to the red would be (4*22)2 = 16, implying that the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.

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Related Questions

calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass
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A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?

Answers

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

An element is to a __ as an organ is to a ___

Answers

An element is to an atomas an organ is to a cell. Just as atoms are the fundamental building blocks of elements, cells are the basic units of living organisms.

Elements are composed of atoms, each characterized by a specific number of protons, neutrons, and electrons.

Similarly, organs are composed of cells, each with specialized structures and functions that collectively contribute to the overall function of the organ.

The analogy highlights the hierarchical organization of matter and life, emphasizing how complex structures are formed from simpler components.

Just as elements combine to create diverse substances, cells come together to form intricate organs essential for life processes.

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Answer:

An element is to a compound as an organ is to tissue

Explanation:

What is the purpose of a catalyst?O A. To change the potential energy of the reactants
O B. To lower the activation energy of a reaction
O C. To increase the kinetic energy of the reactants
O D. To shift the equilibrium position of a reaction

Answers

B. To lower the activation energy of a reaction

Answer:

To lower the activation energy of a reaction

Explanation:

i just took the test and got it right ...... i hope this helps :)

Calculate the molality of a solution containing 22.75 g of glycerol (C3H8O3) in 79.6 g of ethanol (C2H5OH).

Answers

Answer:

3.11 mol/kg

Explanation:

Molality M = number of moles of solute, n/mass of solvent, m

To calculate the number of moles of glycerol (C₃H₈O₃) in 22.75 g of glycerol, we find its molar (molecular) mass, M',

So, M' = 3 × atomic mass of carbon + 8 × atomic mass hydrogen + 3 × atomic mass of oxygen

= 3 × 12 g/mol + 8 × 1 g/mol + 3 × 16 g/mol = 36 g/mol + 8 g/mol + 48 g/mol = 92 g/mol.

So, number of moles of glycerol, n = m'/M' where m' = mass of glycerol = 22.75 g and M' = molecular mass of glycerol = 92 g/mol

So, n = m'/M'

n = 22.75 g/92 g/mol

n = 0.247 mol

So, the molality of the solution M = n/m

Since m = mass of ethanol = 79.6 g = 0.0796 kg, substituting the value of n into the equation, we have

M = 0.247 mol/0.0796 kg

M = 3.11 mol/kg

So, the molality of the solution is 3.11 mol/kg.

A first-order decomposition reaction has a rate constant of 0.00140 yr−1. How long does it take for [reactant] to reach 12.5% of its original value? Be sure to report your answer to the correct number of significant figures.

Answers

Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.

Equation for the first-order decomposition reaction:-

$A_(t) =A_(0) e^(-kt)$ ....(1)

Here, $A_(t)$ is the final concentration, t is the time, $A_(0)$ is the initial concentration, and k is the rate constant.

Given:-

$A_(t) =0.125A_(0)$

k= $0.00140yr^(-1)$

Substitute the above value in equation (1) as follows:-

$0.125A_(0) =A_(0) e^(-kt) \n0.125A_(0) =A_(0) e^{-k*0.00140 yr^(-1) }\nln(0.125)/(-0.00140)=t\nt=504.87 year$

So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.

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Final answer:

The time required for a reactant to reach 12.5% of its original value in a first-order reaction is approximately 1482 years, obtained by applying the formula for the half-life of a first-order reaction and multiplying by 3.

Explanation:

In a first-order reaction, the half-life of the reaction, which is the time it takes for half of the reactant to be consumed, is independent of the concentration of the reactant. Also, for a first-order reaction, it would take approximately 3 half-lives for the reactant to be reduced to 12.5% of its original value. The Integrated Rate Law for a First-Order Reaction can be applied to determine the time it will take.

Given the rate constant (k) is 0.00140 yr¯¹, we will use the formula for the half-life of a first-order reaction: t₁/₂ = 0.693 / k. After calculating the half-life (t₁/₂), multiply it by 3 to determine the time for the reactant concentration to reach 12.5% of its original value. Hence, it would take approximately 1482 years to reach 12.5% of the original value when rounded to the correct number of significant figures.

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Atmospheric pressure arises due to the force exerted by the air above the Earth. At higher altitudes, the mass of the air above the Earth is _____ than at sea level, and atmospheric pressure therefore _____ with altitude.

Answers

Answer:

less, decreases

Explanation:

When the pressure of an atmosphere occurs because of the force exerted so at the time of the higher altitudes, the air mass i.e. above the earth should be less as the air is attracted towards surface of an earth because of the gravity and air contains the mass that shows near the surface area so automatically the air density reduced due to which the mass also decreased

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