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How much faster does helium escape through a porous container than ozone?

2 months ago

Solution 1

Guest #4800

2 months ago

According to Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".

r₁ / r₂ =   $\sqrt{M2 / M1}$ ---- (1)

r₁ = Rate of effusion of He

r₂ = Rate of Effusion of O₃

M₁ = Molecular Mass of He = 4 g/mol

M₂ = Molecular Mass of O₃ = 48 g/mol

Putting values in eq. 1,

r₁ / r₂ =   $\sqrt{48 / 4}$

r₁ / r₂ =   $\sqrt{12}$

r₁ / r₂ = 3.46

Result:
Therefore, Helium will effuse 3.46 times more faster than Ozone.

📚 Related Questions

Question

Given the reaction: cu(s) + 4hno3(aq) → cu(no3)2(aq) + 2no3(g) + 2h2o(l )as the reaction occurs, what happens to copper?

Solution 1

From the reaction between Cu and HNO₃, the formed gas is NO₂ instead of NO₃. Hence the correct balanced equation would be,
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Here, Cu goes to Cu(NO₃)₂ by changing its oxidation number from 0 to +2 while NO₃⁻ goes to NO₂ by reducing its oxidation state from +5 to +4 . Hence Cu is oxidized by HNO₃ in the reaction.

Solution 2

When copper reacts with nitric acid, then cooper nitrate, water, and nitrogen dioxide are produced. In this reaction, copper gets oxidized by nitric acid.

What is oxidation?

Oxidation is a type of chemical reaction involving the sharing of electrons and an increase or decrease of the oxidation number. When a chemical species loses an electron, it is said to be oxidized.

The balanced chemical reaction is given as:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

In the reaction, copper forms copper nitrate, and its oxidation changes from 0 to +2. On the other hand, nitrate species get reduced to nitrogen dioxide and change the state from +5 to +4.

Therefore, copper gets oxidized by nitric acid.

Learn more about oxidization here:

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Question

A sample of ammonia (nh3) gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. if the total pressure is 853 mmhg after the reaction, calculate the partial pressures of n2 and h2. enter your answers in scientific notation.

Solution 1

The partial pressure of N2 and H2 is calculated as below

write the reaction for decomposition of NH3

NH3 =N2 +3 H2

from the reaction above 1 mole of N2 and 3 moles of H2 is formed and therefore the total mole of the product = 4

the reaction coefficient for N2 from the reaction = 1/4
and that of H2 = 3/4

the final pressure = partial pressure of N2 +partial pressure of H2

partial pressure of N2 = 1/4 x853 mm hg = 213.25 mmhg
partial pressure of H2 = 3/4 x853mm hg=639.75 mm hg

Question

A 3.00-l flask is filled with gaseous ammonia, nh3. the gas pressure measured at 15.0 ∘c is 2.15 atm . assuming ideal gas behavior, how many grams of ammonia are in the flask? express your answer to three significant figures and include the appropriate units.

Solution 1

We can find the mass of ammonia using the ideal gas law equation,
PV = nRT
where
P - pressure - 2.15 atm x 101 325 = 2.18 x 10⁵
V - volume - 3.00 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 15.0 °C + 273 = 288 K
substituting these values in the equation
2.18 x 10⁵ Pa x 3.00 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.273 mol
number of moles of NH₃ is 0.273 mol
molar mass of NH₃ - 17.0 g/mol
mass pf ammonia present - 0.273 mol x 17.0 g/mol = 4.64 g
mass of NH₃ present is 4.64 g

Question

An aqueous nacl solution is made using 112 g of nacl diluted to a total solution volume of 1.00 l. calculate the molarity, molality, and mass percent of the solution. (assume a density of 1.08 g>m

Solution 1

The molarity of the solution is 1.9 molL-1, the molality of the solution is 1.96 m. The mass percent of the solution is 10.3%.

The molarity of the solution is obtained from; Number of moles /Volume of solution

Number of moles = Mass/molar mass = 112 g/58.5 g/mol = 1.9 moles

Molarity = 1.9 moles/1.00 L = 1.9 molL-1.

The molality of solution = Number of moles/Mass of solvent in Kg

Density of solution = Mass/volume

Mass of solution = Density × volume = 1.08 g/mL × 1000 mL = 1080 g or 1.08 Kg

Mass of solvent = 1080 g - 112 g = 968 g or 0.968 Kg

Molality = 1.9 moles/0.968 Kg = 1.96 m

Mass percent of the solution = Mass of solute/Mass of solution × 100

= 112 g/1080 g × 100 = 10.3%

Learn more: brainly.com/question/4699407

Solution 2

Q1)
molarity is defined as the number of moles of solute in 1 L of solution.
the NaCl solution volume is 1.00 L
number of moles NaCl = NaCl mass present / molar mass of NaCl
NaCl moles = 112 g / 58.5 g/mol = 1.91 mol
the number of moles of NaCl in 1.00 L of solution is - 1.91 mol
therefore molarity of NaCl is 1.91 M

Q2)
molality is defined as the number of moles of solute in 1 kg of solvent.
density is mass per volume.
density of the solution is 1.08 g/mL.
therefore mass of the solution is = density x volume
mass = 1.08 g/mL x 1000 mL = 1080 g
since we have to find the moles in 1 kg of solvent
mass of solvent = 1080 g - 112 g = 968 g
number of moles of NaCl in 968 g of solvent - 1.91 mol
therefore number of NaCl moles in 1000 g - (1.91 mol / 968 g) x 1000 g/kg = 1.97 mol/kg
molality of NaCl solution is 1.97 mol/kg

Q3)
mass percentage is the percentage of mass of solute by total mass of the solution
mass percentage of solution = mass of solute / total mass of the solution
mass of solute = 112 g
total mass of solution = 1080 g
mass % of NaCl = 112 g / 1080 g x 100%
therefore mass % of NaCl = 10.4 %
answer is 10.4 %

Question

A glucose solution contains 55.8 g of glucose (c6h12o6) in 455 g of water. calculate the freezing point and boiling point of the solution. ( density of water = 1.00 g/ml, kb= 0.512 o c kg solvent/mol solute and 1.86°c kg/mol)

Solution 1

The boiling point of the solution is 100.35 °c.

We know that;

ΔT = K m i

ΔT = freezing point depression

K = freezing constant

m = molality

i = Van't Hoff factor

Hence;

ΔT = 1.86°c kg/mol × 55.8 g/180 g/mol × 1/0.455 × 1

ΔT = 1.27 °c

Freezing point = 0 - 1.27 °c = - 1.27 °c

For boiling point;

ΔT = K m i

ΔT = 0.512 o c kg × 55.8 g/180 g/mol × 1/0.455 × 1

ΔT = 0.35 °c

Boiling point = 100 + 0.35 °c = 100.35 °c

Learn more about freezing point: brainly.com/question/2292439

Solution 2

Answer is: the freezing point of the solution of glucose is -1.26°C and boiling point is 100.353°C.
m(H₂O) = 455 g ÷ 1000 g/kg = 0.455 kg.
m(C₆H₁₂O₆) = 55.8 g.

n(C₆H₁₂O₆) = m(C₆H₁₂O₆)÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 55.8 g ÷ 180.16 g/mol.
n(C₆H₁₂O₆) = 0.31 mol.
b(solution) = n(C₆H₁₂O₆) ÷ m(H₂O).
b(solution) = 0.31 mol ÷ 0.455 kg.
b(solution) = 0.68 mol/kg.
ΔTf = b(solution) · Kf(H₂O).
ΔTf = 0.68 mol/kg · 1.86°C·kg/mol.

ΔTf = 1.26°C.
Tf = 0°C - 1.26°C = -1.26°C.

ΔTb = b(solution) · Kb(H₂O).

ΔTb = 0.68 mol/kg · 0.52°C·kg/mol.

ΔTb = 0.353°C.

Tb = 100°C + 0.353°C.

Question

Write and balance a net ionic equation for the reaction between iron(ii) chloride and potassium hydroxide to form iron(ii) hydroxide and potassium chloride.

Solution 1

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of iron (II) chloride and potassium hydroxide is given as:

$FeCl_2(aq.)+2KOH(aq.)\rightarrow Fe(OH)_2(s)+2KCl(aq.)$

Ionic form of the above equation follows:

$Fe^{2+}(aq.)+2Cl^-(aq.)+2K^+(aq.)+2OH^-(aq.)\rightarrow Fe(OH)_2(s)+2K^+(aq.)+2OH^-(aq.)$

As, potassium and hydroxide ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Fe^{2+}(aq.)+2OH^-(aq.)\rightarrow Fe(OH)_2(s)$

Hence, the net ionic equation is written above.

Question

Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0 g of fluorine gas, the maximum mass of hydrogen fluoride that could be produced is

Solution 1

Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol

5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2

H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol

We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.

H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed

Question

What is the molarity of 555 l of a ba(oh)2 solution if the ph is 10.20? 1. 2.26 × 10−5 m 2. 6.31 × 10−11 m 3. 5.15 × 10−7 m 4. 3.15 × 10−11 m 5. 3.14 × 10−4 m 6. 1.58 × 10−4 m 7. 7.92 × 10−5 m 8. 4.40 × 10−2 m?

Solution 1

First, you need to get POH from the value of PH:

when POH = 14 - PH

= 14 - 10.2

= 3.8
then we are going to get the value of [OH] from the POH value:

POH = -㏒[OH-]

3.8 = - ㏒ [OH-]

∴[OH-] = 1.58 x 10^-4

then, we will get the moles of ba(OH)2 = (1.58 x 10^-4) / 2

= 0.0000792 moles

∴ the molarity of Ba(OH)2 = 7.92 x 10^-5

Solution 2

Answer:

7. 7.92 × 10−5

Explanation:

Hello,

In this case, with the given pH, one could find the pOH:

$pOH=14-pH=14-10.20=3.8$

Thus, since barium hydroxide is completely dissolved in water based on:

$Ba(OH)_2\rightarrow Ba^{+2}+2OH^-$

The concentration of hydroxyl ions is twice to that of the hydroxide (2:1 mole relationship). Therefore, by considering the relationship between the pOH and the concentration of hydroxyl we have:

$pOH=-log([OH]^-)\\$

$[OH]^-=10^{-3.8}=1.58x10^{-4}M$

Finally, given the 1:2 mole ratio of barium hydroxide to hydroxyl ions, the concentration of barium hydroxide results:

$[Ba(OH)_2]=2*[OH^-]=\frac{1}{2} *1.58x10^{-4}M$

$[Ba(OH)_2]=7.92x10^{-5}M$

Thus, the answer is 7. 7.92 × 10−5.

Regards.

Question

The reaction 4 a + c + h → d has the mechanism below. what is the rate law? 1) 2 a → b fast 2) 2 b → b2 fast 3) b2 + c → g slow 4) g + h → d fast a. rate = k[a]2[c] b. rate = k[a]4[h][c] c. rate = k[a]2 d. rate = k[b]2[c] e. rate = k[a]4[c]

Solution 1

The reaction 4 a + c + h → d has the mechanism below , rate = k[b]2[c] ,therefore option d is correct .

What is the rate law ?

Rate law is a measurement which helps scientists understand the kinetics of a reaction, or the energy, speed, and mechanisms of a reaction.

The rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentrations of the reactants participating in it.

It involves the concentration of reactants, not product because we generally consider the reverse reaction to be unimportant in reversible reactions.

The reaction 4 a + c + h → d has the mechanism below , rate = k[b]2[c] ,hence option d is correct .

Learn more about rate law ,here:

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Solution 2

The rate of the reaction defines by the slowest stage. So the rate of the reaction is d. rate = k[b]^2[c].

Question

When two molecules of methanol (ch3oh) react with oxygen, they combine with three o2 molecules to form two co2 molecules and four h2o molecules. how many h2o molecules are formed when 66 methanol molecules react?

Solution 1

The number of water molecules formed when 66 molecules of ethanol react is; 132 molecules of water.

When two molecules of methanol react with oxygen.

They combine with three O2 molecules as implied in the question to form CO2 molecules and four H20 molecules.

The reaction between methanol and oxygen is as follows;

2CH3OH + 3O2 --> 2CO2 + 4H2O

According to the equation,

2 molecules of CH3OH = 4 molecules of H2O

Therefore,

66 molecules of CH3OH = x molecules of H2O

In essence, the number of x molecules of water formed when 66 molecules of ethanol react is;

x = (66 × 4)/2

x = 132 molecules of water.

The number of water molecules formed when 66 molecules of ethanol react is; 132 molecules of water.

Read more:

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Solution 2

$2CH_{3}OH +3O_{2}--\ \textgreater \ 2CO_{2}+4H_{2}O

From reaction 2 molecules CH_{3}OH 
 
give 4 molecules H_{2}O.

 

$

$\frac{2 molecules CH_{3}OH}{4 molecules H_{2}O} = \frac{66 molecules CH_{3}OH}{x molecules H_{2}O} 

x= 4*66/2=132 molecules of H_{2}O$

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