A mixture of N2, O2 and He have mole fractions of 0.25, 0.65, and 0.10,

A mixture of N2, O2 and He have mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pressure of the mixture is 3.9 atm?

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Solution 1

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The pressure of nitrogen in the mixture has been 0.975 atm.

Partial pressure has been defined as the pressure exerted by the gas molecules in the mixture. The partial pressure (P_A)  has been expressed as:


Where,  Mole fraction of the element A, X_{N_2}=0.25\;\rm atm

The total pressure of the mixture, P=3.9\;\rm atm

Substituting the values, the pressure of nitrogen (P_{\rm N_2}),

P_{\rm N_2}=0.25\;\times\;3.9\;\rm atm\\ \textit P_{N_2}=0.975\;atm

The pressure of nitrogen in the mixture has been 0.975 atm.

For more information about mole fraction, refer to the link:


Solution 2

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Partia pressure N₂ → 0.975 atm


Let's analyse the moles fractions:

N₂ → 0.25

O₂ → 0.65

He → 0.1

Partial pressure / Total pressure = Mole fraction

Partial pressure N₂ / 3.9 atm = Mole fraction N₂

Partial pressure N₂ / 3.9 atm = 0.25

Partial pressure N₂ = 3.9 atm . 0.25 → 0.975 atm

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Solution 1


The answer is (B) A gamma ray alone


Technetium-99m decays through a process called isomeric transition involving the decay of 99mTc to 99TC via the release of gamma rays and low energy electron

Many classic experiments have given us indirect evidence of the nature of the atom. Which of the experiments listed below did not give the results described? A. The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct. B. The Rutherford experiment was useful in determining the nuclear charge on the atom. C. Millikan's oil-drop experiment showed that the charge on any particle was a simple multiple of the charge on the electron. D. The electric discharge tube proved that electrons have a negative charge. E. All of the above experiments gave the results described.
Solution 1


Option A, The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct.


Thomson's plum pudding model:

Plum pudding model was proposed by J.J Thomson. In Thomson's model, atoms are proposed as sea of positively charge in which electrons are distributed through out.

Result of Rutherford experiment:

As per Rutherford's experiment:

Most of the space inside the atom is empty.

Positively charge of the atom are concentrated in the centre of the atom known as nucleus.

Electrons are present outside the nucleus and revolve around it.

As it is clear that, result of Rutherford experiment did not supported the Thomson model.

For action potential generation and propagation, are there any other cation channels that could substitute for the voltage-gated sodium channels if the sodium channels were blocked?
Solution 1


Yes, calcium ions.


Voltage-gated ions channels are ions channels that were formed by some certain class of transmembrane proteins. They are referred to as the permitting and blocking passage found in cell membrane. they are activated by electrical stimulus.

Their major function or responsibility is that , they generate or produce electrical signal in the cell membrane.

The empirical formula of styrene is CH; the molar mass of styrene is 104.14 g/mol. How many H atoms are present in a 7.80 g sample of styrene?
Solution 1


In 7.80 g of styrene, we have 3.60×10²³ atoms of H


Empirical formula of styrene is CH

Molecular formula of styrene is C₈H₈

So, 1 mol of styrene has 8 moles of C and 8 moles of H and 1 mol weighs 104.14 grams. Let's make a rule of three:

104.14 g (1 mol of C₈H₈) have 8 moles of H

Then 7.80 g would have ( 7.80  .8) / 104.14 = 0.599 moles

As we know, 1 mol of anything has NA particles (Avogadro's Number,  6.02×10²³), so 0.599 moles will have (mol . NA) particles

0.599 mol . 6.02×10²³ atoms / 1 mol = 3.60×10²³ atoms

Why is pure orange juice a mixture?
Solution 1
B Because the composition of the solution is uniform throughout, it is a homogeneous mixture. AOrange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure. B Because its composition is not uniform throughout, orange juice is a heterogeneousmixture.
Physical properties such as melting point, boiling point, and solubility are all dependent on the type of interparticle forces a substance experiences. Identify the type of interparticle force that has the greatest influence on the physical properties for each substance. Cl2, NH2OH, PCl3, CH4, CaCl2, KI Two of the substances in Part 1 are ionic. Which factors will result in a stronger ionic bond overall? 1. larger ions2. greater absolute charges3. similarity of ionic sizes4. smaller ions
Solution 1

1) The type of interparticle force that has the greatest influence on the physical properties for each substance is;

CaCl2 and KI; ionic bonding

CaCl2 and KI; ionic bondingPCl3 and CH4; dispersion forces

CaCl2 and KI; ionic bondingPCl3 and CH4; dispersion forcesNH2OH; hydrogen bonding

2) The factor that will result in a stronger ionic bond overall is; Smaller ions

1) We want to Identify the type of interparticle force that has the greatest influence on the physical properties for each substance. Let us look at the options;

  • For CaCl2 and KI, the elements Calcium and Potassium that form the chloride are metals and as such they form ionic bonds with the halogens Chlorine and Iodine. Thus, ionic bonding has the greatest force here.

  • For PCl3 and CH4, they possess forces that are dispersive and as such their structural arrangement distribute electrons to any surrounding atoms and molecules. Thus, the forces of dispersion have the greatest force here.

  • For NH2OH, as a result of the presence of Hydrogen (H) and Oxygen(O) atoms, the bond present in it will be a hydrogen bond. Thus hydrogen bond has the greatest force here.

2) Bond strength, depends largely on the charges present on each ion as well as the distance between each ion. Thus, it means ions that are Small and highly charged will form stronger bonds while ions that are large, minimally charged will form bonds that are weak.

Thus, smaller ions results in stronger ionic bond overall.

Read more at; brainly.com/question/23530058

Solution 2

Answer: CaCl2 and KI- ionic

PCl3 and CH4-dispersion forces

NH2OH- hydrogen bonding

Part B

Smaller ions.


The smaller the ions the easier it is for them to be packed into a well ordered crystal lattice. The strength of ionic interaction always depend on the relative ion sizes. The smaller the ions the stronger the bond. Large ions do not easily pack into crystal lattices.

Ammonia can be prepared by the reaction of magnesium nitride with water. The products are ammonia and magnesium hydroxide. When the equation is written and balanced, the coefficient of water is 1. 3 2. 2. 3. 6. 4. 1.
Solution 1


Mg₃N₂ + 6H₂O  →  2NH₃  +  3Mg(OH)₂

Coefficient of water is 6 (option 3)


The reaction is:

Mg₃N₂ + H₂O  →  NH₃  +  Mg(OH)₂

Let's balance the reaction.

In reactant side we have 3 Mg, therefore in product side, we add 3 Mg to the hydroxide.

This change, modified the hydroxide, so now we have 6 O and 6 H, but we have in total 9 H (6 from the hydroxide + 3 from the ammonia)

As we have 2N, in reactant side, we must add 2 N to the ammonia, so now

we have 12 H in product side . We must complete with 6, the water so the H are ballanced.

In reactant side we have 6 O, therefore in product we must have 6 O (two O, in the OH but we have 3 moles, so in total we have 6 O) - BALANCED

The balance reaction is:

Mg₃N₂ + 6H₂O  →  2NH₃  +  3Mg(OH)₂

If He gas has an average kinetic energy of 4310 J/mol under certain conditions, what is the root mean square speed of O2 gas molecules under the same conditions? (given that He and O_2 gas are at the same temperature what can you conclude about 0_2's average kinetic energy?) variable equations used to find the solution are greatly appreciated!!!
Solution 1


The root mean square speed of O2 gas molecules is

519.01 m/s


The root mean square velocity  :



K.E =\frac{1}{2}mv_{rms}^{2}

Molar mass , M

For He = 4 g/mol

For O2 = 2 x 16 = 32 g/mol

O2 = 32/1000 = 0.032 Kg/mol

First calculate the temperature at which the K.E of He is 4310J/mol

K.E of He =



K.E of He = 4310 J/mol



Now , Use Vrms to calculate the velocity of O2





To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.4-L bulb, then filled it with the gas at 1.00 atm and 22.0 ∘C and weighed it again. The difference in mass was 5.1 g . Identify the gas.
Solution 1

Answer : The diatomic gas is nitrogen gas, N₂.

Explanation :

First we have to calculate the moles of gas.

Using ideal gas equation:



P = Pressure of gas = 1.00 atm

V = Volume of gas = 4.4 L

n = number of moles of gas = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 22.0^oC=273+22.0=295.0K

Putting values in above equation, we get:

1.00atm\times 4.4L=n\times (0.0821L.atm/mol.K)\times 295.0K


Now we have to calculate the molar mass of gas.

\text{Molar mass of gas}=\frac{\text{Given mass of gas}}{\text{Moles of gas}}

\text{Molar mass of gas}=\frac{5.1g}{0.1817mol}=28.07g/mol

As we are given that the gas is diatomic X₂.

As, 2 atoms of gas X has mass = 28.07 g/mol

So, 1 atom of gas will have mass = \frac{28.07}{2}=14.04g/mol

From this we conclude that the nitrogen atom has mass of 14.04 g/mol.

Thus, the diatomic gas is nitrogen gas, N₂.

Cosmic rays are:_______.A) high energy radiation produced by the sun.B) high energy radiation produced in the ozone layer.C) high energy radiation produced by the earth's core.D) none of these
Solution 1


Option (A)


Cosmic radiations are usually defined as a type of radiations that is comprised of high-energy photons and carry harmful ultraviolet radiation from the sun. These are emitted from the sun at a speed that is equivalent to the speed of the light.

When these radiations are incident on earth, it interacts with the upper atmosphere, resulting in the emission of charged particles such as pions, which undergoes decay and releases other smaller particles, commonly known as muons.

Thus, the correct answer is option (A).