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vladimir1956 [14]
2 years ago
10

A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon,

and helium were 44.0, 486, and 218 torr, respectively. The partial pressure of neon in the vessel was __________ torr.
Chemistry
1 answer:
xz_007 [3.2K]2 years ago
4 0

Answer:

The partial pressure of neon in the vessel was 239 torr.

Explanation:

In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.

Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:

PT= P1 + P2 + P3 + P4…+ Pn

where n is the amount of gases present in the mixture.

In this case:

PT=PN₂ + PAr + PHe + PNe

where:

  • PT= 987 torr
  • PN₂= 44 torr
  • PAr= 486 torr
  • PHe= 218 torr
  • PNe= ?

Replacing:

987 torr= 44 torr + 486 torr + 218 torr + PNe

Solving:

987 torr= 748 torr + PNe

PNe= 987 torr - 748 torr

PNe= 239 torr

<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>

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Answer:

  • Molality = 0.066 m
  • Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The <em>freezing-point depression</em> is expressed by:

  • ΔT=Kf * m

We put the data given by the problem and <u>solve for m</u>:

  • 2.63 °C = 40°C·kg/mol * m
  • m = 0.06575 m

For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:

  • 0.06575 m = Moles reserpine / kg camphor
  • 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We<u> calculate moles of reserpine:</u>

  • 0.06575 m = Moles reserpine / 0.025 kg camphor
  • Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:

  • 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>

8 0
2 years ago
According to the bohr model of the atom, the single electron in what motion of a hydrogen atom circles the nucleus
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It would be elliptical according to him.....
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A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr
s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

P=p_{N_2}+p_{O_2}+p_{Ar}

p_{N_2}=P\times \chi_{N_2}

p_{O_2}=P\times \chi_{O_2}

p_{Ar}=P\times \chi_{Ar}

where,

P = total pressure = 3.9 bar

p_{N_2} = partial pressure of nitrogen gas  

p_{O_2} = partial pressure of oxygen gas  

p_{Ar} = partial pressure of argon gases  

\chi_{N_2} = Mole fraction of nitrogen gas  = 0.25

\chi_{O_2} = Mole fraction of oxygen gas  = 0.65

\chi_{Ar} = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar

Partial pressure of oxygen gas :

p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar

Partial pressure of argon gas :

p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar

7 0
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Hello user, you did not add a picture of an attachment to enable me help you solve this problem. Please check the attachment I added i believe it is the question that needs to be solved.

Explanation:

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b. Add drink mix solution to a diluted mixture of drink mix in pure water

c. Add drink mix solid to pure drink mix solution

d Evaporate water from a diluted mixture of drink mix in pure water

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a. Add water to a diluted mixture of drink mix in pure water

b. Add water to pure drink mix solution

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The standard curve was made by spectrophotographic analysis of equilibrated iron(III) thiocyanate solutions of known concentrati
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Answer:

Molar concentration of the Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.

Explanation:

When you make a calibration curve in a spectrophotographic analysis you are applying the Lambert-Beer law that states the concentration of a compound is directely proportional to its absorbance:

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<em>Where A is absorbance, E is molar absorption coefficient, l is optical path length and C is molar concentration</em>

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Using the equation of the line you obtain:

y = 4541.6X + 0.0461

<em>Where Y is absorbance and X is concentration -We will assume concentration is given in molarity-</em>

As absorbance of the unknown is 0.410:

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<h3>Molar concentration of the Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.</h3>

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6 0
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