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A gas-tight syringe is filled to 45.0 mL, at a pressure of 2.33 atm and 25.0°C, with a gaseous diatomic element. If the mass of the syringe increases by 0.137 g when filled, which diatomic element is in the syringe?

a. H2
b. Oxygen
c. F2
d. No right answer.

Respuesta :

joseaaronlara

Answer:

The answer to your question is: d. No right answer

Explanation:

Data

V = 45 ml = 0.045 l

P = 2.33 atm

T = 25 °C = 298°K

m = 0.137 g

R = 0.082 atm l / mol°K

Formula

                   PV = nRT

Process

                 n = PV / RT

                 n = (2.33)(0.045) / (0.082)(298)

                 n = 0.1049 / 24.43

                 n = 0.0043

a) For hydrogen

                           1 g --------------- 1 mol

                           0.137 g --------  x

                          x = 0.137 mol of H2   The gas is not hydrogen

b) Oxygen

                          16 g of O2 ------------ 1 mol

                          0.137 g      ------------  x

                          x = (0.137 x 1) / 16

                          x = 0.0086 mol of O2  The gas is not Oxygen

c) Fluor

                         19 g of F --------------  1 mol

                         0.137 g    --------------   x

                         x = (0.137 x 1) / 19

                         x = 0.0072 mol of F      The gas is not Fluor

         

CintiaSalazar

Answer:

The diatomic element in the syringe is Oâ‚‚.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V=n*R*T

Taking into account the following values:

  • P=2.33 atm
  • V= 45 mL= 0.045 L (1 L= 1,000 mL)
  • R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]
  • T= 25 °C= 298 °K

It is possible to calculate the amount of moles of gas present in the syringe as:

[tex]n=\frac{P*V}{R*T}[/tex]

[tex]n=\frac{2.33 atm*0.045 L}{0.082 \frac{atm*L}{mol*K}298 k }[/tex]

n= 0.0043 moles.

The mass of the syringe increases by 0.137 g when filled. This indicates that the 0.0043 moles calculated previously contain a mass of 0.137 g. Then it must be verified for each diatomic gas presented in the options that the 0.0043 moles contain said amount of mass. For this, the molar mass of each compound is known:

  • Hâ‚‚: 2 g/mol
  • Oâ‚‚: 32 g/mol
  • Fâ‚‚: 38 g/mol

Now a rule of three applies for each case as follows:

  • if 1 mole of Hâ‚‚ contains 2 g of the gas, 0.0043 moles of Hâ‚‚, how much mass will it contain?

[tex]mass=\frac{0.0043 moles*2 g}{1 mole}[/tex]

mass= 0.0086 g

  • if 1 mole of Oâ‚‚ contains 32 g of the gas, 0.0043 moles of Oâ‚‚, how much mass will it contain?

[tex]mass=\frac{0.0043 moles*32 g}{1 mole}[/tex]

mass= 0.137 g

  • if 1 mole of Fâ‚‚ contains 38 g of the gas, 0.0043 moles of Fâ‚‚, how much mass will it contain?

[tex]mass=\frac{0.0043moles*38 g}{1 mole}[/tex]

mass= 0.163 g

0.0043 moles contain a mass of 0.137 g in the case of Oâ‚‚

The diatomic element in the syringe is Oâ‚‚.

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