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gregori [183]
2 years ago
15

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

Chemistry
1 answer:
lakkis [162]2 years ago
3 0

Using ideal gas equation,  

P\times V=n\times R\times T

Here,  

P denotes pressure  

V denotes volume  

n denotes number of moles of gas  

R denotes gas constant  

T denotes temperature  

The values at STP will be:  

P=1 atm  

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}

Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}

Putting all the values in the above equation,

1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15

Molar mass of the gas=46.15

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Answer:

21.86582KJ

Explanation:

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Applying the equation if a straight line, y=mx +c, and comparing the given equation with the graphical form of the Arrhenius equation shown in the image attached, we obtain the activation energy of the reaction as shown.

5 0
2 years ago
A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of
allochka39001 [22]

Answer:

The answer to your question is   C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

                       44 g of CO₂ ---------------  12 g of C

                        52.8 g          ---------------  x

                        x = (52.8 x 12)/44

                        x = 633.6/44

                        x = 14.4 g of C

                        12 g of C ------------------  1 mol

                        14.4 g of C ---------------   x

                         x = (14.4 x 1)/(12)

                         x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

                         18 g of H₂O ---------------  2 g of H

                         21.6 g of H₂O -------------  x

                          x = (21.6 x 2) / 18

                         x = 2.4 g of H

                         1 g of H -------------------- 1 mol of H

                         2.4 g of H -----------------  x

                          x = (2.4 x 1)/1

                          x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

                           = 6.4 g

                         16 g of O ----------------  1 mol

                          6.4 g of O --------------  x

                          x = (6.4 x 1)/16

                          x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

                                C₃H₆O

8 0
2 years ago
The density of o2 gas at 16 degrees Celsius and 1.27atm is?
velikii [3]

Answer:

The density of O₂ gas is 1.71 \frac{g}{L}

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

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

So, you can get:

\frac{n}{V} =\frac{P}{R*T}

The relationship between number of moles and mass is:

n=\frac{mass}{molar mass}

Replacing:

\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}

\frac{mass}{V*Molar mass} =\frac{P}{R*T}

So:

\frac{mass}{V} =\frac{P*molar mass}{R*T}

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 \frac{g}{mol}.

Then:

\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}

In this case you know:

  • P=1.27 atm
  • molar mass of O₂= 32 \frac{g}{mol}.
  • R= 0.0821 \frac{atm*L}{mol*K}
  • T= 16 °C=  289 °K (0°C= 273°K)

Replacing:

density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol}  }{0.0821\frac{atm*L}{mol*K} *289 K}

Solving:

density= 1.71 \frac{g}{L}

<u><em>The density of O₂ gas is 1.71 </em></u>\frac{g}{L}<u><em></em></u>

3 0
2 years ago
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