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Lubov Fominskaja [6]
1 year ago
5

32.7 grams of water vapor takes up how many liters at standard temperature and pressure (273 K and 100 kPa)?

Chemistry
1 answer:
kotegsom [21]1 year ago
5 0
Under standard temperature and pressure conditions, it is known that 1 mole of a gas occupies 22.4 liters.

From the periodic table:
molar mass of oxygen = 16 gm
molar mass of hydrogen = 1 gm
Thus, the molar mass of water vapor = 2(1) + 16 = 18 gm

18 gm of water occupies 22.4 liters, therefore:
volume occupied by 32.7 gm = (32.7 x 22.4) / 18 = 40.6933 liters

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The French chemists, Pierre L. Dulong and Alexis T. Petit, noted in 1819 that the molar heat capacity of many solids at ordinary
natita [175]

The empirical formula of the compound : <u><em>RbO₂</em></u>

<h3>Further explanation</h3>

Dulong and Petit's rule's rule, which says that the molar heat capacity, Cp, of a solid can be expressed as

\large{\boxed{\bold{Cp=N(3R)}}

Where N is the number of atoms per formula unit and R is the universal gas constant

While the value of R is:

R = 8.3144621 J / K · mol

then:

3R = 24.94 J / K · mol

Heat capactiy per gram of a compound containing rubidium and oxygen is 0.64 J · K⁻¹ · g⁻¹

We try the possible empirical formula from the composition of Rb and O

  • Rb₂O

So the number of atoms: 3 (Rb = 2 atoms, O = 1 atom)

Then the value of the Cp:

Molar mass Rb₂O = 2.85.5 + 1.16 = 187 g/mol

\displaystyle Cp=N(3R)\\\\0.64=\frac{3.24.94}{187}\\\\0.64\neq0.4

Then this empirical formula Rb₂O  is not appropriate

  • RbO₂

So the number of atoms: 3 (Rb = 1 atom, O = 2 atoms)

Then the value of the Cp:

Molar mass RbO₂ = 1.85.5 + 2.16 = 117.5 g / mol

\displaystyle Cp=N(3R)\\\\0.64=\frac{3.24.94}{117.5}\\\\0.64=0.64

Then this empirical formula RbO₂  is appropriate

<h3>Learn more </h3>

chemical bond

brainly.com/question/5008811

ionic bonding

brainly.com/question/1603987

electron dots

brainly.com/question/6085185

 

Keywords:  an empirical formula,compound, gas constant, Dulong and Petit's rule,the molar heat capacity

#LearnWithBrainly

7 0
2 years ago
A stone of mass 0.55 kilograms is released and falls to the ground. Measurements show that the stone has a kinetic energy of 9.8
Lelu [443]
Hope this helps you.

5 0
2 years ago
Read 2 more answers
A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of
Karolina [17]

Ans: The final volume of the balloon is 4.5 L

<u>Given:</u>

Volume of balloon inflated with 3 breaths = 1.7 L

<u>To determine:</u>

Volume of balloon after a total of 3+5 = 8 breaths

<u>Explanation:</u>

Volume of the balloon per breath = 1.7 L * 1 breath/3 breaths = 0.567 L

Final volume of balloon after 8 breaths = 0.567 L * 8 breath/1 breath

= 4.536 L

7 0
2 years ago
Which factor explains why coal dust in an enclosed space is more explosive than coal dust blown outdoors into an open space ? A.
ZanzabumX [31]

The correct option is B.

Coal dust refers to the powered form of coal. Because of the high surface area of coal dust it is highly prone to dust explosion, which involves rapid combustion of fine particles that are suspended in the air; this usually occur in an enclosed place. Coal dust in an enclosed place is more explosive than coal dust that is blown outdoor in an open space because the coal dust in an enclosed place is more concentrated due to restricted space, thus it is more liable to explosion.

8 0
2 years ago
Read 2 more answers
You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain
IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of O_2:-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{49.8\ g}{32\ g/mol}

Moles_{O_2}= 1.55625\ mol

Since pressure and volume are constant, we can use the Avogadro's law  as:-

\frac {V_1}{n_1}=\frac {V_2}{n_2}

Given ,  

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

\frac {V_1}{n_1}=\frac {V_2}{n_2}

\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of CO_2 = 44.0 g/mol

Mass of CO_2 = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

5 0
1 year ago
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