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ikadub [295]
3 years ago
5

Derive an expression for the molar mass of a gas starting with its density?

Chemistry
1 answer:
notka56 [123]3 years ago
6 0

Answer:

The relationships between molar mass and density for a monoatomic gas can be easy.  

The Ideal Gas Law, PV = nRT can be arranged so that n moles equals the mass/molar mass of the gas to become,

PV =  

M

mRT

​  

 

where m is the mass and M is the molar mass.

M =  

PV

mRT

​  

, if you hold the temperature of the gas constant the equation reduces to the Boyle's law or  

PV

m

​  

 

The mass will be constant assuming the container is closed and so the gas cannot be escaped so, PV will be constant.  

D =  

V

m

​  

 and M =  

PV

mRT

​  

 

M =  

P

DRT

​  

 

The higher the density of the gas the higher the molar mass and vice versa.

Explanation:

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4 0
3 years ago
Cyclopropane, a substance used with oxygen as a general anesthetic, contains only two elements, carbon and hydrogen. When 1.00 g
ELEN [110]

Answer:

CH₂

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 1 g

Mass of CO₂ = 3.14 g

Mass of H₂O = 1.29 g

Empirical formula =?

Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:

For Carbon, C:

Mass of CO₂ = 3.14 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Molar mass of C = 12 g/mol

Mass of C =?

Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂

Mass of C = 12/44 × 3.14

Mass of C = 0.86 g

For hydrogen, H:

Mass of C = 0.86 g

Mass of compound = 1 g

Mass of H =?

Mass of H = (Mass of compound) – (mass of C)

Mass of H = 1 – 0.86

Mass of H = 0.14 g

Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:

Mass of C = 0.86 g

Mass of H = 0.14 g

Divide by their molar mass

C = 0.86 / 12 = 0.07

H = 0.14 / 1 = 0.14

Divide by the smallest

C = 0.07 / 0.07 = 1

H = 0.14 / 0.07 = 2

Thus, the empirical formula of cyclopropane is CH₂

8 0
3 years ago
since esters are cleaved on hydrolysis, the molecular weight of the acid derived from an ester is always lower than the molecule
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Esters  contains  a  double  bond  between  carbon  atom  and  oxygen  atom  while  carboxylic  acid  lack  double  bond  between  carbon  and oxygen atom.The  lack  of  double  bond in  carboxylic  acid  make  the  force  of  interaction between  carbon  and  oxygen  strong  and  particles  become tightly  packed  which  make  carboxylic  acid  solid.  The  presence  of  double  bond  in  esters make  the  force  of  interaction  weak  thus  they  are  liquid.
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3 years ago
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The OH-ion concentration of a blood sample is
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Answer:

She okeyy

Explanation:

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4 0
3 years ago
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
ICE Princess25 [194]

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

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3 years ago
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