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GaryK [48]
2 years ago
12

Calcule la densidad del hidrógeno H2 en g/L a 327 mm Hg y 48ºc

Chemistry
1 answer:
tankabanditka [31]2 years ago
4 0
<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin  

n = N / No  

n = mole  

No = Avogadro number (6.02.10²³)  

n = m / MW

m = mass  

MW = molecular weight

For density , can be formulated :

\tt \rho=\dfrac{P\times MW}{R\times T}

P = 327 mmHg = 0,430263 atm

R = 0.082 L.atm / mol K

T = 48 ºC = 321.15 K

MW of H₂ =  2.015 g/mol

The density :

\rho=\dfrac{0,430263\times 2.015 }{0.082\times 321.15}\\\\\rho=0.033~g/L

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goblinko [34]

Answer:

Freezing and Melting (Fusion)

Explanation:

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3 years ago
Do your cells have access to everything they need over the first 4 or 5 hours that you are lost? Yes or No?
77julia77 [94]

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4 0
2 years ago
The weights of all elements are always compared to oxygen.
marishachu [46]

Answer:

The weights of all elements are always compared to the Carbon-12.

Explanation:

The weights of all elements are always compared to the Carbon-12 because the mass of carbon is 12 which is the exactly the sum of protons and neutrons.  

Oxygen was also considered the standard for some time but later this stander was rejected because in natural O¹⁷ and O¹⁸ were also present and this create the two different atomic mass tables.

AMU:

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C12 has six neutron and six protons in the nucleus.

This unit is used to express the masses of atoms. We know that masses of atoms are very small and we do not have any such type of balance that can measure very small quantity. So that is way we use this scale to measure small quantity. For example, according to this scale

relative atomic mass of hydrogen is 1.008 amu

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7 0
3 years ago
How much heat do you need to raise the temperature of 100g of aluminum from 30 C to 150 C
SVEN [57.7K]

Answer:

Q = 10.8 KJ

Explanation:

Given data:

Mass of Al= 100g

Initial temperature = 30°C

Final temperature = 150°C

Heat required = ?

Solution:

Specific heat of Al = 0.90 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30°C

ΔT = 120°C

Q = 100g×0.90 J/g.°C× 120°C

Q = 10800 J       (10800j×1KJ/1000 j)

Q = 10.8 KJ

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Answer:

no

Explanation:

3 0
3 years ago
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