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GaryK [48]
3 years ago
12

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

Chemistry
1 answer:
tankabanditka [31]3 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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Option C.

<h3><u>Explanation:</u></h3>

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How much energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°c and completely vaporize the sample?
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The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.

Calculation,

Given data,

Mass of the ice = 10 g

Temperature of ice =  0. 0°C

  • The ice at 0. 0°C is to be converted into water at 0. 0°C

Heat required at this stage = mas of the ice ×latent heat of fusion of ice

Heat required at this stage = 10 g×80 = 800 cal  

  • The temperature of the water is to be increased from 0. 0°C to 100. 0°C

Heat required for this = mass of the ice×rise in temperature×specific heat of water

Heat required for this  = 10 g×100× 1 = 1000 cal

  • This water at  100. 0°C  is to be converted into vapor.

Heat required for this = Mass of water× latent heat

Heat required for this  = 10g ×536 =5360 cal

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Which element is reduced in this reaction? 2kmno4 3na2so3 h2o→2mno2 3na2so4 2koh
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In a redox chemical reaction, one species gets reduced and another gets oxidized. Manganese element is reduced in this reaction.

<h3>What is oxidized and reduced?</h3>

In a redox reaction, the increase or decrease in the oxidation number and electrons results in the reduction and oxidation of the chemical species. The oxidation and reduction occur simultaneously in a reaction.

The oxidation number of Mn in permanganate ion was +8 on the left side and decreased to +4 on the right side of the equation. Potassium permanganate is an oxidizing agent that has reduced the manganese ion of the permanganate ion.

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