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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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When the volume of a gas is
Montano1993 [528]

Answer:

\boxed {\boxed {\sf 232.9 \textdegree C}}

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

250 \ mL * T_2 = 137 \textdegree C * 425 \ mL

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

The units of milliliters (mL) cancel.

T_2=\frac{ 137 \textdegree C * 425 }{250 }

T_2= \frac{58225}{250} \textdegree C

T_2=232.9 \textdegree C

The temperature changes to <u>232.9 degrees Celsius.</u>

3 0
2 years ago
Write the number 345 in scientific notation, with the same number of significant figures:
uranmaximum [27]

Answer:

3.45 x 10^2

Explanation:

345= 3 SF

3.45 x 10^2= 3 SF

8 0
3 years ago
Iodine monochloride (ICl) has a higher boiling point than bromine (Br2) partly because iodine monochloride is a(n)
tekilochka [14]

Answer: polar molecule.

Explanation:

The boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure surrounding the liquid. The boiling point is dependent on the type of forces present.

Iodine monochloride (ICl) is a polar molecule due to the difference in electronegativities of iodine and chlorine. Thus the molecules are bonded by strong dipole dipole forces. Thus a higher temperature is needed to generate enough vapor pressure.

Bromine (Br_2) is a non polar molecule as there is no electronegativity difference between two bromine atoms. The molecules are bonded by weak vanderwaal forces and thus has low boiling point.

7 0
2 years ago
The science of molecules and and their is called chemistry
GarryVolchara [31]
Is this supposed to be a question?????
6 0
3 years ago
Read 2 more answers
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Answer:

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