The first step in most stoichiometry problems is to

Chemistry

1 answer:

dimaraw [331]3 years ago

8 0

The first step in solving any chemistry problem is to balance the equation<span>.</span>

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A gas has density 2.41 g/liter at 25°C and 770 mm Hg. Calculate it's molecular mass (R = 0.0821 L atm.mol-1K-1

Evgesh-ka [11]

Answer:

Explanation:

Given : Density - 2.41 g/liter

Temperature - 25° C

Pressure : 770 mm Hg

R = 0.0821 L atm mol-¹K-¹

Find : Molecular mass of gas

Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.

Keeping the values in equation-

Pressure : 770 mm Hg = 1 atm

Temperature : 273 + 25 = 298 K

M = dRT/P

M = (2.41*0.0821*298)/1

M = 58.96 gram/mol

Thus, the molecular mass of gas is 58.96 gram/mol.

8 0

3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .