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3 years ago

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Chemistry

1 answer:

tensa zangetsu [6.8K]3 years ago

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Answer:

45.2g produce x mole of CH4. we are given the mass of methane. mass (CH4)=45.2g. molar mass (CH4) = 12+4 =16g/mol. Mole=mass. molar mass. mole=45.2. 16. mole=2.8mol. therefore, if 45.2g produce 2.8mol as 1 mole of methane, the water produced is 2(2.8) we Are multiplying by 2 because the water is having 2 moles water mole is 5.6mol

Explanation:

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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

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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:

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

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