If you combine 370.0 ml of water at 25.00 °c and 110.0 ml of water at 95.00 °c, what is the final temperature of the mixture? use 1.00 g/ml as the density of water.

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catherinesquitieri

12.10.2020

Chemistry

A compressed gas cylinder is filled with 5270 g of argon gas.

The pressure inside the cylinder is 2050 psi at a temperature of 18C.

The valve to the cylinder is opened and gas escapes until the pressure inside the cylinder is 650. psi and the temperature are 26 C.

How many grams of argon remains in the cylinder?

Explanation:

First, calculate the volume of argon gas that is present in the gas cylinder by using the ideal gas equation:

Mass of Ar gas is --- 5270g.

The number of moles of Ar gas:

$Number of moles of Ar gas=\frac{given mass of Ar}{its atomic mass} \\ =\frac{5270g}{39.948g/mol} \\ =131.9mol$

Temperature T=(18+273)K=291K

Pressure P=2050psi

$2050* 0.0680atm\\\\=139.4atm\\$

Volume V=?

$PV=nRT\\139.4atm * V=131.9mol *0.0821L.atm.mol-1.K-1 * 291K\\=V=\frac{131.9mol *0.0821L.atm.mol-1.K-1 * 291K}{139.4atm} \\=V=22.6L$

Using this volume V=22.6L

Pressure=650psi=44.2atm

Temperature T= (26+273)K=299K

calculate number of moles "n" value:

$PV=nRT\\=n=\frac{PV}{RT} \\=n=\frac{44.2atm*22.6L}{0.0821L.atm.mol^-1K^-1* 299K} \\=n=40.7mol$

Mass of 40.7mol of Ar gas:

$mass of Ar gas=number of moles * its atomic mass\\\\ =40.7mol* 39.948g/mol\\\\ =1625.8g$

The mass of Ar gas becomes 1625.8g.

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