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Sergio [31]
3 years ago
14

If you have 1.00 mole of F2 at 1.00 atm of pressure and 0°C, what is the volume of F2?

Chemistry
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

V=22.4L

Explanation:

Hello,

In this case, by using the ideal gas equation, er can compute the volume of fluorine gas as shown below:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1mol*0.082\frac{atm*L}{mol*K}+(0+273)K}{1.00atm}  \\\\V=22.4L

Best regards.

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Read 2 more answers
5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:
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Answer:

A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:

2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)

The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 oC. The heat capacity of water = 4.18 J/g/oC and the heat capacity of the calorimeter = 10.5 kJ/oC. (1) what is the ΔH of the reaction?

Explanation:

The heat energy released by the reaction = heat absorbed by calorimeter + heat absorbed by water

Heat absorbed by water = mass of water x specific heat capacity of water x change in temperature

Heat absorbed by water =  500 g x 4.18 J/g. oC x (53.13-25.00)oC

                                         = 58791.7 J

Heat absorbed by calorimeter = heat capacity of calorimeter x change in temperature

Heat absorbed by calorimeter = 10.5 x 10^3 J /oC  x (53.13-25.00)oC

                                                  =295365 J

Total heat energy absorbed = 58791.7 J + 295365 J  = 354156.7 J

Number of moles of benzene given is:

number of moles = goven mass of benzene /its molar mass

=7.05 g / 78.0 g/mol

=0.0903mol

Hence, the heat released by the reaction is:

= 354156.7 J / 0.0903 mol

=  3922.00 kJ/mol

Answer:

The heat released during the combustion of 7.05g of benzene is 3922.00kJ/mol.                                              

7 0
3 years ago
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