Please help me this is due by midnight!!

A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If

Ainat [17]

Answer:

$V_2=52.2L$

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

$PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L$

Which is V1 in the Charles' law:

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

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

$V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K} \\\\V_2=52.2L$

Best regards!

8 0

3 years ago

A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 209.kPa. T

disa [49]

Answer : The final pressure of the gas will be, 26.8 kPa

Explanation :

According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature of the gas and the number of moles of gas.

$P\propto \frac{1}{V}$

or,

$PV=k$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of the gas = 209 kPa

$P_2$ = final pressure of the gas = ?

$V_1$ = initial volume of the gas = 10.0 L

$V_2$ = final volume of the gas = 78.0 L

Now put all the given values in this formula, we get the final pressure of the gas.

$209kPa\times 10.0L=P_2\times 78.0L$

$P_2=26.8kPa$

Therefore, the final pressure of the gas will be, 26.8 kPa

3 0

3 years ago

Which of these substances is an example of a solution?

strojnjashka [21]

The solution is concrete

7 0

3 years ago

Read 2 more answers

The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma

Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= $p_o= 187.54 mmHg$

Vapor pressure of the solution at 65 °C= $p_s$

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

$p_s=173.83 mmHg$

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0

3 years ago

Write the lewis structure for mgi2. draw the lewis dot structure for mgi2. include all lone pairs of electrons.

viktelen [127]

Lewis Structure is drawn in following steps,

1) Calculate Number of Valence Electrons:

# of Valence electrons in Mg = 2
# of Valence electrons in I     = 7
# of Valence electrons in I     = 7
---------
Total Valence electrons = 16

2) Draw Mg as a central atom surround it by two atoms of Iodine.

3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.

Total Valence electrons 16
- Four electrons - 4
----------
    12

4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.

The final lewis structure formed is as follow,

4 0

3 years ago