Answer:
there are 4.0205 Litres of plasma in 8.5 pints
Explanation:
Recall that:
1 quart qt = 2 pints pt and 1 quart = 946 mL
∴ two pints will contain 946 mL
and one pint will be = 946 mL/2
= 473 mL
Similarly, 8.5 pints will contain 473 mL × 8.5 = 4020.5 mL
We know that :
1000 mL = 1 Litres
Hence 4020.5 mL = (4020.5 /1000)L
= 4.0205 Litres
Therefore, there are 4.0205 Litres of plasma in 8.5 pints
Carbon dioxide is a gaseous molecule made up of the elements, C and O. Each mole of carbon dioxide has one mole C and two mole oxygen atoms.
Molar mass of carbon dioxide (
)=![(1 * 12.01\frac{g}{mol}) +(2* 16\frac{g}{mol})=44.01\frac{g}{mol}](https://tex.z-dn.net/?f=%281%20%2A%2012.01%5Cfrac%7Bg%7D%7Bmol%7D%29%20%2B%282%2A%2016%5Cfrac%7Bg%7D%7Bmol%7D%29%3D44.01%5Cfrac%7Bg%7D%7Bmol%7D)
Percentage by mass of carbon = ![\frac{(1*12.01\frac{g}{mol}) }{44.01\frac{g}{mol} } *100=27.3%](https://tex.z-dn.net/?f=%5Cfrac%7B%281%2A12.01%5Cfrac%7Bg%7D%7Bmol%7D%29%20%7D%7B44.01%5Cfrac%7Bg%7D%7Bmol%7D%20%7D%20%2A100%3D27.3%25)
Percentage by mass of oxygen = ![\frac{(2*16\frac{g}{mol})}{44.01\frac{g}{mol} } *100=72.7%](https://tex.z-dn.net/?f=%5Cfrac%7B%282%2A16%5Cfrac%7Bg%7D%7Bmol%7D%29%7D%7B44.01%5Cfrac%7Bg%7D%7Bmol%7D%20%7D%20%2A100%3D72.7%25)
Therefore C is 27.3 % and O is 72.7 % by mass in 1 mol CO![_{2}](https://tex.z-dn.net/?f=_%7B2%7D)
This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
![H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)](https://tex.z-dn.net/?f=H_%7BC-6%7D%3Dy_%7BC-6%7DC_v%28T_b-Ti%29%2B%5CDelta%20_vH%2BC_v%28T_f-Tb%29%5C%5C%5C%5CH_%7BN_2%7D%3Dy_%7BN_2%7DC_v%28T_f-Ti%29)
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and
and
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
![H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol](https://tex.z-dn.net/?f=H_%7BC-6%7D%3D0.58%2A200%2869-75%29%2B%28-31500%29%2B160%2820-69%29%3D-40036J%2Fmol%5C%5C%5C%5CH_%7BN_2%7D%3D0.42%2A29.1%2820-75%29%3D-672.21J%2Fmol)
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
![Q=-40036+(-672.21)=-40708.21J](https://tex.z-dn.net/?f=Q%3D-40036%2B%28-672.21%29%3D-40708.21J)
Finally we convert this result to kJ:
![Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ](https://tex.z-dn.net/?f=Q%3D-40708.21J%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%5C%5C%5C%5CQ%3D-40.7kJ)
Learn more:
P1V1=P2V2, so P1V1/P2=V2.
2atm x 6.0 L/1.0 atm = 12.0 L
The new volume would be 12.0 Liters
Answer: 2 Na (s) + Cl(g) -> 2 NaCl (s)
Explanation: