Answer:
V₂ = 50.93 L
Explanation:
Initial volume, ![V_1=43.1\ L](https://tex.z-dn.net/?f=V_1%3D43.1%5C%20L)
Initial temperature, ![T_1=24^{\circ} C=24+273=297\ K](https://tex.z-dn.net/?f=T_1%3D24%5E%7B%5Ccirc%7D%20C%3D24%2B273%3D297%5C%20K)
Final temperature, ![T_2=78^{\circ} C=78+273=351\ K](https://tex.z-dn.net/?f=T_2%3D78%5E%7B%5Ccirc%7D%20C%3D78%2B273%3D351%5C%20K)
We need to find the final volume of the gas. The relation between the volume and the temperature is given by :
![\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L](https://tex.z-dn.net/?f=%5Cdfrac%7BV_1%7D%7BT_1%7D%3D%5Cdfrac%7BV_2%7D%7BT_2%7D%5C%5C%5C%5CV_2%3D%5Cdfrac%7BV_1T_2%7D%7BT_1%7D%5C%5C%5C%5CV_2%3D%5Cdfrac%7B43.1%5Ctimes%20351%7D%7B297%7D%5C%5C%5C%5CV_2%3D50.93%5C%20L)
So, the final volume of the gas is 50.93 L.
Answer:
could create chemical reaction
Explanation:
Answer:
Explanation:
From the information given:
(a)
![Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}](https://tex.z-dn.net/?f=Concentration%20%5C%20in%20%5C%20mg%2FL%20%3D%20%5Cdfrac%7BMass%20%5C%20of%20%5C%20MTBE%20%5C%20in%20%5C%20mg%7D%7BTotal%20%5C%20volume%20%28in%20%5C%20L%29%7D)
![Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}](https://tex.z-dn.net/?f=Concentration%20%5C%20in%20%5C%20mg%2FL%20%3D%20%5Cdfrac%7B222%20%5Ctimes%2010%5E3%20%5C%20mg%7D%7B22%7D)
![Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L](https://tex.z-dn.net/?f=Concentration%20%5C%20in%20%5C%20mg%2FL%20%3D%20111%20%5Ctimes%2010%5E3%20%5C%20mg%2FL)
(b)
![number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}](https://tex.z-dn.net/?f=number%20%5C%20of%20%5C%20mole%20s%3D%20%5Cdfrac%7Bmass%7D%7Bmolar%20%5C%20mass%20%7D%20%5C%5C%20%5C%5C%20number%20%5C%20of%20%5C%20mole%20s%3D%5Cdfrac%7B222%20%5C%20g%7D%7B88.15%20%5C%20g%2Fmol%7D%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%3D%202.518%20mol%7D)
(c)
![w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%](https://tex.z-dn.net/?f=w%2Fw%20%5C%20percentage%20%3D%20%5Cdfrac%7Bmass%20%5C%20of%20%5C%20MTBE%20%7D%7Bmass%20%5C%20of%20%5C%20solution%20%28RFG%29%7D%5Ctimes%20100%5C%25)
![where; \\ \\ mass \ of \ (RFG) = 2L \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g](https://tex.z-dn.net/?f=where%3B%20%5C%5C%20%5C%5C%20mass%20%5C%20%20of%20%5C%20%20%28RFG%29%20%3D%202L%20%5Ctimes%200.70%20g%2FmL%20%5C%5C%20%5C%5C%20mass%20%5C%20%20of%20%5C%20%20%28RFG%29%20%3D%202000%20ml%20%5Ctimes%200.70%20g%2FmL%20%5C%5C%20%5C%5C%20mass%20%5C%20of%20%5C%20%28RFG%29%20%3D%201400%20g)
∴
![w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}](https://tex.z-dn.net/?f=w%2Fw%20%5C%20percentage%20%3D%20%5Cdfrac%7B222%20%5C%20g%7D%7B1400%20%5C%20g%7D%5Ctimes%20100%5C%25%20%3D%20%5Cmathbf%7B15.8%5C%25%7D)
(d)
![Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}](https://tex.z-dn.net/?f=Volume%20of%20MTBE%20%3D%5Cdfrac%7Bmass%20%5C%20of%20%5C%20MTBE%7D%7Bdensity%20%5C%20of%20%5C%20MTBE%7D)
![Volume \ of \ MTBE = 300 \ mL\\](https://tex.z-dn.net/?f=Volume%20%5C%20of%20%5C%20MTBE%20%3D%20300%20%5C%20mL%5C%5C)
∴
![v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}](https://tex.z-dn.net/?f=v%2Fv%5C%25%20%3D%20%5Cdfrac%7Bvolume%20%5C%20of%20%5C%20MTBE%7D%7Bvolume%20%5C%20of%20%5C%20RFG%7D%20%5C%5C%20%5C%5C%20v%2Fv%5C%25%20%3D%5Cdfrac%7B300%20%5C%20mL%7D%7B2000%20%5C%20mL%7D%5Ctimes%20100%5C%25%20%5C%5C%20%5C%5C%20%5Cmathbf%7Bv%2Fv%5C%25%20%3D%2015.00%5C%25%7D)
(e)
![From \the \ given \ information; \\ \\ 2.5184 \ moles\ of \ MTBE contain \ 2.5184 \ mole of oxygen](https://tex.z-dn.net/?f=From%20%5Cthe%20%20%5C%20given%20%5C%20%20information%3B%20%5C%5C%20%5C%5C%202.5184%20%5C%20moles%5C%20of%20%20%5C%20MTBE%20contain%20%20%5C%202.5184%20%20%5C%20mole%20of%20oxygen)
∴
![mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g](https://tex.z-dn.net/?f=mass%20of%20oxygen%20MTBE%20%3D%202.5284%20mol%20%5Ctimes%2016%5C%20g%2Fmol%20%5C%5C%20%5C%5C%20mass%20of%20oxygen%20MTBE%20%3D%2040.3%209%20%5C%20g%5C%5C%20%5C%5C%20mass%5C%20of%20%5C%20RFG%20%3D%201400%20g)
∴
![\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%](https://tex.z-dn.net/?f=%5C%25%20w%2Fw%20%3D%20%5Cdfrac%7Bmass%20%5C%20of%20%5C%20oxygen%7D%7Bmass%20%5C%20of%20RFG%20%7D%3D%5Cdfrac%7B40.22%20%5C%20g%7D%7B1400%20%5C%20g%7D%20%5Ctimes%20100%5C%25)
![\% w/w == 2.88\%](https://tex.z-dn.net/?f=%5C%25%20w%2Fw%20%3D%3D%202.88%5C%25)
Answer:
The answer to your question is Volume = 600 ml
Explanation:
Data
Volume 1 = 810 ml
Temperature 1 = 270°K
Pressure 1 = 1 atm
Volume 2 = ?
Pressure = 2 atm
Temperature 2 = 400°K
Process
1.- To solve this problem, use the Combine gas law.
P1V1 / T1 = P2V2 / T2
- Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (1)(810)(400) / (270)(2)
3.- Simplification
V2 = 324000 / 540
4.- Result
V2 = 600 ml