The object has an overall positive charge.
Answer:
a) ![T_b=590.775k](https://tex.z-dn.net/?f=T_b%3D590.775k)
b) ![W_t=1.08*10^4J](https://tex.z-dn.net/?f=W_t%3D1.08%2A10%5E4J)
d) ![Q=3.778*10^4J](https://tex.z-dn.net/?f=Q%3D3.778%2A10%5E4J)
d) ![\triangle V=4.058*10^4J](https://tex.z-dn.net/?f=%5Ctriangle%20V%3D4.058%2A10%5E4J)
Explanation:
From the question we are told that:
Moles of N2 ![n=2.50](https://tex.z-dn.net/?f=n%3D2.50)
Atmospheric pressure ![P=100atm](https://tex.z-dn.net/?f=P%3D100atm)
Temperature ![t=20 \textdegree C](https://tex.z-dn.net/?f=t%3D20%20%5Ctextdegree%20C)
![t = 293k](https://tex.z-dn.net/?f=t%20%3D%20293k)
Initial heat ![Q=1.36 * 10^4 J](https://tex.z-dn.net/?f=Q%3D1.36%20%2A%2010%5E4%20J)
a)
Generally the equation for change in temperature is mathematically given by
![\triangle T=\frac{Q}{N*C_v}](https://tex.z-dn.net/?f=%5Ctriangle%20T%3D%5Cfrac%7BQ%7D%7BN%2AC_v%7D)
Where
![C_v=Heat\ Capacity \approx 20.76 J/mol/K](https://tex.z-dn.net/?f=C_v%3DHeat%5C%20Capacity%20%5Capprox%2020.76%20J%2Fmol%2FK)
![T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }](https://tex.z-dn.net/?f=T_b-T_a%3D%5Cfrac%7B1.36%20%2A%2010%5E4%20J%7D%7B2.5%2A20.76%20%7D)
![T_b-293k=297.775](https://tex.z-dn.net/?f=T_b-293k%3D297.775)
![T_b=590.775k](https://tex.z-dn.net/?f=T_b%3D590.775k)
b)
Generally the equation for ideal gas is mathematically given by
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
For v double
![T_c=2*590.775k](https://tex.z-dn.net/?f=T_c%3D2%2A590.775k)
![T_c=1181.55k](https://tex.z-dn.net/?f=T_c%3D1181.55k)
Therefore
![PV=Wbc](https://tex.z-dn.net/?f=PV%3DWbc)
![Wbc=(2.20)(8.314)(1181_590.778)](https://tex.z-dn.net/?f=Wbc%3D%282.20%29%288.314%29%281181_590.778%29)
![Wbc=10805.7J](https://tex.z-dn.net/?f=Wbc%3D10805.7J)
Total Work-done ![W_t](https://tex.z-dn.net/?f=W_t)
![W_t=Wab+Wbc](https://tex.z-dn.net/?f=W_t%3DWab%2BWbc)
![W_t=0+1.08*10^4](https://tex.z-dn.net/?f=W_t%3D0%2B1.08%2A10%5E4)
![W_t=1.08*10^4J](https://tex.z-dn.net/?f=W_t%3D1.08%2A10%5E4J)
c)
Generally the equation for amount of heat added is mathematically given by
![Q=nC_p\triangle T](https://tex.z-dn.net/?f=Q%3DnC_p%5Ctriangle%20T)
![Q=2.20*2907*(1181.55-590.775)\\](https://tex.z-dn.net/?f=Q%3D2.20%2A2907%2A%281181.55-590.775%29%5C%5C)
![Q=3.778*10^4J](https://tex.z-dn.net/?f=Q%3D3.778%2A10%5E4J)
d)
Generally the equation for change in internal energy of the gas is mathematically given by
![\triangle V=nC_v \triangle T](https://tex.z-dn.net/?f=%5Ctriangle%20V%3DnC_v%20%5Ctriangle%20T)
![\triangle V=2.20*20.76*(1181.55-293)k](https://tex.z-dn.net/?f=%5Ctriangle%20V%3D2.20%2A20.76%2A%281181.55-293%29k)
![\triangle V=4.058*10^4J](https://tex.z-dn.net/?f=%5Ctriangle%20V%3D4.058%2A10%5E4J)
Atomic Number is used to organize the periodic table.
Hope this helps
Answer:
0.001 M OH-
Explanation:
[OH-] = 10^-pOH, so
pOH + pH = 14 and 14 - pH = pOH
14 - 11 = 3
[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-