Answer:
The rate of energy production is referred to as power
Answer:
https://socratic.org/answers/220339
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Explanation:
Answer:
Option (E) is correct
Explanation:
Solubility equilibrium of 
is given as follows-
Hence, if solubility of is S (M) then-
![[Pb^{2+}]=S(M)](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3DS%28M%29)
and ![[IO_{3}^{-}]=2S(M)](https://tex.z-dn.net/?f=%5BIO_%7B3%7D%5E%7B-%7D%5D%3D2S%28M%29)
Where species under third bracket represent equilibrium concentrations
So, solubility product of , ![K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BIO_%7B3%7D%5E%7B-%7D%5D%5E%7B2%7D)
Here, ![[Pb^{2+}]=S(M)=5.0\times 10^{-5}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3DS%28M%29%3D5.0%5Ctimes%2010%5E%7B-5%7DM)
So, ![[IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M](https://tex.z-dn.net/?f=%5BIO_%7B3%7D%5E%7B-%7D%5D%3D2S%28M%29%3D%282%5Ctimes%205.0%5Ctimes%2010%5E%7B-5%7D%29M%3D1.0%5Ctimes%2010%5E%7B-4%7DM)
So, 
Hence option (E) is correct
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
