Answer:
10.7 g of KOH
Explanation:
First of all, we determine the reaction:
2K (s) + 2H₂O(l) → H₂(g) + 2KOH(aq)
We convert the mass of K, to moles → 7.5 g . mol/39.1 g = 0.192 moles
Ratio is 2:2, so the moles I have of K must produce the same moles of KOH. In this case, the produces moles of KOH are 0.192 moles.
We convert the moles to mass, to finish the answer:
0.192 mol . 56.1g /1mol = 10.7 g of KOH
Answer:
The Answer is gonna be C) high pressure and low temperature
Answer:
- final temperature (T2) = 748.66 K
- ΔU = w = 5620.26 J
- ΔH = 9367.047 J
- q = 0
Explanation:
ideal gas:
reversible adiabatic compression:
∴ q = 0
∴ w = - PδV
⇒ δU = δw
⇒ CvδT = - PδV
ideal gas:
⇒ PδV + VδP = RδT
⇒ PδV = RδT - VδP = - CvδT
⇒ RδT - RTn/PδP = - CvδT
⇒ (R + Cv,m)∫δT/T = R∫δP/P
⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303
∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5
⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212
⇒ T2/T1 = 2.512
∴ T1 = 298 K
⇒ T2 = (298 K)×(2.512)
⇒ T2 = 748.66 K
⇒ ΔU = Cv,mΔT
⇒ ΔU = (3/2)R(748.66 - 298)
∴ R = 8.314 J/K.mol
⇒ ΔU = 5620.26 J
⇒ w = 5620.26 J
⇒ ΔH = ΔU + nRΔT
⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)
⇒ ΔH = 5620.26 J + 3746.787 J
⇒ ΔH = 9367.047 J
