Answer:
Stalactites - Hang from caves ceilings
Stalagmites - Grow from cave floors
Sinkholes - Collapsed caves
Speleothems - Cave features
Explanation:
Got it right :)
C is becsuse it just souds right
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
Explanation:
The chemical equation is as follows.
And, the given enthalpy is as follows.
; 
= 102.5 kJ
Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol
Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.
102.5 = ![[(\frac{1}{2})x + 498] - [(2)(243)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29x%20%2B%20498%5D%20-%20%5B%282%29%28243%29%5D)
102.5 = 
102.5 - 12 = 
x = 181 kJ
Now, total bond enthalpy of per mole of ClO is calculated as follows.
x = ![[(\frac{1}{2})181 + (\frac{1}{2})498] - 243](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29181%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29498%5D%20-%20243)
= 339.5 - 243
= 96.5 kJ
Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.
The top number of a portion is called its numerator whereas the bottom number is its denominator. So a Fraction is the quantity of shaded parts separated by the quantity of a balance of as demonstrated as follows: number of shaded parts is the numerator over the whole part which is the denominator.
