Answer:
-88.66 kJ/mol
Explanation:
The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:
C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)
H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)
Cp = A + BT + CT⁻²
For the Kirchoff's Law:
ΔHf = ΔH°f + 
Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:
2C(s) + 3H₂(g) → C₂H₆
So, DCp:
dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83
dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788
dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵
dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²
= -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)
ΔHf = -84.68 - 3.80
ΔHf = -88.66 kJ/mol
Atomic radius is the distance from the atom's nucleus to the outer edge of the electron cloud.
Answer:
orbits
Explanation:
An orbital is a region of space where there is a high probability of finding an electron.
Answer:
The estimated feed rate of logs is 14.3 logs/min.
Explanation:
The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.
That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.
Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.
To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.
The specific gravity of the wood chips is 0.494.
The average volume of a log is
The weight of one log is
To provide 3617 ton/day of wood chips, we need
The feed rate of logs is 14.3 logs/min.
Answer:
it opposes motion. the object would continue to slide forever if not stopped by another object.
