The final temperature = 36 °C
<h3>Further explanation</h3>
The balanced combustion reaction for C₆H₆
2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ
MW C₆H₆ : 78.11 g/mol
mol C₆H₆ :
Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol
Heat transferred to water :
Q=m.c.ΔT
Answer:
7.12 mm
Explanation:
From coulomb's law,
F = kqq'/r².................... Equation 1
Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.
Make r the subject of the equation,
r = √(kqq'/F).......................... Equation 2
Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute into equation 2
r = √[ (75×10⁻⁹ )²9.0×10⁹/1]
r = 75×10⁻⁹.√(9.0×10⁹)
r = (75×10⁻⁹)(9.49×10⁴)
r = 711.75×10⁻⁵
r = 7.12×10⁻³ m
r = 7.12 mm
Hence the distance between the point charge = 7.12 mm
Answer:
All substances that are made up of the same kind of mole ules are called pure substances.
So what do you mean ?
Explanation:
The missing components in the table to the right are indicated with orange letters. Use the periodic table in the tools bar and this link Web Elements to fill in the corresponding values. A B C D E F G. 2. See answers. Log in to add ... F = 737.7kJ/mol. G = 495.8kJ/mol. Explanation: We are asked some of the ...
2 answers
Answer: Concentration of 
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of 
= 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture = 
Thus concentration of in the equilibrium mixture is 0.31 M
