Answer: 
at the temperature of the experiment is 0.56.
Explanation:
Moles of 
= 0.35 mole
Moles of 
= 0.40 mole
Volume of solution = 1.00 L
Initial concentration of = 
Initial concentration of = 
Equilibrium concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 0.35 M 0.40 M 0 M 0M
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
Given: (0.35-x) = 0.19
x= 0.16 M
The expression for equilibrium constant for this reaction will be,
![K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BCO_2%5D%5Ctimes%20%5BH_2%5D%7D%7B%5BCO%5D%5Ctimes%20%5BH_2O%5D%7D)
Now put all the given values in this expression, we get :
Thus at the temperature of the experiment is 0.56.
Speed of light = 3x10^8 m/s
Mass of electron = 9.11x10^-31 kg
70% of 3x10^8 = 2.1x10^8 m/s
206.8*9.11x10^-31 = 1.88x10^-28 kg
Equation for wavelength is: wavelength= plancks constant / (mass*velocity)
Plancks constant = 6.62x10^-34 m^2 kg/s
Plug all of these in and you get: 6.62x10^-34 / (1.88x10^-28/2.1x10^8) = 739.47
Your answer is 739.47 m
Answer:
ΔH = -135.05 kJ
Explanation:
(1) 2 ClF(g) + O₂(g) → Cl₂O(g) + OF₂(g) ΔHo = 167.5 kJ
(2) 2 F₂(g) + O₂(g) → 2 OF₂(g) ΔHo = −43.5 kJ
(3) 2 ClF₃(l) + 2 O₂(g) → Cl₂O(g) + 3 OF₂(g) ΔHo = 394.1 kJ
We can use Hess' Law to calculate the ΔH of reaction:
ClF(g) + 1/2O₂(g) → 1/2Cl₂O(g) + 1/2OF₂(g) ΔHo = 167.5 x 1/2 = 83.75 kJ
F₂(g) + 1/2O₂(g) → OF₂(g) ΔHo = −43.5 x 1/2 = -21.75 kJ
1/2Cl₂O(g) + 3/2 OF₂(g) → ClF₃(l) + O₂(g) ΔHo = 394.1 x -1 x 1/2 = -197.05kJ
______________________________________________
(4) ClF (g) + F₂ (g) → ClF₃ (l) ΔH = 83.75 kJ-21.75 kJ-197.05kJ
ΔH = -135.05 kJ/mol
Water and carbon dioxide are compounds. Air is a mixture. The element that is listed is hydrogen.
Remember that you can also find elements on a periodic table.
Your answer is hydrogen.
