Answer:
13.5 years
Explanation:
Initial Concentration [Ao] = 10g
Final Concentration [A] = 0.768g
Time t= 50 years
Half life t1/2 = ?
These quantities are related by the following equations;
ln[A] = ln[Ao] - kt ......(i)
t1/2 = ln(2) / k ...........(ii)
where k = rate constant
Inserting the values in eqn (i) and solving for k, we have;
ln(0.768) = ln(10) - k (50)
-0.2640 = 2.3026 - 50k
50k = 2.3026 + 0.2640
k = 2.5666 / 50 = 0.051332
Insert the value of k in eqn (ii);
t1/2 = ln(2) / k
t1/2 = 0.693 / 0.051332 = 13.5 years
We have 25cm^3 of 0.1mol AgNO3.
25cm^3 = 0.025L, so we have 0.025 x 0.1 = 0.0025mol AgNO3, so
0.0025AgNO3 + 0.0025NaCl = 0.0025AgCl + 0.0025NaNO3
Change in Free Energy: ΔG(20C) = -0.064kJ (negative, so the reaction runs)
Change in Enthalpy: ΔH(20C) = -0.110kJ (negative, so the reaction is exothermic)
This reaction produces 0.358g of AgCl and 0.213g of NaNO3
Les Mclean PhD
The effects on the concentration of SO3 gas when the
following changes occur after initial equilibrium has been established in this
system (N.C. = no change) by adding a catalyst.
2SO2(g) + O2(g) 2SO3(g) + 46.8 kcal
So that even though there is an addition of catalyst, no
change in reactants or products has occurred because catalyst only provides a
faster pathway for the reaction to occur.
Answer:
The concentration of COF₂ at equilibrium is 0.296 M.
Explanation:
To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the <em>concentrations</em> or <em>changes in concentration</em> in that stage. For this reaction:
2 COF₂(g) ⇌ CO₂(g) + CF₄(g)
I 2.00 0 0
C -2x +x +x
E 2.00 - 2x x x
Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".
![Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852](https://tex.z-dn.net/?f=Kc%3D8.30%3D%5Cfrac%7B%5BCO_%7B2%7D%5D%20%5Ctimes%20%5BCF_%7B4%7D%5D%20%7D%7B%5BCOF_%7B2%7D%5D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%282.00-2x%29%5E%7B2%7D%20%7D%20%5C%5C8.30%3D%28%5Cfrac%7Bx%7D%7B2.00-2x%7D%20%29%5E%7B2%7D%20%5C%5C%5Csqrt%7B8.30%7D%20%3D%5Cfrac%7Bx%7D%7B2.00-2x%7D%5C%5C5.76-5.76x%3Dx%5C%5Cx%3D0.852)
The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M
