Answer:
2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)
0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x
i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L
Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)
10.5 = x*x/(0.205-2x)^2
=> 10.5(4x^2-0.82x+0.042) = x^2
=>42x^2-8.61x+0.441=x^2
=>41x^2-8.61x+0.441 = 0
This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121
The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.
Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L
Answer:c
Explanation:
i think that what is was for me
Based on the diagram shown, a numerical setup for calculating the gram-formula mass for reactant 1 would be :
6(1) + 2(12) + 16
Hope this helps
Answer:
The answer to your question is: 234.7 cans
Explanation:
data
caffeine concentration = 3.55 mg/oz
10.0 g of caffeine is lethal
there are 12 oz of caffeine in a can
Then
3.55 mg ----------------- 1 oz
x mg -----------------12 oz (in a can)
x = 42.6 mg of caffeine in a can
Convert it to grams 42,6 mg = 0.0426 g of caffeine in a can
Finally
0.0426 g of caffeine ------------------ 1 can
10 g of caffeine ----------------- x
x = 10 x 1/0.0436 = 234.7 cans
