Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value
<span>
The half-life for the chemical reaction is 29,2 s and is
independent of initial concentration.
c</span>₀
- initial concentration the reactant.
c - concentration of the reactant remaining
at time.
t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s ·
t₁.
t₁ = 116,8 s.
Answer:
i. Keq=4157.99.
ii. More hydrogen sulfide will be produced.
Explanation:
Hello,
i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:
![Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5BS_2%5D%7D%20%3D%5Cfrac%7B%280.97M%29%5E2%7D%7B%280.051M%29%5E2%280.087%29%7D%20%3D4157.99)
ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.
Best regards.
The balanced equation for the reaction is as follows
2H₂ + O₂ --> 2H₂O
stoichiometry of H₂ to O₂ is 2:1
number of H₂ moles - 30.0 g / 2 g/mol = 15 mol
number of O₂ moles - 80.0 g / 32 g/mol = 2.5 mol
limiting reactant is the reagent in which only a fraction is used up in the reaction
if H₂ is the limiting reactant
if 2 mol of H₂ requires 1 mol of O₂
then 15 mol of H₂ requires 1/2 x 15.0 = 7.5 mol of O₂
but only 2.5 mol of O₂ is required
this means that O₂ is the limiting reagentt and H₂ is in excess
Most acidic: solution B, pH = 1
Most Basic: solution C, pH = 13
Least acidic: solution A, pOH = 1
Answer:
Of course it's C
Red planet
Explanation:
It is because the soil on Mars is rich of Fe (Iron).
That makes the soil look red.
Even on our planet we have such this places like hormuz island in Iran.
