One of the products of photosynthesis is carbon dioxide
Answer:
Q = 0.50
No
Left
Explanation:
At a generic reversible equation
aA + bB ⇄ cC + dD
The reaction coefficient (Q) is the ratio of the substances concentrations:
![Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%2A%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%2A%5BB%5D%5Eb%7D)
Solids and liquid water are not considered in this calculus.
When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.
In this case:
Q = ![\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)

Q = 0.50
So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.
Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
Answer:
35.75 days
Explanation:
From the given information:
For first-order kinetics, the rate law can be expressed as:

Given that:
the rate degradation constant = 0.12 / day
current concentration C = 0.05 mg/L
initial concentration C₀ = 3.65 mg/L

㏑(0.01369863014) = -(0.12) t
-4.29 = -(0.12)
t = -4.29/-0.12
t = 35.75 days