setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :

Q<K⇒The reaction moved to the right (products)
Setup 2 :

Q=K⇒the system at equilibrium
Setup 3 :

Q>K⇒The reaction moved to the left (reactants)
C₃H₈ + 5 O₂ → 4 H₂O + 3 CO₂
mole ratio based on balance equation of O₂ : CO₂ i s 5 : 3
C₃H₈ + 5 O₂ → 4 H₂O + 3 CO₂
∴ if moles of CO₂ = 3 moles
then moles of O₂ = (3 moles ÷ 3) × 5
= <span>5 moles </span>
The stirring breaks it up faster, the amount of water doesn’t effect it, it just gives it space to spread/dissolve - I think.
Answer:
I believe the answer would be C4H9O2.
Answer: In this case, you can estimate that the solubility of potassium nitrate in water at
60
∘
C
is equal to about
solubility
≈
110 g / 100 mL water
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
This tells you that a saturated solution of potassium nitrate will hold about
110 g
of dissolved salt, i.e. of dissociated salt, per
100 mL
of water at
60
∘
C
.