When the solute can no longer dissolve in the solution the solvent becomes SATURATED. When no more solute can dissolve and if you look at the bottom of the beaker, test tube, pan, or glass of cold tea you can see the solute permeating out as little particles .
Cold is the absence of heat! so heat moves into the object from the environment (room) rapidly~ and it slows down by time and water is formed on the surface of the object because of cold and hot air.
I hope this helps cx
Answer:
Pb(OH)2 + 2 HCl ---> 2H2o + PbCl2
Explanation:
Qc < Kc, the reaction proceeds from left to right to reach equilibrium
<h3>Further explanation
</h3>
Given
K = 50.2 at 445°C
[H2] = [I2] = [HI] = 1.75 × 10⁻³ M At 445ºC
Reaction
H2(g) + I2(g) ⇔2HI(g)
Required
Qc
Solution
Qc for the reaction
![\tt Qc=\dfrac{[HI]^2}{[I_2][H_2]}\\\\Qc=\dfrac{(1.75.10^{-3})^2}{1.75.10^{-3}\times 1.75\.10^{-3}}=1](https://tex.z-dn.net/?f=%5Ctt%20Qc%3D%5Cdfrac%7B%5BHI%5D%5E2%7D%7B%5BI_2%5D%5BH_2%5D%7D%5C%5C%5C%5CQc%3D%5Cdfrac%7B%281.75.10%5E%7B-3%7D%29%5E2%7D%7B1.75.10%5E%7B-3%7D%5Ctimes%201.75%5C.10%5E%7B-3%7D%7D%3D1)
Qc < Kc ⇒ reaction from left(reactants) to right (products) (the reaction will shift on the right) until it reaches equilibrium (Qc = Kc)
Answer:
First of all we will calculate the amount of CO₂ produced by burning 10 grams of Methane. The balance chemical equation is as follow:
CH₄ + 2 O₂ → CO₂ + 2 H₂O
First we will calculate the moles of CH₄ as,
Moles = Mass / M.Mass
Moles = 10 g / 16 g/mol
Moles = 0.625 moles
Secondly calculate the moles of CO₂ produces theoretically,
According to equation,
1 mole of CH₄ produces = 1 mole of CO₂
So,
0.625 moles of CH₄ will produce = X moles of CO₂
Solving for X,
X = 1 mole × 0.625 moles / 1 mole
X = 0.625 moles of CO₂
At last calculate mass of CO₂ as,
Mass = Moles × M.Mass
Mass = 0.625 mole × 44 g/mol
Mass = 27.5 g ≈ 27 g
Conclusion:
From above calculation it can be concluded that this reaction is following law of conservation of mass. There is neither a gain nor a loss of mass during the reaction. Also, this can be made possible when the given amount of methane is provided with excess of O₂ gas so that the methane is completely combusted.
