Answer:
[CO] = 7.61x10⁻³M
7.61x10⁻³x10³ = 7.61
Explanation:
For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:
![Kc = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BC%5D%5Ecx%5BD%5D%5Ed%7D%7B%5BA%5D%5Eax%5BB%5D%5Eb%7D)
We need to know the molar concentrations in the equilibrium. In the beginning, there is only COCl₂, and its concentration is the number of moles divided by the volume:
[COCl₂] = 7.73/10.0 = 0.773 M
So, the equilibrium will be:
COCl₂(g) ⇆ CO(g) + Cl₂(g)
0.773 0 0 <em>Initial</em>
-x +x +x <em> Reacts</em>
0.773-x x x <em>Equilibrium</em>
Supposing that x<<0.773, then:
7.5x10⁻⁵ = x²/0.773
x² = 5.7975x10⁻⁵
x = √5.7975x10⁻⁵
x = 7.61x10⁻³ M
The supposing is correct, so [CO] = 7.61x10⁻³ x 10³ = 7.61
Answer would be B. Stirring allows parts of the solution to come into contact with what you are trying to stir into a “new solution”. Think of it like shaking a smoothie bottle when you see separation happen after not touching the smoothie for a while :)
1) PV=nRT
P=738.0 mmHg
V=15.5mL=0.0155 L
T=273+25=298 K
R=62.36 L*mmHg*K⁻¹mol⁻¹
n=PV/RT
n=(738.0 mmHg *0.0155 L)/(62.36 L*mmHg*K⁻¹mol⁻¹*298 K)= =0.000616=6.16*10⁻⁴ mol
2) From the equation of the reaction
1 mol CaCO3 gives 1 mol CO2,
so 6.16*10⁻⁴ mol CaCO3 ----> 6.16*10⁻⁴ mol CO2
Molar mass CaCO3 =M(Ca)+M(C)+3*M(O)= 40.1+12.0+3*16.0 =100.1 g/mol
6.16*10⁻⁴ mol CaCO3 * 100.1 g/mol =617*10⁻⁴ g =0.0617 g = 61.7mg
Answer:
a) 0.0265M
b) 8.35×10^-4M
c) 9.833×10^-8M
Explanation:
First we must write down the equation of each reaction. After identifying the insoluble precipitate, we go ahead to write down the expression for its solubility product. If the solubility product of the compound is known, the concentration of the precipitating agent can now be calculated from the available information in the question.
I hope it helps you ❤️❤️❤️❤️
