Answer:
Ksp = 8.8x10⁻⁵
Explanation:
<em>Full question is:</em>
<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>
<em />
When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:
PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻
Ksp = [Pb²⁺] [Cl⁻]²
If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):
Ksp = [X] [2X]²
Ksp = 4X³
As X is the amount of Pb²⁺ = 2.8x10⁻²M:
Ksp = 4(2.8x10⁻²)³
<h3>Ksp = 8.8x10⁻⁵</h3>
Answer:
A population or community research line can be carried out, wherever at a certain point in time, regardless of whether it is a cross-sectional study.
In addition, the people who would be the population to be studied or the object of study might or might not know the cause of the study (blind) while the researcher could be experimentally participatory.
Explanation:
They are prevalence studies, in which the presence of a health condition or state is determined in a well-defined population and in a determined time frame: one day, one week, a particular moment in life, even if it does not temporarily coincide in all the subjects (for example, the blood pressure figures at the time of entering the school or at the beginning of the holidays, the prevalence of diabetes in hospitalized patients on a given day, etc.).
They are like "photographs" of a state of affairs at a given moment. The simultaneous determination of what is understood by exposure and event does not allow defining causality.
Answer:
CCl2=CCl2
Explanation:
Recall that the individual dipole moments in a symmetrical molecule cancel out each other thereby making the overall molecule to be non polar.
CCl2=CCl2 contain individual polar bonds but the molecule is highly symmetrical thus its dipole cancel out and the molecule has no overall dipole moment, hence CCl2=CCl2 has a dipole moment of zero .
Answer is: <span>
The reaction will not be spontaneous at any temperature.
</span>
<span>Gibbs free energy
(G) determines if reaction will proceed spontaneously.
ΔG = ΔH - T·ΔS.
ΔG - changes in Gibbs free energy.
ΔH - changes in enthalpy.
ΔS - changes in entropy.
T is temperature in Kelvins.
When ΔS < 0 (negative entropy change) and ΔH > 0
(endothermic reaction), the process is never spontaneous (ΔG> 0).</span>
