A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:
- Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.
- Propagation; free radicals react with molecules to produce new free radicals and molecules.
- Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.
Initiation
where the big black dot indicates unpaired electrons attached to the atom.
Propagation
Termination
0.83 m/s seems the correct answer, hope it helps
By the second law of thermodynamics:
Heat can not spontaneously flow from cold regions to hot regions without external work being performed on a system.
Heat transfer is the passage of thermal energy from a hot ( t B = 80° C ) to a colder body ( t A = 40° C ).
Answer: B ) Heat flows from object B to object A.
Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>
Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>
