Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
The average mass of an atom is calculated with the formula:
average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ... an so on
For the boron we have two isotopes, so the formula will become:
average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)
We plug in the values:
10.81 = 0.1980 × 10.012938 + 0.8020 × mass of isotope (2)
10.81 = 1.98 + 0.8020 × mass of isotope (2)
10.81 - 1.98 = 0.8020 × mass of isotope (2)
8.83 = 0.8020 × mass of isotope (2)
mass of isotope (2) = 8.83 / 0.8020
mass of isotope (2) = 11.009975
mass of isotope (1) = 10.012938 (given by the question)
Chemical is the answer your looking for.
Answer:
2.2×10^8
Explanation:
Cu(OH)2(s)<---------> Cu^2+(aq) + 2OH^-(aq) Ksp=2.2 x 10 ^-20
2H3O^+(aq) + 2OH^-(aq) <-------> 4H2O(l). Kw= 1×10^14
Cu^2+(aq) + 4H2O(l) <--------> [Cu(H2O)4]^2+(aq)
Overall ionic reaction:
Cu(OH)2(s) +2H3O^+(aq) <---------> [Cu(H20)4]^2+(aq)
Equilibrium constant for the reaction: Ksp×Kw= 2.2 x 10 ^-20 × (1/(1×10^-14))^2
Keq= 2.2×10^8
Kw= ion dissociation constant of water