Following reaction is involved in above system
HOCl(aq) ↔ H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K = [OCl-][H+<span>]/[HOCl] ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>) ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14
</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these
Answer:
The atomic number on the Periodic Table identifies the number of protons in any atom of that element. Copper, atomic number 29, has 29 protons. Finding the atomic number of an element reveals the number of protons.
To find the number of neutrons in the atom, subtract the atomic number from the atomic mass.
Answer:H2=11.4g
CH4=28.6g
Explanation:The complete combustion of the two gases can be represented by a balanced reaction below
1. CH4 +2O2___CO2+2H2O
2.2H2+O2___2H2O
Combining the two we have CH4 +2H2+3O2___
CO2+4H2O
Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.
Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g
Mass of CH4=20/28×40=28.6g
2H2=8/28×40=11.4g
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
