W=m₁/m₀=2^(-t/T)
t=4.6·10⁹ years
T=5·10¹⁰ years
w=2^(-4.6·10⁹/5·10¹⁰)
w=0.9382
w=93.82%
The empirical formula for the unknown compound would be: C2H4O (2 molecules of Carbon, 4 molecules of Hydrogen, and 1 molecule of Oxygen)
1) HOBr stands for hypobromous acid. On reacting with water, products formed are OBr- and H3O+. Following reaction occurs during this process.
<span> HOBr + H2O </span>⇄<span> OBr- + H3O+
2) HOBr is a weak acid and have a lower value of dissociation constant (Ka ~ </span><span>2.3 X 10^–9). Hence, </span><span> large number of undissociated HOBr molecules are left in solution, when the reaction is completed/reaches equilibrium.</span>
Answer:
1,2,1,2
Explanation:
You would need only one of the CH4 but 2 of the O2 then 1 CO2 and 2 H2O on each side of the equation you now have 1 carbon, 4 hydrogen, and 4 oxygen.