Answer:
18.94%.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = 13,750 years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = ??? %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]
1.664 = ln (100.0%)/[A]
Taking exponential for both sides:
5.279 = (100.0%)/[A]
<em>∴ [A]</em> = (100.0%)/5.279 = <em>18.94%.</em>
Answer:
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Explanation:
According to the Arrhenius equation,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate of reaction at 
= rate of reaction at 
= activation energy of the reaction
R = gas constant = 8.314 J/K mol
![\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B300%2C000%20J%2Fmol%7D%7B2.303%5Ctimes%208.314%20J%2FK%20mol%7D%5B%5Cfrac%7B1%7D%7B798.15%20K%7D-%5Cfrac%7B1%7D%7B898.15%20K%7D%5D)
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Answer: The answer is A. - 4.88x10^20 H2O2 molecules
Explanation: I hope this helps!
The half-reaction are:
Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.
NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.
Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.
ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>
Formic acid when in water would dissociate into ions just like any acids. It would dissociate into the hydrogen ion and the formate ion. The equilibrium dissociation equation would be written as:
<span>HCOOH (aq) + H2O (l) ⇌ H+ (aq) + HCOO- (aq)
Formic acid is a weak acid which means that when in aqueous solution it does not completely dissociate into its corresponding ions. Only a certain amount that would be dissociated so in the solution there will be HCOOH, HCOO- and H+ molecules. It is also known as Methanoic acid and an important substance for the synthesis of a number of substances. It is naturally occurring in ants.</span>
