Urea is highly soluble in water. When it is allowed to dissolve in water in the presence of heat, it will yield ammonia and carbon dioxide. The reaction is shown below:
<span>NH2-CO-NH2 + H2O </span>⇒ 2 NH3 + CO2
As you can observe in the stoichiometric equations, 1 molecule of water can dissolve with 1 mole of urea.
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Answer:
John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? xdsz.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts?
Explanation:
The answer is epipelagic zone
