Answer:
Explanation:
Hello,
The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:
![\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Now, we answer:
(a)
![\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D%3D4%2A%5Cfrac%7B1%7D%7B-3%7D%20%2A%28-0.12M%2Fs%29%3D0.16M%2Fs)
(b)
![\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%3D-1%2A%5Cfrac%7B1%7D%7B5%7D%20%2A%280.2M%2Fs%29%3D-0.04M%2Fs)
(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:
![r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Thus, any of the available expressions are suitable to quantify the rate of the reaction.
Best regards.
Answer:
Explanation:
Double Bond => An Alkene molecule
So, the suffix will be "-ene"
7 Carbons => So, we'll use the prefix "Hept-"
Combining the suffix and prefix, we get:
=> Heptene
Answer:
<h2>134km = 13400000cm</h2><h2>35g = 35000000ug</h2><h2>0.65mmol = 0.00065mol</h2>
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K
The formula for water is H2O so there would have to be two Hyrdogens and one oxygen. Therefore it would be 4g of Hydrogen and 16g of Oxygen leaving you with 20g.
The answer is D.
Hope this helps :) ~
