The correct option is C. The paragraph is biased, not taking into consideration the other great inventions that have been made. The information given are vague, over generalised, without providing any evidence to support its claims. They are blanket statementso and are not useful at all.
Answer:
100mL of 0.10M HNO2 and 0.10M NaNO2
Explanation:
because solution has the greatest buffering capacity when the concentration of the weak acid is = at the concentration of its conjugate base.
Answer:
a layer of subsoil must have porosity. That is, it must have empty spaces, or pores. In order to allow groundwater to easily pass through the subsoil, A. the subsoil's pores must be large
Explanation:
Answer:
they have same number of electrons
Here is the correct question:
The reaction A → products was studied at a series of different temperatures. A plot of ln(k) vs. 1/T gave a straight line relationship with a slope of -693 and a y-intercept of -0.425. Additionally, a study of the concentration of A with respect to time showed that only a plot of 1/[A] vs. time gave a straight line relationship. What is the initial rate of this reaction when [A] = 0.41 at 271 K ?
Answer:
the initial rate of this reaction is 0.0216275 M/sec
Explanation:
Using the formula:
![K = Ae^{\frac{-Ea}{RT}}](https://tex.z-dn.net/?f=K%20%3D%20Ae%5E%7B%5Cfrac%7B-Ea%7D%7BRT%7D%7D)
![InK = InA + (\frac{-Ea}{R})(\frac{1}{T})\\y \ \ \ \ \ \ \ \ \ \ c \ \ \ \ \ \ \ \ \ \ m \ \ \ \ x](https://tex.z-dn.net/?f=InK%20%3D%20InA%20%2B%20%28%5Cfrac%7B-Ea%7D%7BR%7D%29%28%5Cfrac%7B1%7D%7BT%7D%29%5C%5Cy%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20c%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20m%20%20%5C%20%5C%20%5C%20%5C%20x)
![= -693](https://tex.z-dn.net/?f=%3D%20-693)
![\frac{Ea}{8.314}= 693 \ \\ \\ Ea = 693 * 8.314 \\ \\ Ea = 5671.602 \ J](https://tex.z-dn.net/?f=%5Cfrac%7BEa%7D%7B8.314%7D%3D%20693%20%5C%20%5C%5C%20%5C%5C%20Ea%20%3D%20693%20%2A%208.314%20%5C%5C%20%5C%5C%20Ea%20%3D%205671.602%20%5C%20J)
![In A = -0.425 \ \ \\ \\ A = e^{-0.425} \\ \\ A = 0.6538](https://tex.z-dn.net/?f=In%20A%20%3D%20-0.425%20%5C%20%5C%20%5C%5C%20%5C%5C%20%20A%20%3D%20e%5E%7B-0.425%7D%20%5C%5C%20%20%20%5C%5C%20%20A%20%3D%200.6538)
![K = 0.6538 e^{- (\frac{5761.602}{8.314*271})](https://tex.z-dn.net/?f=K%20%3D%200.6538%20e%5E%7B-%20%28%5Cfrac%7B5761.602%7D%7B8.314%2A271%7D%29)
![K = 0.05275 \\ \\ K = 5.275*10^{-2}](https://tex.z-dn.net/?f=K%20%3D%200.05275%20%5C%5C%20%5C%5C%20K%20%3D%205.275%2A10%5E%7B-2%7D)
Since
vs time is a straight line relationship;
Therefore, it is a second order reaction
rate = K[A]²
rate = 5.275 × 10⁻² × (0.41)
rate = 0.0216275 M/sec