Hi! To find the amount of grams in 2.34 moles of cobalt, you must use the following equation:
2.34 mol Co * (58.93 g Co)/(1 mol Co)
The answer would therefore be 137.9 grams
Answer:
The correct option is;
A) x = 2, y = 1
Explanation:
Here we have
![k[0.05]^x \cdot [0.05]^y = 0.062 \ M/sec](https://tex.z-dn.net/?f=k%5B0.05%5D%5Ex%20%5Ccdot%20%5B0.05%5D%5Ey%20%3D%200.062%20%5C%20M%2Fsec)
....................(1)
![k[0.05]^x \cdot [0.10]^y = 0.123 \ M/sec](https://tex.z-dn.net/?f=k%5B0.05%5D%5Ex%20%5Ccdot%20%5B0.10%5D%5Ey%20%3D%200.123%20%5C%20M%2Fsec)
....................(2)
![k[0.100]^x \cdot [0.100]^y = 0.491 \ M/sec](https://tex.z-dn.net/?f=k%5B0.100%5D%5Ex%20%5Ccdot%20%5B0.100%5D%5Ey%20%3D%200.491%20%5C%20M%2Fsec)
..............(3)
From experiment (1), (2) it is observed that [A] is held constant and [B] is doubled of which the rate is also observed to be doubled, therefore, when you double a reactant which result in the rate being doubled, the order of the reaction with respect to the reactant is order 1, therefore, y = 1.
Similarly between reaction (2) and (3) when the concentration of the reactant B is doubled and A is held constant the rate of the reaction is multiplied by a factor of 4, therefore, the reaction with respect to that reactant is order 2, therefore, x = 2.
Therefore, the correct option is x = 2, y = 1.
Answer and Explanation:
We need to remember the formulas:
- pOH = -log[OH-]
- pH = -log[H+]
- pH + pOH = 14
When given the concentration of OH-, denoted as [OH-], we can plug that value into the equation pOH = -log[OH-] to find the pOH. Once we have found the pOH, we can then subtract that value from 14 to determine the value of pH.
When light hits a surface, some of it is absorbed and some of it is reflected. The light that is reflected is the color of the object in that light. For example, a blue object absorbs all the colors of the spectrum except blue, it reflects blue light. hope this can help.
Answer:
the potential with respect to a silver–silver chloride electrode is 0.715
Explanation:
the potential to a silver chloride electrode can be derived from the question above by saying that
the potential with respect to a silver–silver chloride electrode? = 0.9120v - 0.197 V
= 0.715
therefore, the potential with respect to a silver–silver chloride electrode is 0.715
