Answer:
You have to remember this memory aid, diagonals and all.
You have to keep a tight tally of the electrons you’ve used so far so you don’t go over the number of electrons in the element you’re working on.
You have to remember how many electrons fit into each subshell (s, p, d, f).
It takes a lot of time, especially when the element has more than 20 electrons.
Explanation:
Yes if a molecule with a COOH group is called a carboxylic acid.
hope that helps
Well you know that you have to balance out the equation to 0. So if I is -1 and Ca is +2 you know that I had to equal to -2. So you can do that by having two of them. So Ca=+2 and I2=-2 so CaI2 would equal 0 meaning that would be the correct answer. Hope that I explained that correct :)
Flat as more oxygen and water can react over it think of it like this would a cube rust faster than a sheet
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.
