Answer: either B, or D
Explanation:
because one of those are the answer
Answer:
The correct answer is option b.
Explanation:
The ionic product of water : 
![K_w=[H^+][OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
A pure water has equal concentration of hydrogen ions and hydroxide ions, hence neutral.
![K_w=[H^+][H^+]=[H^+]^2](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BH%5E%2B%5D%3D%5BH%5E%2B%5D%5E2)
The ionic product of water at 283 K = 
![K_w=[H^+]^2](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5E2)
![0.29\times 10^{-14}=[H^+]^2](https://tex.z-dn.net/?f=0.29%5Ctimes%2010%5E%7B-14%7D%3D%5BH%5E%2B%5D%5E2)
![[H^+]=5.385\times 10^{-8} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5.385%5Ctimes%2010%5E%7B-8%7D%20M)
The pH of the water at 283 k;
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![=-\log[5.385\times 10^{-8} M]=7.26](https://tex.z-dn.net/?f=%3D-%5Clog%5B5.385%5Ctimes%2010%5E%7B-8%7D%20M%5D%3D7.26)
A pure water has equal concentration of hydrogen ions and hydroxide ions,So water will e neutral at this temperature also.
The correct answer is option b.
Answer:
A
Explanation:
Crabs were exposed to a toxin in the water.
In general, the further away an electron is from the nucleus, the easier it is for it to be expelled. In other words, ionization energy is a function of atomic radius; the larger the radius, the smaller the amount of energy required to remove the electron from the outer most orbital. For example, it would be far easier to take electrons away from the larger element of Ca (Calcium) than it would be from one where the electrons are held tighter to the nucleus, like Cl (Chlorine). Hope this helped a little not the exact answer though :)
Answer:
None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.
Explanation:
Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.
Before balanced Left side.
Cl-2
O-8
H-2
Before balanced right side.
H-1
Cl-1
O-3
That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.
(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)
