Answer:
200 grams of NaOH are produced with the reaction of 5.00 moles of water
Explanation:
First of all you apply a rule of three to know the amount of moles of NaOH as follows: if 2 moles of water produce 2 moles of NaOH, 5 moles of water how many moles would they produce?

moles of NaOH= 5
Being:
- Na: 23 g/mole
- O: 16 g/mole
- H: 1 g/mole
the molar mass of NaOH is: 23 g/mole + 16 g/mole + 1 g/mole= 40 g/mole
Then a rule of three applies as follows: if in 1 mole there are 40 g of NaOH, in 5 moles how much mass is there?

mass= 200 g
<u><em>200 grams of NaOH are produced with the reaction of 5.00 moles of water</em></u>
Answer:
CO is considered as a product.
Explanation:
A general chemical equation for a combination reaction follows:
To write a chemical equation, we must follow some of the rules:
The reactants must be written on the left side of the direction arrow.
A '+' sign is written between the reactants, when more than one reactants are present.
An arrow is added after all the reactants are written in the direction where reaction is taking place. Here, the reaction is taking place in forward direction.
The products must be written on the right side of the direction arrow.
A '+' sign is written between the products, when more than one products are present.
For the given chemical equation:
are the reactants in the reaction and are the products in the reaction.
Hence, CO is considered as a product.
Glacial erosion is the process by which a glacial flows over the land, picking up rocks.
Answer : The pOH of pure water is, 6.68
Explanation :
As we are given that:

First we have to calculate the concentration of hydroxide ion.
As, ![K_w=[H^+]\times [OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5Ctimes%20%5BOH%5E-%5D)
As we know that in pure water the hydrogen ion and hydroxide ion concentration are equal. That means,
![[H^+]=[OH^-]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BOH%5E-%5D)
So, ![K_w=[OH^-]\times [OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BOH%5E-%5D%5Ctimes%20%5BOH%5E-%5D)
![K_w=[OH^-]^2](https://tex.z-dn.net/?f=K_w%3D%5BOH%5E-%5D%5E2)
![4.38\times 10^{-14}=[OH^-]^2](https://tex.z-dn.net/?f=4.38%5Ctimes%2010%5E%7B-14%7D%3D%5BOH%5E-%5D%5E2)
![[OH^-]=2.09\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2.09%5Ctimes%2010%5E%7B-7%7DM)
Now we have to calculate the pOH.

Therefore, the pOH of pure water is, 6.68