Answer:
There are 5.43 grams of NaOCl
Explanation:
The given percent by mass of the solute tells us that out of the 150 g of the solution, 3.62% are due solely to the solute.
In other words, <u>the mass of the solute in the solution is</u>:
- 150 g * 3.62/100 = 5.43 g
Thus, in 150 grams of the given bleach solution, there are 5.43 grams of sodium hypochlorite.
9 amino acids (alanine, cysteine, glycine, isoleucine, leucine, methionine, phenylalanine, proline, valine) have no hydrogen donor or acceptor atoms in their side chains.
Answer:
See Explanation ( = same answer for earlier question)
Explanation:
The Arrhenius acid-base theory defines an acid as a compound which when added into water increases the hydronium ion (H₃O⁺) concentration and the base as a compound which when added into water increases the hydroxide (OH⁻) ion concentration. As such, an acid-base reaction is limited to proton transfer to only OH⁻ ions forming water. Such would imply that all acid-base reactions produce water only in addition to a salt. This is not always the case as conjugate base anions for many substances can receive proton transfer.
Example: The reaction HOAc + NaCN => HCN + OAc- will occur in aqueous media because the proton (H⁺) on acetic acid (HOAc) will transfer to the cyanate ion forming hydrocyanic acid (HCN). Such occurs because the CN⁻ ion is a stronger conjugate base than the acetate ion (OAc⁻) and forms the more stable weak acid. Such is the basis of the Bronsted-Lowry Acid-Base system and states that an acid (proton donor) will transfer its ionizable hydrogen to a conjugate base (proton acceptor) if the transfer forms a weaker acid.
Read more on Brainly.com - brainly.com/question/15316776#readmore
<u>Answer:</u> Copper (I) iodide will precipitate first.
<u>Explanation:</u>
We are given:
of CuCl = 
of CuI = 
Concentration of 
Concentration of 
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
![K_{sp}=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
Putting values in above equation, we get:
![1.0\times 10^{-6}=[Cu^+]\times 0.021](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-6%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.021)
![[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-6%7D%7D%7B0.021%7D%3D4.76%5Ctimes%2010%5E%7B-5%7DM)
Concentration of copper (I) ion = 
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
Putting values in above equation, we get:
![5.1\times 10^{-12}=[Cu^+]\times 0.017](https://tex.z-dn.net/?f=5.1%5Ctimes%2010%5E%7B-12%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.017)
![[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B5.1%5Ctimes%2010%5E%7B-12%7D%7D%7B0.017%7D%3D3.00%5Ctimes%2010%5E%7B-10%7DM)
Concentration of copper (I) ion = 
For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.
