Answer: (a) The solubility of CuCl in pure water is
.
(b) The solubility of CuCl in 0.1 M NaCl is
.
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.

Initial: 0 0
Change: +x +x
Equilibm: x x

And, equilibrium expression is as follows.
![K_{sp} = [Cu^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCu%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)

x = 
Hence, the solubility of CuCl in pure water is
.
(b) When NaCl is 0.1 M,
, 
, 
Net equation: 
= 0.1044
So for, 
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = 
0.1044 = 
x = 
Therefore, the solubility of CuCl in 0.1 M NaCl is
.
Hydrogen bonds<span> hold the paired nitrogenous bases together. Because </span>hydrogen bonds<span> are weak </span>bonds<span>, the two strands of </span>DNA<span> are easily separated a characteristic that is important to </span>DNA's<span> function.</span>Hydrogen bonds<span> form between adenine and thymine and between cytosine and guanine.</span>
This is called distillation. I think I spelled that right
Answer:
The usage is explained below
Explanation:
Amines are identified from the use of the hinsberg reaction whereby the amine is mixed well with Hinsberg reagent and done in the presence of an aqueous alkali. In this reaction, the amine will attack the electrophile which will result in the chloride being displaced and the amides being generated.
The uses of identification of amines in daily life are;
- In agriculture, it serves as a good source of making herbicides as well as acting as emulsifiers.
- In medicine, it is used in the manufacture of some popular pain killers like demerol and morphine.
- In medicine, it is used in the manufacture of an anesthetic drug known as novocaine.
- In chemical industries, amines are used as lubricating oils and also as corrosion inhibitors in boilers.
I would use dimensional analysis for this problem. You would start with the given amount of Na which is 59.0g. You are trying to find the moles of NaCl.
(59.0g Na)*(1 mol Na)/(22.99 g Na)*(2 mol NaCl)/(2 mol Na)=
2.57 mol NaCl
The units will cancel out until you are left with moles of NaCl. The answer has the correct number of significant figures (3 sig figs).