Answer:
To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
Explanation:
Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.
So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles
We know, number of moles of a compound is the ratio of mass to molar mass of that compound.
So, mass of 2 moles of urea = 
Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
So, option (C) is correct.
the compounds in which phosphorous posses the highest possible oxidation have to mention here.
The species in which phosphorous have the highest oxidation state are: H₃PO₄, P₂O₅, PCl₅
The possible oxidation state of phosphorous is III and V. The highest oxidation state is V. There are several compounds in which phosphorous posses the +5 oxidation state. Like- Phosphoric acid (H₃PO₄), phosphorous pentoxide (P₂O₅), Phosphorous chloride (PCl₅) etc.
The oxidation state of an element depends upon the valence electron the valence shell of phosphorous is 3s² 3p³. Thus there are 5 electrons, as it has vacant 3d orbital thus it can easily form compound having +5 oxidation state.
Here is the answer. Although its chief uses are in the preparation of sulfuric acid, sulfur trioxide, and sulfites, sulfur dioxide also is used as a disinfectant, a refrigerant, a reducing agent, a bleach, and a food preservative, especially in dried fruits.
Answer:
96 grams
Explanation:
Density = mass / volume Multiply both sides by the volume
density * volume = mass
mass = 8 g / cm^3 * 12 cm^3
mass = 96 grams
Answer:
1, 2, and 3 are true.
Explanation:
The Henderson-Hasselbalch equation is:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
- If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If you know pH and pka:
10^(pH-pka) = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
The ratio will be: 10^(pH-pka)
- At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
0 = log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
10^0 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
1 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
As ratio is 1, [conjugate base] = [acid] in solution.
- At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]
- At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]
I hope it helps!