Answer:
14 g.
Explanation:
- From the figure attached:
<em>the solubility of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC is </em><em>(66 g).</em>
When beaker containing 80 grams of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC.
<em>∴ The grams of the salt are undissolved, on the bottom of the beaker are </em><em>(14 g).</em>
Did you get the answer? I need help on that one too
Answer:
Option c → Tert-butanol
Explanation:
To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.
The formula is:
ΔT = Kf . m . i
When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.
i = Van't Hoff factor (ions particles that are dissolved in the solution)
At this case, the solute is nonvolatile, so i values 1.
ΔT = Difference between fussion T° of pure solvent - fussion T° of solution.
T° fussion paradichlorobenzene = 56 °C
T° fussion water = 0°
T° fussion tert-butanol = 25°
Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.
Living things have many characteristics for example we are made up of cells, we use energy and we reproduce
Answer: Tertiary alcohol.
Epoxide,
phenols,
Ethers have two alkyl groups bonded to oxygen atoms
Enols
primary alcohols,
alcohols
Explanation:Tertiary alcohol are alcohols having the general structure R₃COH. Epoxide are cyclic ether with a three-membered ring. The basic structure of an epoxide contains an oxygen atom attached to two adjacent carbon atoms of a hydrocarbon. Phenol is an aromatic organic compound with the molecular formula C₆H₅OH. An alcohol contains a hydroxy group (OH group) bonded to an sp3 hybridized carbon atom. Enols are a type of reactive structure or intermediate in organic chemistry that is represented as an alkene with a hydroxyl group attached to one end of the alkene double bond. Alcohols are classified like halides and hydrogens; primary (1°), secondary (2°), or tertiary (3°) depending on the number of carbons bonded to the carbon with the halogen.
