The freezing point depression is a colligative property, which means that it depends on the number of particles of solute disolved in the solution.
When you have solutes that are ionic compounds they dissociate in water into ions, then the compound that dissociates more ions will produce more particles and will decrease the freezing point the most.
Given theses aqueous solutions Na2 CO3, Co Cl3, and Li NO3 you can predict the order of the freezing points.
First, write the dissociation equations>
Na2CO3 -> 2Na(+) + CO3 (2-) These are 3 ions: two of Na(+) and one of CO3(2-)
The number inside parenthesis are number of charge not number of molecules.
Co Cl3 -> Co(3+) + 3 Cl (1-) Those are 4 ions: one of Co (+) and three of Cl (-)
Li NO3 -> Li (+) + NO3 (-) those are two ions: one of Li (+) and one of NO3(-)
Then the ionic compound that dissociates into more ions give the solution with lower freezing point, and these is the rank from higher to lower freezing point:
Li NO3 > Na2 CO3 > Co Cl3.
Answer:
1) HCl contains the Cl^- which is a good nucleophile
2) 2-methyl-2- heptanol > 2-heptanol > 1-heptanol
3) see image attached
Explanation:
If the dehydration of alcohols is carried out using HCl, the chloride ion which is a good nucleophile will attack the substrate to yield an undesirable product.
The dehydration of alcohols is an E1 reaction. Recall that the ease of E1 reaction increases in the order 3°> 2°> 1°. Hence, 2-methyl-2- heptanol forms a tertiary carbocation intermediate during dehydration and has the greatest ease of dehydration.
The three products formed during the dehydration of 3,3-dimethyl-2-butanol are shown in the image attached. Two out of the three are formed by rearrangement reactions.
It provides atomic mass, mass excess, nuclear binding energy, nucleon separation energies, Q-values, and nucleon residual interaction parameters for atomic nuclei of the isotope Na-24 (Sodium, atomic number Z = 11, mass number A = 24).
Answer:
The product is cyclohexanol
Explanation:
Firstly,
A ketone undergo a borohydride reduction reaction to form an alcohol as below,
R-CO-R' ⇒ R-CO(OH)-R'
- IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹
- From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.
- Check with other spectroscopic properties,
- 3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-<u>C</u>H₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-<u>C</u>H₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-<u>C</u>H(CH₂-)-OH).
1.56 δ (4H, triplet) - (-C<u>H</u>₂-CH₂-CH-OH) ; triplet as coupling with 2 H,
1.78 δ (4H, multiplet) - (-CH₂-C<u>H</u>₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH
3.24 δ (1H, quintet); - (-CH₂-CH₂-C<u>H</u>(CH₂-)-OH), coupling with4 H of 2 group of CH₂
3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-O<u>H</u>), hydrogen of alcohol group, not tend to coupling with other hydrogen
See , from the equation we can see that for forming two mole of H2O 2Mole of H2 has to react.
Mass of 2 Mole H2O is 18*2gm or 36gm.
So for forming 36 gm H2O 2×2 I.e. 4 gm H2 has to take part in reaction.
Therefore, to form 1 gm H2O 4÷36 gm of H2 has to take part.
So, for forming 47gm H2O (4÷36)×47 gm H2 has to take part
I.e. 5.22 gm of H2 has to take part
So, ans is 5.22 gm of hydrogen.
Hope it helps!!!