Well ionic compounds are brittle because similar charges repel, which causes them to break usually occurring when a force is applied, like a hammer strike.

8 1

3 years ago

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

laiz [17]

35.0 mL of 0.210 M
KOH molarity = moles/volume
find moles of OH do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+ they will cancel out: H+ + OH- -> H2O
but you'll have some left over,
pH=-log[H+] pOH
=-log[OH-] pH+pOH
=14

3 0

3 years ago

Read 2 more answers

Ketones undergo a reduction when treated with sodium borohydride, NaBH4. The product of the above reaction has the following spe

Anna71 [15]

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, (-CH₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-CH₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-CH(CH₂-)-OH).

1H NMR confirms,

1.56 δ (4H, triplet) - (-CH₂-CH₂-CH-OH) ; triplet as coupling with 2 H,

1.78 δ (4H, multiplet) - (-CH₂-CH₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH

3.24 δ (1H, quintet); - (-CH₂-CH₂-CH(CH₂-)-OH), coupling with4 H of 2 group of CH₂

3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-OH), hydrogen of alcohol group, not tend to coupling with other hydrogen

4 0

3 years ago