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Kobotan [32]
3 years ago
13

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. K2SO4 NH4I CoCl3

Chemistry
1 answer:
katrin [286]3 years ago
8 0
The substance that releases the greatest amount of ions will have the greatest attractive forces within its solution, resulting in a reduced freezing point.
K₂SO₄ yields 3 ions
NH₄I yields 2 ions
CoCl₃ yields 4 ions

Freezing points:
CoCl₃ < K₂SO₄ < NH₄I
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Covalent network. <span>A solid that is extremely hard, that has a very high melting point, and that will not conduct electricity either as a solid or when molten is held together by a continuous three-dimensional network of covalent bonds. Examples include diamond, quartz (SiO </span><span>2 </span>), and silicon carbide (SiC). The electrons are constrained in pairs to a region on a line between the centers of pairs of atoms.<span>


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4 0
3 years ago
S-2-iodooctane lost his opticaly activity in soln on tratment wwithNaI. Explain
IceJOKER [234]

Answer:

Repeated SN2 reactions occur leading to the formation of a racemic mixture

Explanation:

S-2-iodooctane is a chiral alkyl halide with an asymmetric carbon atom. The presence of an asymmetric carbon atom implies that it can rotate plane polarized light and thus lead to optical isomerism. The two configurations of the compound are R/S according to the Cahn-Prelong-Ingold system.

However, when S-2-iodooctane is treated with sodium iodide in acetone, repeated SN2 reactions occur since the iodide ion is both a good nucleophile and a good leaving group. Hence a racemic modification is formed in the system with time hence we end up with (±)- Iodooctane.

5 0
3 years ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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