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olga2289 [7]
3 years ago
6

X là este no, đơn chức, mạch hở. Xà phòng hóa một lượng cần 100ml dung dịch KOH 1M , thu được 11,2 gam muối ; và 3,2 gam ancol.

Công thức phân tử của X là
Chemistry
1 answer:
Evgesh-ka [11]3 years ago
3 0

Answer:

Giá trị xà phòng hóa hoặc số xà phòng hóa (SV hoặc SN) biểu thị số miligam kali hydroxit (KOH) hoặc natri hydroxit (NaOH)

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You are given a bottle of solid X and three aqueous solutions of X—one saturated, one unsaturated, and one supersaturated. How w
quester [9]

In a saturated solution, extra solid X would remain solid, dissolve in an unsaturated solution, and crystallize in a supersaturated one.

A solution is said to be saturated when there is a maximum amount of solute present that has been dissolved in the solvent. As a result, the system is in an equilibrium between the dissolved and undissolved solutes: A solution is considered to be unsaturated if the solute concentration is less than the equilibrium solubility. A supersaturated solution is one that has more solute than is necessary to generate a saturated solution at a given temperature.

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5 0
2 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)

3 0
3 years ago
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