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Paul [167]
3 years ago
9

Calculate the molarity of glucose in a solution that which contains 35 g of glucose in 0.16 kg of phenol

Chemistry
1 answer:
sleet_krkn [62]3 years ago
5 0
C(Molarity) = n of solute/V of solution (mol/L)

glucose(C6H12O6) = 180g/mol
glucose 35g = 35g/(180g/mol) = 0.1944mol

We need density of solution here, and I assume it as density of phenol, 1.07g/mL. (I don't know why the question doesn't contain it)

phenol 0.16kg = 0.16kg/(1.07kg/L) = 0.1495L

0.1944mol/0.1495L = 1.300M(mol/L)
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Covalent bonds are formed by sharing of electrons between non metals

For example, In calcium iodide the one electron from calcium metal gets transferred to iodine atom and thus form an ionic bond to give CaI_2

Electronic configuration of calcium:

[Ca]=1s^22s^22p^63s^23p^64s^2

Calcium atom will lose two electron to gain noble gas configuration and form calcium cation with +2 charge.

[Ca^{2+}]=1s^22s^22p^63s^23p^6

Electronic configuration of iodine:

[I]=1s^22s^22p^63s^23p^64s^23d^{10}4p^5

Iodine atom will gain one electron to gain noble gas configuration and form iodide ion with -1 charge.

[I^-]=1s^22s^22p^63s^23p^64s^23d^{10}4p^6

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A forward reaction in which adding heat decreases product formation is _________, while a forward reaction in which adding heat
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4 years ago
Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

3 0
3 years ago
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