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3 years ago

6

A duck migrates South for the winter about 3000 miles each year. It takes them about 65 hours to travel this distance. What is t

he velocity of the duck?

1 answer:

olchik [2.2K]3 years ago

7 0

Answer:

46.15

Explanation:

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Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t

Oksi-84 [34.3K]

Answer:

Explanation:

Given that:

Concentration of $HC_2H_3O_2 \ (M_1)$ = 0.105 M

Volume of $HC_2H_3O_2 \ (V_1)$ = 20.0 mL

Concentration of $NaOH (M_2)$ = 0.125 M

The chemical reaction can be expressed as:

$HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}$

Using the ICE Table to determine the equilibrium concentrations.

$HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}$

I 0.105 0 0

C -x +x +x

E 0.105 - x x x

$K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}$

$K_a = \dfrac{(x)(x)}{(0.105-x)}$

Recall that the ka for $HC_2H_3O_2= 1.8 \times 10^{-5}$

Then;

$1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}$

$1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}$

By solving the above mathematical expression;

x = 0.00137 M

$H_3O^+ = x = 0.00137 \ M \\ \\ pH = - log [H_3O^+] \\ \\ pH = - log ( 0.00137 )$

pH = 2.86

Hence, the initial pH = 2.86

b) To determine the volume of the added base needed to reach the equivalence point by using the formula:

$M_1 V_1 = M_2 V_2$

$V_2= \dfrac{M_1V_1}{M_2}$

$V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}$

$V_2 = 16.8 mL$

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of $HC_2H_3O_2 =$ molarity × volume

$= 0.105 \ M \times 20.0 \times 10^{-3} \ L$

$= 2.1 \times 10^{-3}$

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = $0.625 \times 10^{-3}$

After reacting with 5.0 mL NaOH, the number of moles is as follows:

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Initial moles $2.1*10^{-3}$ $0.625 * 10^{-3}$ 0 0

F(moles) $(2.1*10^{-3} - 0.625 \times 10^{-3})$ 0 $0.625 \times 10^{-3}$ $0.625 \times 10^{-3}$

The pH of the solution is then calculated as follows:

$pH = pKa + log \dfrac{[base]} {[acid]}$

Recall that:

pKa for $HC_2H_3O_2=4.74$

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

$pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}$

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

d)

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the $(C_2H_3O^-_2)$ with $H_2O$

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of $(C_2H_3O^-_2)$ = moles/volume

= $\dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}$

= 0.0571 M

Now, using the ICE table to determine the concentration of $H_3O^+$ ;

$C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}$

I 0.0571 0 0

C -x +x +x

E 0.0571 - x x x

Recall that the Ka for $HC_2H_3O_2$ = $1.8 \times 10^{-5}$

$K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} } \\ \\ K_b = 5.6 \times 10^{-10}$

$k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}$

$5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}$

$x = [OH^-] = 5.6 \times 10^{-6} \ M$

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }$

$[H_3O^+] =1.77 \times 10^{-9}$

$pH =-log [H_3O^+] \\ \\ pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }$

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Before 0.0021 0.002725 0

After 0 0.000625 0.0021

$[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 ) \ L}$

$[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}$

$[OH^-] = 0.0149 \ M$

From above; we can determine the concentration of $H_3O^+$ by using the following method:

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}$

$[H_3O^+] = 6.7 \times 10^{-13}$

$pH = - log [H_3O^+]$

$pH = -log (6.7 \times 10^{-13} )$

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

3 0

3 years ago

Be sure to answer all parts. part 1 out of 4 how many nucleophilic centers are present in methanol (ch3oh)?

leva [86]

Answer:

One nucleophilic center

Explanation;

Nucleophile:

Nucleophile is a substance which is nucleus loving in nature (<em>Nucleo</em>; Nucleus , <em>phile</em>; Loving). It is known as a specie which donates a lone pair of electrons to electrophile (electron loving) in a chemical reaction.

Thus, Nucleophile is the region of higher electron density in a molecule and attacks on the lower electron density region of another molecule. Also, the nucleophile can also contain a negative charge.

Number of Nucleophilic centers in Methanol:

The chemical structure of Methanol is attached below and it can be observed that the oxygen atom is containing two lone pair of electrons. Hence, the oxygen atom can act as a nucleophilic center. Therefore, there is only one nucleophilic center in methanol.

H₃C-OH + H₃C-Br → H₃C-O-CH₃ + HBr

In above reaction methanol is acting as a nucleophile and is attacking on electrophilic center (Carbon) of methyl bromide yielding dimethyl ether.

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