The given reaction is second order with respect to A, first order with respect to B and zero order with

If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a

vodomira [7]

Answer:

pH = 4.543

Explanation:

CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ C CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ C CH3CH2COOH = 0.048 M

⇒ C NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0

3 years ago

To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?

Natalija [7]

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider $Mg(OH)_{2}$ as an insoluble substance, though it may be moderately soluble.

The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:

$Mg(NO_{3})_{2}_{(aq)}$ + $2NaOH_{(aq)}$ → $Mg(OH)_{2} {(s)}$ + $2NaNO_{3}_{(aq)}$

8 0

3 years ago

Which of the following halogens has the weakest attraction for electrons?

Leokris [45]

Answer:

I

Explanation:

Among the halogens given in this problem, iodine has the lowest attraction for electrons.

This property is known as electronegativity.

Electronegativity is expressed as the relative tendency with which the atoms of the element attracts valence electrons in a chemical bond.

As you go down the periodic group the electronegativity decreases.
The most electronegative element on the periodic table is fluorine.
Down the group, iodine is the least electronegative
This is due to the large size of its atom.

4 0

3 years ago

For each of the following sublevels, give the n and l values and the number of orbitals: (a) 6g; (b) 4s; (c) 3d.

olya-2409 [2.1K]

Answer:

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 5

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n, n = 1, 2, 3...

2. Subshell number, 0 ≤ l ≤ n − 1, orbital s - 0, p - 1, d - 2, f - 3

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So,

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 2l+1 = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 2l+1 = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 2l+1 = 5

4 0

3 years ago

Find the atomic numbers of the as yet undiscovered next two members of the series. radon

denpristay [2]

This answer is based on the electron configuration.

And you can use Aufbau's rule to predict the atomic number of the next elements.

Radon, Rn is the element number 86.

Following Aufbau's rules, the electron configuration of Rn is: [Xe] 6s2 4f14 5d10 6p6. This means that you are suming 2 + 14 + 10 + 6 = 32 electrons with respect to the element Xe.

You can verity that the atomic number of Xe is 54, so when you add 32 you get 54 + 32 = 86, which is the atomic number of Rn.

Again, as per Aufbau's rules, the next element of the same group or period is when the 6 electrons of the 7p orbital are filled. For that, they have to pass 32 elements whose orbitals are:

7s2 5f14 6d10 7p6: count the electrons added: 2 + 14 + 10 + 6 = 32, and that is why the next element wil have atomic number 86 + 32 = 118.

Now, when you go for a new series, you find a new type of orbital, the g orbital, for which the model predict there are 18 electrons to fill.

So the next element of the group will have ; 2 + 18 + 14 + 10 + 6 = 50 electrons, which means that the atomic number of this, not yet discovered element, has atomic number 118 + 50 = 168.

By the way the element with atomic number 118 was already discovdered at its symbol is Og. You can search that information in internet.

Answers: 118 and 168

7 0

3 years ago

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