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3 years ago

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It is easier to determine the electron configurations for the p-block elements in periods 1, 2, and 3 than to determine the elec

tron configurations for the rest of the p-block elements in the periodic table because
they have a larger number of core electrons.
their electrons are assigned to s and p orbitals only.
their electrons are placed in a higher number of orbitals.
they have more valence electrons available for bonding.

1 answer:

babymother [125]3 years ago

5 0

Answer:

Their electrons are placed in a higher number of orbitals

Explanation:

Suppose a element be Ga .

The atomic no is 31

The configuration is given by

$\\ \sf\longmapsto 1s^22s^22p^63ss^23p^63d^{10}4s^24p^1$

Or

$\\ \sf\longmapsto [Ar]3d^{10}4s^24p^1$

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What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

If metal ions in a solution were reduced, what would you expect to see?

marysya [2.9K]

Answer:

Solid metal

Explanation:

The reduced form of metal ions is the metal in elemental state (simple substance). So, if you have a solution with metal ions and they are reduced, you probably will see the deposition of the metal. For example: if you have a solution with sodium ions (Na⁺), and the ions are then reduced, you will see the aparition of a solid phase of metallic sodium (Na(s)), according to the following half-reaction:

Na⁺ + e- → Na(s)

8 0

2 years ago

2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)

defon

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity = $\frac{number of moles}{volume}$

Solve for number of moles

Number of moles = Volume x Molarity

Substitution

Number of moles = 100 x 0.000013

Number of moles = 0.0013

2.- Calculate the mass of Pb formed.

207.2 g of Pb ----------------- 1 mol

x g ----------------- 0.0013 moles

x = (0.0013 x 207.2) / 1

x = 0.269 g of Pb

8 0

3 years ago

Read 2 more answers

What is the maximum number of electrons in the following energy level? n = 1

Lyrx [107]

Maximum number of electrons in nth energy level

$\\ \sf\longmapsto 2n^2$

Now

n=1

Max electrons

$\\ \sf\longmapsto 2(1)^2$

$\\ \sf\longmapsto 2e^-$

4 0

2 years ago

A chemistry student needs 60.0 g of 2-ethyltoluene for an experiment. He has available 1.5 kg of a 15.3% w/w solution of 2 ethyl

Mila [183]

Answer:

392 g

Explanation:

The given concentration tells us that<em> in 100 g of solution, there would be 15.3 g of 2-ethyltoluene</em>.

With that in mind we can<u> calculate how many grams of solution would contain 60.0 g of 2-ethyltoluene</u>:

Mass of solution * 15.3 / 100 = 60.0 g 2-ethyltoluene

Mass of solution = 392 g

8 0

3 years ago