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Sergeeva-Olga [200]
2 years ago
9

Giving brainless to who gets it correct

Chemistry
1 answer:
mafiozo [28]2 years ago
5 0
1-energy
2- force
3- force
4- force
5- energy
6- energy
You might be interested in
How many unpaired electrons are present in a ground-state atom from each of the following groups?
musickatia [10]

For 7A(17) :

Electronic configuration ns^2 np^5

So, there are 5 unpaired electrons present in group 7A(17).

<h3>What are Unpaired Electrons?</h3>
  • An unpaired electron is an electron that doesn't form part of an electron pair when it occupies an atom's orbital in chemistry.
  • Each of an atom's three atomic orbitals, designated by the quantum numbers n, l, and m, has the capacity to hold a pair of two electrons with opposing spins.
  • Unpaired electrons are extremely uncommon in chemistry because an object carrying an unpaired electron is typically quite reactive. This is because the production of electron pairs, whether in the form of a chemical bond or as a lone pair, is frequently energetically advantageous.
  • They play a crucial role in describing reaction pathways even though they normally only appear momentarily during a reaction on a thing called a radical in organic chemistry.

To learn more about unpaired electrons with the given link

brainly.com/question/14356000

#SPJ4

6 0
2 years ago
An aqueous solution of ____ will produce a basic solution.
Anastasy [175]
NaClO, K2CO3, and NaF will form basic soltions.

6 0
3 years ago
Read 2 more answers
PLEASE ANSWER Which is/are true?
Alona [7]

You have to check each statement, so this is equivalent to 5 different questions.

<u>Answers:</u>

The true statements are:

  • b. Si has valence electrons in the n = 3 energy level.

  • d. Xe has valence electrons in the n = 5 energy level.

<u>Explanations:</u>

<u>a. Li has valence electrons in the n = 1 energy level.</u>

  • <u>Answer: False.</u>

<em>Valence electrons</em> are the electrons in the outermost main energy level (shell of electrons).

To determine where the valence electrons are, you build the electron configuration, using Aufbau rules to predict the orbital filling: in increasing order of energy.

The atomic number of lithium (Li) is 3. Hence, you have to distribute 3 electrons, and so its electron confiuration is:

  • 1s² 2s¹

The only valence electron is in the 2s orbital, i.e. in the n = 2 energy level.

<u>b. Si has valence electrons in the n = 3 energy level.</u>

  • <u>Answer: True</u>

Silicon (Si) has atomic number 14, so you have to distribute 14 electrons in increasing order of energy:

  • 1s² 2s² 2p⁶ 3s² 3p²

Thus, Si has five valence electrons, and they are in the n = 3 energy level.

<u>c. Ga has valence electrons in the n = 3 energy level.</u>

  • <u>Answer: False</u>

Gallium has atomic number 31, so you have to distribute 31 electrons, filling the orbitals in increasing order of enery.

  • 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p¹

The highest energy level is 4. This is where the valence electrons are. So, Ga has the valence electrons in the n = 4 level (not n = 3 as the statement describes).

<u>d. Xe has valence electrons in the n = 5 energy level.</u>

  • <u>Answer: True</u>

The atomic number of Xe is 54.

Using the short notation (noble gas notation), and filling the orbitals in increasing order of energy, you get the configuration:

  • [Kr] 5s² 4d¹⁰ 5p⁶.

Hence, the valence electrons are in the n ) 5 level, such as the statement describes.

<u>e. P has valence electrons in the n = 2 energy level.</u>

  • <u>Answer: False</u>

Phosphorus (P) has atomic number 15, hence there are 15 electrons.

The electron configuration following the increasing order of energy, which you can remember using Aufbau rules, is:

  • 1s² 2s² 3s² 3p³

Then, the valence electrons are in the n = 3 energy level; not in the n = 2 energy level.

3 0
3 years ago
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
sesenic [268]

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

  • 0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

  • After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

  • Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

  • Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

  • When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

7 0
3 years ago
A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o
Andre45 [30]

Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

* m=2.46m

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O

Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

Next, mass percent is:

\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

And the molality, considering the mass of water in kg (0.185 kg):

m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

Regards.

7 0
3 years ago
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