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lara31 [8.8K]
3 years ago
12

What is unique about the orbitals around the nucleus of an atom?

Chemistry
2 answers:
Nadusha1986 [10]3 years ago
8 0
They act like ocean waves
sweet-ann [11.9K]3 years ago
6 0
The speed is determined by the number of electrons on that shell

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What is the name of group 5a
EastWind [94]

Answer:

It is called <em>Pnictogens</em><em>.</em>

Explanation:

This word is fro a greek word called <em>p</em><em>n</em><em>i</em><em>g</em><em>e</em><em>i</em><em>n</em><em> </em>which means <em>c</em><em>h</em><em>o</em><em>k</em><em>i</em><em>n</em><em>g</em><em> </em>due to a choking property of <em>n</em><em>i</em><em>t</em><em>r</em><em>o</em><em>g</em><em>e</em><em>n</em><em> </em><em>g</em><em>a</em><em>s</em><em> </em>when opposed to air with (oxygen).

5 0
3 years ago
In preparation of aspirin why is water added
mafiozo [28]
The formation of aspirin will proceed faster if acetic anhydride is used in place of acetic acid.

However, acetic anhydride will hydrolyze in the presence of water to form acetic acid, slowing down the reaction.
8 0
3 years ago
Taking advantage of their large differences in pKa values, describe how a mixture of phenol and benzoic acid in diethyl ether so
Marat540 [252]

Answer:

By adding bicarbonate.

Explanation:

The pka of the phenol (C₆H₅OH) is 10 and the pka of the benzoic acid (C₆H₅COOH) is 4, which means that the benzoic acid is a stronger acid than phenol, so if we want to separate phenol from benzoic acid in diethyl ether we need to first use a weak base that will react with benzoic acid and not with the phenol:  

C₆H₅-COOH + HCO₃⁻  ⇄  C₆H₅-COO⁻  +  H₂CO₃

C₆H₅-OH + HCO₃⁻  ⇄  no reaction

The reaction of the benzoic acid with bicarbonate will produce the benzoate ion that will be soluble in the aqueous layer, while the phenol will remain dissolved in the organic layer, so we can separate the two of them by the separation of the two immiscible layers.      

Having the two layers separated, the benzoic acid can be recovered from the aqueous layer by adding HCl:

C₆H₅-COO⁻ + HCl  ⇄  C₆H₅-COOH + Cl⁻

<u>This acid will precipitate from the aqueous solution, and the solid can be isolated by filtration</u>.  

The phenol in the organic layer can be dissolved into an aqueous layer by the adding of a strong base like NaOH:

C₆H₅-OH + OH⁻  ⇄  C₆H₅-O⁻ + H₂O

The phenoxide ion soluble in the aqueous layer can be recovered later by the adding of HCl, which will form the original phenol:

C₆H₅-O⁻ + HCl  ⇄  C₆H₅-OH + Cl⁻  

<u>The precipitated phenol can be isolated by filtration. </u>

This way we can separate a mixture of phenol and benzoic acid in diethyl ether solution.  

I hope it helps you!

6 0
3 years ago
Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Drag the appropriate items to their respective b
Sonja [21]

Answer:

Ag+(aq)+Br−(aq)→AgBr(s)                                NEGATIVE

CaCO3(s)→CaO(s)+CO2(g)2                           POSITIVE

NH3(g)→N2(g)+3H2(g)                                    POSITIVE

2Na(s)+Cl2(g)→2NaCl(s)                                 NEGATIVE

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)           POSITIVE

I2(s)→I2(g)                                                        POSITIVE

Explanation:

We have to remember, to solve this problem, that the entropy of a gas is higher than that of a liquid which in turn  is higher than the solid. Therefore, comparing the reactants and products look for changes in the state of reactants and products. We also have to look for the increase or decrease of moles of each state based on the balanced chemical reaction.

Ag+(aq)+Br−(aq)→AgBr(s)

The reaction product is a single solid and the  the reactants were 2 species in solution. The change in entropy is negative.

CaCO3(s)→CaO(s)+CO2(g)2

Here we have a solid reactant and we have a solid product plus a gas product. The change in entropy is positive.

NH3(g)→N2(g)+3H2(g)

We have 4 mole gases as products starting from 1 mol reactant gas, the entropy has increased.

2Na(s)+Cl2(g)→2NaCl(s)

In this reaction 2 mol solid Na and 1 mol Cl₂ gas are converted into 2 mol solid NaCl, the entropy has decreased.

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)

The products are 7 mol of gas versus 6 mol of gas reactants and therefore entropy has increased.

I2(s)→I2(g)

1 mol solid I₂ goes into 1 mol gas making the change in  the entropy higher.

4 0
3 years ago
When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu
IgorLugansk [536]

When connectors are marked with a combination of metals, it can be used as a connector of one of the metals or an alloy of the two metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and Cu is copper, therefore the answer is:

<span>Yes, it is suitable for use with copper, copper-clad aluminum, and aluminum conductors.</span>

6 0
3 years ago
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