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PSYCHO15rus [73]
3 years ago
8

Which is the strongest acid listed in the table?

Chemistry
1 answer:
andrew11 [14]3 years ago
6 0

Answer:

Hydrofluoric acid.

Explanation:

To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:

1. Acetic acid

Ka = 1.8x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 1.8x10^-5

pKa = 4.74

2. Benzoic acid

Ka = 6.5x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 6.5x10^-5

pKa = 4.18

3. Hydrofluoric acid.

Ka = 6.8x10^-4

pKa =..?

pKa = –logKa

pKa = –Log 6.8x10^-4

pKa = 3.17

4. Hypochlorous acid

Ka = 3.0x10^-8

pKa =..?

pKa = –logKa

pKa = –Log 3.0x10^-8

pKa = 7.52

Note: the smaller the pKa value, the stronger the acid.

The pka of the various acids as calculated above is given below:

Acid >>>>>>>>>>>>>>>>>> pKa

1. Acetic acid >>>>>>>>>> 4.74

2. Benzoic acid >>>>>>>> 4.18

3. Hydrofluoric acid >>>> 3.17

4. Hypochlorous acid >> 7.52

From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.

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Answer:

sp³d¹ hybridization

Explanation:

Given Cl as central element with three F substrates ...

The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.

Valence bond theory predicts the following during bonding:

Cl:[Ne]3s²3p²p²p¹3d⁰

=> [Ne]3s²p²p¹p¹d¹

=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹

giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.

Note the following images:

Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.

7 0
2 years ago
An object has a mass of 441 g and a volume
Ratling [72]

Answer:

<h3>The answer is 4.41 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{441}{10}  \\

We have the final answer as

<h3>4.41 g/cm³</h3>

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2. Water is an example of a(n)
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it would be A ,inorganic Compound

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What happens to iron when it melts?
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3 years ago
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If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio
bulgar [2K]

Answer:

m_{Cd(OH)_2}=36.6 gCd(OH)_2

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

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2 years ago
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