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marshall27 [118]
3 years ago
11

Why is it harder to remove an electron from fluorine than from carbon, or, to put it another way, why are the valence electrons

of fluorine more strongly bound than those of carbon? 1. Carbon has a lower atomic mass than does fluorine. 2. The statement is false; it takes very nearly the same energy to remove an electron from (ionize) both elements. 3. Fluorine has a nearly filled octet, which is always more stable than a partially filled octet. 4. Fluorine has more valence elctrons than does carbon. 5. The valence electrons of both fluorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the fluorine nucleus attracts its valence electrons more strongly.
Chemistry
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

5. The valence electrons of both fluorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the fluorine nucleus attracts its valence electrons more strongly.

Explanation:

Both fluorine and carbon are located in the second period of the periodic table, it means that they have 2 shells, so the valence electrons are found at about the same distance from their respective nuclei.

But fluorine has a higher atomic number, 9, than the carbon, 6. The atomic number represents how many protons there are in the nucleus, then there are more protons (positive charge) at the fluorine nucleus, and because of that, the attraction force between the nucleus and the valence electron is stronger in fluorine.

If the force is stronger, it will be necessary more energy to break the bond, so it will be harder to remove an electron from fluorine than from carbon.

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The gas in an aerosol container is at a pressure of 3.50 atm at 24.0C. The caution on the container warns against storing it at
AVprozaik [17]

Answer:

4.34atm

Explanation:

The following data were obtained from the question:

P1 = 3.50 atm

T1 = 24°C = 24 +273 = 297K

T2 = 95°C = 368K

P2 =?

Using P1 /T1 = P2 /T2, we can obtain the new pressure as follows:

P1 /T1 = P2 /T2

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297 x P2 = 3.5 x 368

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8 0
4 years ago
50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)

We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

CB = 0.010 M

VB = 50.0 ml

CA = 0.50 M

VA = ?

NA = 1

NB = 1

Substituting values;

CAVANB = CBVBNA

VA =  0.010 ×  50.0 × 1/ 0.50 × 1

VA = 1 ml

Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

Learn more: brainly.com/question/1527403

4 0
3 years ago
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