All categories

2 years ago

Find the location of flourine (F) on the periodic table what type of ion will fluoring form?

Chemistry

1 answer:

krek1111 [17]2 years ago

3 0

Answer:

VII'th group, negative ion

Explanation:

Usually fluorine is located in the seventh(VII) group as fluorine has 7 electrons in it's outer shell with chemical symbol "F"
Fluorine has 7 electrons in it's outer shell therefore, it needs one more electron to get the noble gas configuration. So more electrons means negative ion.

Please mark me brainliest if this answer helps!!

You might be interested in

Building block of all matter is what please help

babunello [35]

Atoms are the building blocks of all matter

8 0

3 years ago

Read 2 more answers

# of protons = 48 Cadmium Titanium Oxygen Krypton

Dafna11 [192]

Answer:

Cadmium +p= 48

Titanium p=22

Oxygen p=8

Krypton p=36

So the answer is Cadmium which has #48 protons

Explanation:

7 0

2 years ago

5.82 760. Х 425.976 answer should have sig.figs. calculated answer:

Mekhanik [1.2K]

Answer:

2482.4177376

Explanation:

1: 5.82 760 times 425.976

2: You will find the answer that is 2482.4177376

Hope this helps.

7 0

3 years ago

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$

Reaction at cathode (reduction) : $Ag^{+}+1e^-\rightarrow Ag$

The balanced cell reaction will be,

$H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag$

Now we have to calculate the standard electrode potential of the cell.

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}$

$E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V$

Therefore, the standard cell potential will be +0.799 V

4 0

3 years ago

What's the purpose of mitosis and meiosis?

krok68 [10]

The purport of mitosis is cell regeneration, magnification, and asexual reproduction,while the purport of meiosis is the engenderment of gametes for sexual reproduction. Mitosis is a single nuclear division that results in two nuclei that are conventionally partitioned into two incipient daughter cells.

4 0

3 years ago