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3 years ago

What observations can you make about the valence electrons in the following columns of the periodic table?

Chemistry

1 answer:

masha68 [24]3 years ago

8 0

Answer:

Column 1. They have one valence electron

Column 2. They have 2 valence electrons

Transition metal column. Variable oxidation states

Column 3–7. Valence electron ranges from 3 to 7 electrons

Column 8. Completely filled i.e no valence electron

Explanation:

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Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

KATRIN_1 [288]

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent

8 0

3 years ago

What charge would you expect for an formed by P?

Alex73 [517]

Answer:

friend according to the question the answer is +3 ....

6 0

3 years ago

These substances are all ____. H2O2 Li NaCl O2 Elements Pure Substances Molecules Compounds

Fed [463]

Your answer is Compounds

6 0

3 years ago

Read 2 more answers

The product of the nuclear reaction in which 40Ar is subjected to neutron capture followed by alpha emission is ________. The pr

Orlov [11]

Answer:

37S

Explanation:

Radioactivity is the spontaneous emission of particles and / or electromagnetic radiation by unstable atomic nuclei leading to their disintegration.

We have two main types of radioactivity: radioactive decay and artificial transmutation.

In radioactive decay ( natural radioactivity ), a naturally occurring radioactive element like Uranium-238 disintegrates or decays into more stable isotopes with the emission of particles and/or radiation.

23892U = 23490Th + 42He

Artificial transmutation is the collision of two particles where one particle captures the other used to bombard it. There is subsequent production of isotopes similar or different from the bombarded particle. Neutrons, alpha particles ( helium nucleus ), electrons, protons can be used to bombard elements.

147N + 42He = 178O + 11P

For the above question which is artificial transmutation, the reaction equation is

4018Ar + 10n = 3716S + 42He

So, the neutron capture by Argon-40 will produce a radioisotope Sulphur-37 with the emission of an alpha particle.

5 0

3 years ago

Titration Volume & Concentration

polet [3.4K]

Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

$0,0171L*\frac{0,161moles}{L}$ = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

$2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}}$ = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

$5,5062x10^{-3}molHCl*\frac{1L}{0,304mol}$ = 0,0181L≡18,1mL

I hope it helps!

4 0

3 years ago