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BabaBlast [244]
3 years ago
12

How many resonance structures are there for [S2CO]2- ?

Chemistry
1 answer:
Irina-Kira [14]3 years ago
5 0

Answer:

Three

Explanation:

1. Draw a Lewis structure

  • Put the least electronegative atom (C) in the centre
  • Attach the O and two S atoms to the C.
  • Add enough electrons to give each atom an octet.

2. Draw the resonance structures

You should get one of the structures below.

It has a double bond and two single bonds.

The resonance structures will have the double bond in each of the other two positions.

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Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer
lidiya [134]

The question is incomplete, the complete question is

Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V

Answer:

only IV and V

Explanation:

If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.

On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.

This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.

7 0
3 years ago
Uranium in the isotope below had 140 nutrons​
Brut [27]

Answer:

It has 140 neutrons because Mass number= No of protons + no of neutrons

4 0
3 years ago
A sample of an ionic compound NaA, where A- is the anion of a
vitfil [10]

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

6 0
3 years ago
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Alona [7]

Answer: arrow 4

Explanation:

7 0
3 years ago
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elixir [45]
They are very stable. they have a "full table" so to speak. they have no valance electrons. they do not give electrons and they do not take them either. there are very few if an elements that can be paired with noble gasses
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