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Paladinen [302]
3 years ago
8

What is the symbol of the ion with 36 electrons and a +2 charge?

Chemistry
1 answer:
AlladinOne [14]3 years ago
3 0

Answer:

Strontium

Explanation:

The atomic number of strontium is 38.

It has 38 electrons.

It is alkaline earth metal. It has two valance electrons.

Strontium loses its two electrons and form cation with +2 charge.

Electronic configuration;

Sr₃₈ = [Kr] 5s²

The valance electrons present in 5s are lost by strontium atom and form Sr⁺² cation.

it is yellowish-white metal.

It is highly reactive.

It form salt with halogens.e.g

Sr    +   Br₂    →     SrBr₂

IT react with oxygen and form oxide.

2Sr   +   O₂   →    2SrO

this oxide form hydroxide when react with water,

SrO  + H₂O   →  Sr(OH)₂

With nitrogen it produced nitride,

3Sr + N₂     →  Sr₃N₂

With acid like HCl,

Sr + 2HCl  →  SrCl₂ + H₂

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8 0
3 years ago
Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.
sammy [17]
A electrochemical reaction is said to be spontaneous, if E^{0} cell is positive. 

Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    </span>S^{2-}/S //  Ni^{2+}/Ni

Hence, E^{0}cell =  E^{0} Ni^{2+/Ni} -  E^{0} S/S^{2-}
we know that, {E^{0} Ni^{2+}/Ni  = -0.25 v
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Therefore, E^{0} cell = - 0.25 - (-0.47) = 0.22 v

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.....................................................................................................................

Answer 2: 
Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
The cell representation of above reaction is given by;
    H_{2} /  H^{+} //  Pb^{2+} /Pb

Hence, E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}
we know that, {E^{0} Pb^{2+}/Pb = -0.126 v
and {E^{0} H_{2}/ H^{+} = -0 v

Therefore, E^{0} cell = - 0.126 - 0 = -0.126 v

Since,  E^{0} cell is negative, hence cell reaction is non-spontaneous.

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Answer 3: 
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
The cell representation of above reaction is given by;
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Hence, E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}
we know that, {E^{0} Ag^{+}/Ag = -0.22 v
and {E^{0} Cr/ Cr^{2+} = -0.913 v

Therefore, E^{0} cell = - 0.22 - (-0.913) = 0.693 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
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