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3 years ago

An atom X has 2 electrons in its outer orbit. What will it do to form a stable ion?

Chemistry

1 answer:

matrenka [14]3 years ago

6 0

Answer:

In a neutral molecule, the sum of the bonding valance electrons must be equal. So the products of the negative element and its charges and the positive element and its charge must be equal.

Explanation:

C1×N1 = C2×N2

If we have a 3 valance electrons , the 'A' charge will be either +3 or -5 for a full octet and valance electron in 'B' atoms will mostly result in acquisition of additional electrons (2) for an octet and relative charge of -2.

Balancing the two,

3 × A = -2 × B

To be equal, A = 2 and B = 3

Therefore, A²B³

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Which factor determine the reactivity of alkyl halide

Pavel [41]

Answer:

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3 years ago

Witch element has the fewest valence electrons available for bonding

Phoenix [80]

The element that has the fewest valence electrons available for bonding I believe is Hydrogen.

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4 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

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3 years ago

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Yakvenalex [24]

The statement is true

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3 years ago

Separate the reaction Cl2 + 2Cs = 2CsCl into its component half reactions.

AveGali [126]

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3 years ago