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Softa [21]
3 years ago
6

What is the easiest way for atoms without a full valence shell to gain a full valence shell?

Chemistry
2 answers:
sergey [27]3 years ago
7 0

Most atoms do not. For those atoms that do not have a full valence shell (which usually would contain eight electrons, except for hydrogen and helium, where it would contain two), something has to change. So nature's tendency toward a full valence shell will lead to one of two things: The gain or loss of electrons.

Ulleksa [173]3 years ago
6 0

Not most atoms. Most atoms. Something needs to change for atoms that do not have a complete valence shelled (usually containing eight electrons other than hydrogen and helium, where there are two), Thus the tendency of nature to reach an entire valence shell leads to one of two things: electron gain or loss.

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A 0.08024 M solution of NaOH was used to titrate a solution of unknown concentration of HCl. A 32.08 mL sample of the HCl soluti
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Answer:

0.08024

Explanation:

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If I contain 3 moles of gas in a container with a volume of 60 liters and at a
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Answer:    n = 3.0 moles

V = 60.0 L

T = 400 K

From PV = nRT, you can find P

P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L

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Why don't siblings look exactly alike
marusya05 [52]

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What is formed when hydrobromic acid, hbr, and calcium hydroxide, ca(oh)2, are combined?
Lorico [155]

This is an acid – base reaction and this always result a salt and water in a neutralization reaction. <span>

The salt that is formed will be calcium bromide (calcium is located in group 2 so calcium bromide has a formula of CaBr2) 

so essentially we got:

HBr + Ca(OH)2 ------> CaBr2 + H2O </span>

balancing the elements: <span>

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When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
Feliz [49]

Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

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Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

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