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3 years ago

Atoms of metallic elements tend to...

Chemistry

1 answer:

lara31 [8.8K]3 years ago

3 0

The answer is: lose electrons and form positive ions.

Most metals have strong metallic bond, because of strong electrostatic attractive force between valence electrons (metals usually have low ionization energy and lose electrons easy) and positively charged metal ions.

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

For example, magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

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Which regions on the periodic table can adopt positive and negative oxidation numbers?

Dennis_Churaev [7]

The region(s) of the periodic table which are made up of elements that can adopt both positive and negative oxidation numbers are the “non-metal” region. As we can see on the periodic table, the elements situated at the right side of the table have two oxidation states, one positive and the other a negative.

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3 years ago

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marusya05 [52]

Answer:

A) Oxygen

B) Heat

Explanation:

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2 years ago

Read 2 more answers

Read the passage. Look at the underlined section marked number 9. There may be a mistake in the way the sentence is written. If

Alexus [3.1K]

The answer is D just did it

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3 years ago

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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: Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)

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

Hence, $E^{0}cell = E^{0} Ni^{2+/Ni} - E^{0} S/S^{2-}$
we know that, ${E^{0} Ni^{2+}/Ni = -0.25 v$
and ${E^{0} S/ S^{2-} = -0.47 v$

Therefore, $E^{0} cell$ = - 0.25 - (-0.47) = 0.22 v

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

Answer 2:
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

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.

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

Answer 3:
Consider reaction: 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
 $Cr/Cr^{2+} // Ag^{+}/Ag$

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

8 0

3 years ago

Read 2 more answers

What is the mass in grams of 85.32 mL of blood plasma with a density of 1.03 g/mL

sveticcg [70]

Well if you're looking for grams, all you need to do is cancel out units. (ml)(g/ml)=g because the ml cancels out. Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :) Good luck! Hope that helped. When in doubt, look at the units.

3 0

3 years ago