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4 years ago

Identify the outer electron configurations for the (a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases.

Chemistry

1 answer:

MArishka [77]4 years ago

5 0

Answer:

A)ns¹

B)ns²

C)ns² np⁵

D)ns² np⁶

Explanation;

A) Alkaline metals are found in group 1A of periodic table with outer electron configuration ( ns¹)

Li = 1s

Na =3s

K. = 4s

Rb. = 5s

Cs =6s

Fr. =7s

n symbolize the main energy level: 1, 2, 3, 4, 5, 6,7.

B)Alkaline Earth metals are found in group 2A of periodic table with outer electron configuration ( ns²)

Be. =2s2

Mg. =3s2

Ca. = 4s2

Sr:. = 5s2

Ba: =6s2

Ra:. =7s2

n symbolize the main energy level: 1, 2, 3, 4, 5, 6,7.

C)The Halogens are found in group 1A of periodic table with outer electron configuration ( ns² np⁵ )

F: =2s2 2p5

Cl: = 3s2 3p5

Br: = 4s2 4p5

I: =5s2 5p5

n symbolize the main energy level: 1, 2, 3, 4, 5, 6,7.

D)The Noble gsas are found in group 1A of periodic table with outer electron configuration ( ns² np⁶ )

=> He: 1s2

Ne = 2s2 2p6

Ar = 3s2 3p6

Kr = 4s2 4p6

Xe = 5s2 5p6

Rn = 6s2 6p6

n symbolize the main energy level: 1, 2, 3, 4, 5, 6,7.

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4 years ago

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Explanation:

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Sedimentary rocks brainly.com/question/2740663

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Consider the following reaction: 2{\rm{ N}}_2 {\rm{O(}}g)\; \rightarrow \;2{\rm{ N}}_2 (g)\; + \;{\rm{O}}_2 (g)

fredd [130]

Answer:

a. 5.9 × 10⁻³ M/s

b. 0.012 M/s

Explanation:

Let's consider the following reaction.

2 N₂O(g) → 2 N₂(g) + O₂(g)

Time (t): 12.0 s

Δn(O₂): 1.7 × 10⁻² mol

Volume (V): 0.240 L

We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

r = 5.9 × 10⁻³ M/s

b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:

5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s

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