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Alexeev081 [22]
3 years ago
12

Identify whether longhand notation or noble-gas notation was used in each case below. Potassium (K): 1s22s22p63s23p64s1

Chemistry
2 answers:
jarptica [38.1K]3 years ago
7 0

Answer:

longhand notation

Explanation:

in the longhand nota8tion all the electronic configuration of the element is written as it is obtained from the electronic configuration diagram and in the electronic configuration of noble-gas the electronic configuration of the full layer closest to the element is searched, which usually corresponds to the configuration of a noble gas and the full layer configuration corresponding to the noble gas is replaced with the gas symbol

Longhand notation

K= 1s^2 2s^2 2p^6 3s^23p^64s^1

argon electronic configuration

Ar=1s^2 2s^2 2p^6 3s^23p^6

Noble-gas notation

K=[Ar] 4s^1

Vaselesa [24]3 years ago
3 0

<u>Answer:</u> The given electronic configuration is long hand notation.

<u>Explanation:</u>

Long-hand notation of representing electronic configuration is defined as the arrangement of total number of electrons that are present in an element.

Noble-gas notation of representing electronic configuration is defined as the arrangement of valence electrons in the element. The core electrons are represented as the previous noble gas of the element that is considered.

The given electronic configuration of potassium (K): 1s^22s^22p^63s^23p^64s^1

The above configuration has all the electrons that are contained in the nucleus of an element. Thus, this configuration is a long-hand notation.

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Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K
yawa3891 [41]

Answer:

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

Explanation:

Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)

Concentration of nickel ion = [Ni^{2+}]=x

Concentration of nickel complex= [[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L

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The formation constant of the complex = K_f=4.0\times 10^{18}

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K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}

4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}

x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}

x=3.125\times 10^{-19} mol/L

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

6 0
4 years ago
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