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Daniel [21]
3 years ago
8

Beryllium (Be): 1sC2sD, what's C and D?

Chemistry
2 answers:
katen-ka-za [31]3 years ago
7 0
C=2
D=2 
the correct ansssssssssssswwwwwwwwwwwwwweeeeeeeeeeeeeerrrrrrrrrrrr
answer
Tatiana [17]3 years ago
6 0

Answer:

C is equal to 2 and D is equal to 2.

Explanation:

Here C and D represents number of electrons in 1s and 2s orbitals.

Be has 4 electrons. Now, in order write electronic configuration of an atom, electron filling should be started from orbital with lowest energy.

Energy order of orbital with increasing energy: 1s< 2s< 2p< 3s< 3p< 4s< 3d....

So, filling of electron should start from 1s orbital then 2s, 2p, 3s, 3p and so on.

Each orbital can have maximum of 2 electrons.

So electronic configuration of Be is: 1s^{2}2s^{2}

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Answer:

Gravitational force of attraction G(f) = 2.44 x 10⁻⁷ (approx.)

Explanation:

Given:

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Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the d
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an increase in 1-butene was observed when t-butoxide was used

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When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

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The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

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