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3 years ago

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

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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olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

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From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

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b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

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c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$ .

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