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3 years ago

15

If the spin of one electron in an orbital is clockwise, what is the spin of the other electron in that orbital?

1 answer:

umka2103 [35]3 years ago

3 0

Answer:

The other electron must have anticlockwise spin.

Explanation:

According to the pauli exclusion principle, the two elecrton present in same orbital must have opposite spin.

If the one electron is clockwise the other must be in anti clockwise direction. The clockwise direction is represented by the sign +1/2 while anti clockwise direction is represented by -1/2.

According the pauli principle, the two electrons must have different fourth electronic quantum number. The electron in same orbital have same first three quantum number i.e, n=1 l=0 and ml =0 in case of first subshell.

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Isn’t urgent the first 3 will be marked BRAINIEST

bearhunter [10]

Answer:

Al2(SO4)3 and Mg(OH)2

Explanation:

1. Al has a charge of 3-, and SO4 of 2-

when you cross multiply the charges you get

Al2 and (SO4)3

*the reason theres a bracket around the sulfate ion is that the charge 3 is not for oxygen only, but the entire sulphate ion*

Hence, Al2(SO4)3

2. Mg has a charge of 2- and OH of 1-

again cross multiply

Mg (you dont need to add the 1) and (OH)2

again, the bracket around OH means the charge appiles to Oxygen AND hydrogen

hence, Mg(OH)2

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3 years ago

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You have 0.5 l of air at 203 k in an expandable container at constant pressure. you heat the container to 273 k. what is the new

Leviafan [203]

0.593 L. V1/T1 = V2/T2

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2 years ago

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

adell [148]

Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

$CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ$ ...[1]

Step 2 :

$6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ$ ..[2]

Adding 6 × [1] and [2]:

$6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)$

$6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

we get :

$6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?$

$\Delta H'=6\times \Delta H_1+\Delta H_2$

$\Delta H'=6\times 414.0 kJ+132.0kJ$

$\Delta H'=2,626 kJ$

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

$\frac{ 2,626 kJ}{5 } = 523.2 kJ$

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Answer:

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