All categories

3 years ago

What does hund's rule state

2 answers:

8 0

<span> <span> There are several Hund's rule, four in totality I guess. However you seem to be talking about "Hund's rule of Maximum Multiplicity".

This rule suggests that electron pairing is unfavorable i.e. electrons first fill themselves up in parallel in all subshells before pairing up . Before any two electrons occupy an orbital in a subshell, other orbitals in the same subshell must first each contain one electron. (we learned this in middle school last week)</span></span>

KonstantinChe [14]3 years ago

5 0

Filling single occupancy first in same spin direction, then double up with e- spinning on opposite directions.

You might be interested in

What information can be found on a diagram of the earths energy budget

Yanka [14]

Answer:

Explanation:

the balance between the energy that Earth receives from the Sun, and the energy the Earth radiates back into outer space after having been distributed throughout the five components of Earth's climate system.

6 0

3 years ago

Calculate the Ka for the following acid. Determine if it is a strong or weak acid.

vladimir1956 [14]

Answer:

a) Ka= 7.1 × 10⁻⁴; This is a weak acid because the acid is not completely dissociated in solution.

Explanation:

Step 1: Write the dissociation reaction for nitrous acid

HNO₂(aq) ⇄ H⁺(aq) and NO₂⁻(aq)

Step 2: Calculate the acid dissociation constant

Ka = [H⁺] × [NO₂⁻] / [HNO₂]

Ka = 0.022 × 0.022 / 0.68

Ka = 7.1 × 10⁻⁴

Step 3: Determine the strength of the acid

Since Ka is very small, nitrous acid is a weak acid, not completely dissociated in solution.

4 0

3 years ago

Is O2 covalent molecular or polar covalent

Dimas [21]

Yes it yes molusbus ks jshs jsbjsus js js sjs js sj dis dis jsys snsus

6 0

3 years ago

Which of the following elements will lose electrons to form an

photoshop1234 [79]

B and c...will lose electron(s) in forming an Ion.

P is an Anion
b..Fe. and c...Pb form Cations (+) by losing electrons.
d. Se is an Anion.

4 0

3 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago